16.10 Acid Base Behavior and Chemical Structure We explore the relationship between chemical structure and acid base behavior.

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1 16.6 Weak Acids We learn that the ionization of a weak acid in water is an equilibrium process with an equilibrium constant K a that can be used to calculate the p of a weak acid solution Weak Bases We learn that the ionization of a weak base in water is an equilibrium process with equilibrium constant K b that can be used to calculate the p of a weak base solution Relationship between K a and K b We see that K a and K b are related by the relationship K a K b = K w. ence, the stronger an acid, the weaker its conjugate base Acid Base Properties of Salt Solutions We learn that the ions of a soluble ionic compound can serve as Brønsted Lowry acids or bases Acid Base Behavior and Chemical Structure We explore the relationship between chemical structure and acid base behavior Lewis Acids and Bases Finally, we see the most general definition of acids and bases, namely the Lewis acid base definition. A Lewis acid is an electron-pair acceptor and a Lewis base is an electron-pair donor.

2 672 chapter 16 Acid Base Equilibria O O O O O C C C C O O C C C C O O Tartaric acid O Malic acid O Figure 16.1 Two organic acids: Tartartic acid, 2 C 4 4 O 6, and malic acid, 2 C 4 4 O 5. our lives all critically depend on the acidity or basicity of solutions. We will thus explore in this chapter how we measure acidity and how the chemical reactions of acids and bases depend on their concentrations. We first encountered acids and bases in Sections 2.8 and 4.3, in which we discussed the naming of acids and some simple acid base reactions, respectively. In this chapter we take a closer look at how acids and bases are identified and characterized. In doing so, we consider their behavior both in terms of their structure and bonding and in terms of the chemical equilibria in which they participate Acids and Bases: A Brief Review From the earliest days of experimental chemistry, scientists have recognized acids and bases by their characteristic properties. Acids have a sour taste and cause certain dyes to change color, whereas bases have a bitter taste and feel slippery (soap is a good example). Use of the term base comes from the old English meaning of the word, to bring low. (We still use the word debase in this sense, meaning to lower the value of something.) When a base is added to an acid, the base lowers the amount of acid. Indeed, when acids and bases are mixed in the right proportions, their characteristic properties seem to disappear altogether. (Section 4.3) By 1830 it was evident that all acids contain hydrogen but not all hydrogencontaining substances are acids. During the 1880s, the Swedish chemist Svante Arrhenius ( ) defined acids as substances that produce ions in water and bases as substances that produce O - ions in water. Over time the Arrhenius concept of acids and bases came to be stated in the following way: An acid is a substance that, when dissolved in water, increases the concentration of ions. A base is a substance that, when dissolved in water, increases the concentration of O - ions. ydrogen chloride gas, which is highly soluble in water, is an example of an Arrhenius acid. When it dissolves in water, Cl(g) produces hydrated and Cl - ions: Cl1g2 2O 1aq2 Cl - 1aq2 [16.1] The aqueous solution of Cl is known as hydrochloric acid. Concentrated hydrochloric acid is about 37% Cl by mass and is 12 M in Cl. Sodium hydroxide is an Arrhenius base. Because NaO is an ionic compound, it dissociates into Na and O - ions when it dissolves in water, thereby increasing the concentration of O - ions in the solution. Give It Some Thought Which two ions are central to the Arrhenius definitions of acids and bases?

3 section 16.2 Brønsted Lowry Acids and Bases Brønsted Lowry Acids and Bases The Arrhenius concept of acids and bases, while useful, is rather limited. For one thing, it is restricted to aqueous solutions. In 1923 the Danish chemist Johannes Brønsted ( ) and the English chemist Thomas Lowry ( ) independently proposed a more general definition of acids and bases. Their concept is based on the fact that acid base reactions involve the transfer of ions from one substance to another. To understand this definition better, we need to examine the behavior of the ion in water more closely. The Ion in Water We might at first imagine that ionization of Cl in water produces just and Cl -. A hydrogen ion is no more than a bare proton a very small particle with a positive charge. As such, an ion interacts strongly with any source of electron density, such as the nonbonding electron pairs on the oxygen atoms of water molecules. For example, the interaction of a proton with water forms the hydronium ion, 3 O 1aq2: Go Figure Which type of intermolecular force do the dotted lines in this figure represent? O O 5 O 2 O O [16.2] The behavior of ions in liquid water is complex because hydronium ions interact with additional water molecules via the formation of hydrogen bonds. (Section 11.2) For example, the 3 O ion bonds to additional 2 O molecules to generate such ions as 5 O 2 and 9 O 4 ( Figure 16.2). Chemists use the notations 1aq2 and 3 O 1aq2 interchangeably to represent the hydrated proton responsible for the characteristic properties of aqueous solutions of acids. We often use the notation 1aq2 for simplicity and convenience, as we did in Chapter 4 and Equation The notation 3 O 1aq2, however, more closely represents reality. Proton-Transfer Reactions In the reaction that occurs when Cl dissolves in water, the Cl molecule transfers an ion (a proton) to a water molecule. Thus, we can represent the reaction as occurring between an Cl molecule and a water molecule to form hydronium and chloride ions: O O O 9 O 4 O Figure 16.2 Ball-and-stick models and Lewis structures for two hydrated hydronium ions. Cl(g) Cl 2 O(l) (aq) 3 O (aq) Cl O Cl O [16.3] Acid Base Notice that the reaction in Equation 16.3 involves a proton donor (Cl) and a proton acceptor 1 2 O2. The notion of transfer from a proton donor to a proton acceptor is the key idea in the Brønsted Lowry definition of acids and bases: An acid is a substance (molecule or ion) that donates a proton to another substance. A base is a substance that accepts a proton.

4 674 chapter 16 Acid Base Equilibria Thus, when Cl dissolves in water (Equation 16.3), Cl acts as a Brønsted Lowry acid (it donates a proton to 2 O), and 2 O acts as a Brønsted Lowry base (it accepts a proton from Cl). We see that the 2 O molecule serves as a proton acceptor by using one of the nonbonding pairs of electrons on the O atom to attach the proton. Because the emphasis in the Brønsted Lowry concept is on proton transfer, the concept also applies to reactions that do not occur in aqueous solution. In the reaction between gas phase Cl and N 3, for example, a proton is transferred from the acid Cl to the base N 3 : Cl N Cl N [16.4] Acid Base The hazy film that forms on the windows of general chemistry laboratories and on glassware in the laboratory ( Figure 16.3) is largely solid N 4 Cl formed by the gasphase reaction between Cl and N 3. Let s consider another example that compares the relationship between the Arrhenius and Brønsted Lowry definitions of acids and bases an aqueous solution of ammonia, in which we have the equilibrium: Figure 16.3 Fog of N 4 Cl1s2 caused by the reaction of Cl1g2 and N 3 1g2. N 3 1aq2 2 O1l2 N 4 1aq2 O - 1aq2 [16.5] Base Acid Ammonia is a Brønsted Lowry base because it accepts a proton from 2 O. Ammonia is also an Arrhenius base because adding it to water leads to an increase in the concentration of O - 1aq2. The transfer of a proton always involves both an acid (donor) and a base (acceptor). In other words, a substance can function as an acid only if another substance simultaneously behaves as a base. To be a Brønsted Lowry acid, a molecule or ion must have a hydrogen atom it can lose as an ion. To be a Brønsted Lowry base, a molecule or ion must have a nonbonding pair of electrons it can use to bind the ion. Some substances can act as an acid in one reaction and as a base in another. For example, 2 O is a Brønsted Lowry base in Equation 16.3 and a Brønsted Lowry acid in Equation A substance capable of acting as either an acid or a base is called amphiprotic. An amphiprotic substance acts as a base when combined with something more strongly acidic than itself and as an acid when combined with something more strongly basic than itself. Give It Some Thought In the forward reaction of this equilibrium, which substance acts as the Brønsted Lowry base? 2 S1aq2 C 3 N 2 1aq2 S - 1aq2 C 3 N 3 1aq2 Conjugate Acid Base Pairs In any acid base equilibrium, both the forward reaction (to the right) and the reverse reaction (to the left) involve proton transfer. For example, consider the reaction of an acid A with water: A1aq2 2 O1l2 A - 1aq2 3 O 1aq2 [16.6] In the forward reaction, A donates a proton to 2 O. Therefore, A is the Brønsted Lowry acid and 2 O is the Brønsted Lowry base. In the reverse reaction, the 3 O ion

5 section 16.2 Brønsted Lowry Acids and Bases 675 donates a proton to the A - ion, so 3 O is the acid and A - is the base. When the acid A donates a proton, it leaves behind a substance, A -, that can act as a base. Likewise, when 2 O acts as a base, it generates 3 O, which can act as an acid. An acid and a base such as A and A - that differ only in the presence or absence of a proton are called a conjugate acid base pair.* Every acid has a conjugate base, formed by removing a proton from the acid. For example, O - is the conjugate base of 2 O, and A - is the conjugate base of A. Every base has a conjugate acid, formed by adding a proton to the base. Thus, 3 O is the conjugate acid of 2 O, and A is the conjugate acid of A -. In any acid base (proton-transfer) reaction, we can identify two sets of conjugate acid base pairs. For example, consider the reaction between nitrous acid and water: remove NO 2 (aq) 2 O(l) NO 2 (aq) 3 O (aq) Acid Base Conjugate base Conjugate acid [16.7] add Likewise, for the reaction between N 3 and 2 O (Equation 16.5), we have add N 3 (aq) 2 O(l) N 4 (aq) O (aq) Base Acid Conjugate acid Conjugate base [16.8] remove Sample Exercise 16.1 Identifying Conjugate Acids and Bases (a) What is the conjugate base of ClO 4, 2 S, P 4, CO 3 -? (b) What is the conjugate acid of CN -, SO 4 2 -, 2 O, CO 3 -? Solution Analyze We are asked to give the conjugate base for several acids and the conjugate acid for several bases. Plan The conjugate base of a substance is simply the parent substance minus one proton, and the conjugate acid of a substance is the parent substance plus one proton. Solve (a) If we remove a proton from ClO 4, we obtain ClO - 4, which is its conjugate base. The other conjugate bases are S -, P 3, and 2 CO - 3. (b) If we add a proton to CN -, we get CN, its conjugate acid. The other conjugate acids are SO - 4, 3 O, and 2 CO 3. Notice that the hydrogen carbonate ion 1CO is amphiprotic. It can act as either an acid or a base. Practice Exercise 1 Consider the following equilibrium reaction: SO 4-1aq2 O - 1aq2 SO 4 2-1aq2 2 O1l2 Which substances are acting as acids in the reaction? - (a) SO 4 and O - - (b) SO 4 and 2 O (c) O and SO 4 (d) SO 4 and 2 O (e) None of the substances are acting as acids in this reaction. Practice Exercise 2 Write the formula for the conjugate acid of each of the following: SO 3 -, F -, PO 4 3 -, CO. Once you become proficient at identifying conjugate acid base pairs it is not difficult to write equations for reactions involving Brønsted Lowry acids and bases (proton-transfer reactions). *The word conjugate means joined together as a pair.

6 676 chapter 16 Acid Base Equilibria Sample Exercise 16.2 Writing Equations for Proton-Transfer Reactions The hydrogen sulfite ion 1SO 3-2 is amphiprotic. Write an equation for the reaction of SO 3 - with water (a) in which the ion acts as an acid and (b) in which the ion acts as a base. In both cases identify the conjugate acid base pairs. Solution Analyze and Plan We are asked to write two equations representing reactions between SO 3 - and water, one in which SO 3 - should donate a proton to water, thereby acting as a Brønsted Lowry acid, and one in which SO 3 - should accept a proton from water, thereby acting as a base. We are also asked to identify the conjugate pairs in each equation. Solve (a) SO 3-1aq2 2 O1l2 SO 3 2-1aq2 3 O 1aq2 The conjugate pairs in this equation are SO 3 - (acid) and SO (conjugate base), and 2 O (base) and 3 O (conjugate acid). (b) SO 3-1aq2 2 O1l2 2 SO 3 1aq2 O - 1aq2 The conjugate pairs in this equation are 2 O (acid) and O - (conjugate base), and SO 3 - (base) and 2 SO 3 (conjugate acid). Practice Exercise 1 The dihydrogen phosphate ion, 2 PO - 4, is amphiprotic. In which of the following reactions is this ion serving as a base? (i) 3 O 1aq2 2 PO - 4 1aq2 3 PO 4 1aq2 2 O1l2 (ii) 3 O 2 1aq2 PO - 4 1aq2 2 PO - 4 1aq2 2 O1l2 2 (iii) 3 PO 4 1aq2 PO - 4 1aq2 2 2 PO - 4 1aq2 (a) i only (b) i and ii (c) i and iii (d) ii and iii (e) i, ii, and iii Practice Exercise 2 When lithium oxide 1Li 2 O2 is dissolved in water, the solution turns basic from the reaction of the oxide ion 1O 2-2 with water. Write the equation for this reaction and identify the conjugate acid base pairs. Relative Strengths of Acids and Bases Some acids are better proton donors than others, and some bases are better proton acceptors than others. If we arrange acids in order of their ability to donate a proton, we find that the more easily a substance gives up a proton, the less easily its conjugate base accepts a proton. Similarly, the more easily a base accepts a proton, the less easily its conjugate acid gives up a proton. In other words, the stronger an acid, the weaker its conjugate base, and the stronger a base, the weaker its conjugate acid. Thus, if we know how readily an acid donates protons, we also know something about how readily its conjugate base accepts protons. The inverse relationship between the strengths of acids and their conjugate bases is illustrated in Figure ere we have grouped acids and bases into three broad categories based on their behavior in water: Go Figure If O 2- ions are added to water, what reaction, if any, occurs? Strong acids Weak acids Negligible acidity ACID Cl 2 SO 4 NO 3 3 O (aq) SO 4 3 PO 4 F C 3 COO 2 CO 3 2 S 2 PO 4 N 4 CO 3 PO O O 2 C 4 Acid strength increasing Base strength increasing BASE Cl SO 4 NO 3 2 O SO PO 4 F C 3 COO CO 3 S PO 4 2 N 3 CO 3 2 PO 4 3 O O 2 C 3 Negligible basicity Weak bases Strong bases Figure 16.4 Relative strengths of select conjugate acid base pairs. The two members of each pair are listed opposite each other in the two columns.

7 section 16.2 Brønsted Lowry Acids and Bases A strong acid completely transfers its protons to water, leaving essentially no undissociated molecules in solution. (Section 4.3) Its conjugate base has a negligible tendency to accept protons in aqueous solution. (The conjugate base of a strong acid shows negligible basicity.) 2. A weak acid only partially dissociates in aqueous solution and therefore exists in the solution as a mixture of the undissociated acid and its conjugate base. The conjugate base of a weak acid shows a slight ability to remove protons from water. (The conjugate base of a weak acid is a weak base.) 3. A substance with negligible acidity contains hydrogen but does not demonstrate any acidic behavior in water. Its conjugate base is a strong base, reacting completely with water, to form O - ions. (The conjugate base of a substance with negligible acidity is a strong base.) The ions 3 O 1aq2 and O - 1aq2 are, respectively, the strongest possible acid and strongest possible base that can exist at equilibrium in aqueous solution. Stronger acids react with water to produce 3 O 1aq2 ions, and stronger bases react with water to produce O - 1aq2 ions, a phenomenon known as the leveling effect. Give It Some Thought Given that ClO 4 is a strong acid, how would you classify the basicity of ClO 4 -? We can think of proton-transfer reactions as being governed by the relative abilities of two bases to abstract protons. For example, consider the proton transfer that occurs when an acid A dissolves in water: A1aq2 2 O1l2 3 O 1aq2 A - 1aq2 [16.9] If 2 O (the base in the forward reaction) is a stronger base than A - (the conjugate base of A), it is favorable to transfer the proton from A to 2 O, producing 3 O and A -. As a result, the equilibrium lies to the right. This describes the behavior of a strong acid in water. For example, when Cl dissolves in water, the solution consists almost entirely of 3 O and Cl - ions with a negligible concentration of Cl molecules: Cl1g2 2 O1l2 3 O 1aq2 Cl - 1aq2 [16.10] 2 O is a stronger base than Cl - (Figure 16.4), so 2 O acquires the proton to become the hydronium ion. Because the reaction lies completely to the right, we write Equation with only an arrow to the right rather than using the double arrows for an equilibrium. When A - is a stronger base than 2 O, the equilibrium lies to the left. This situation occurs when A is a weak acid. For example, an aqueous solution of acetic acid consists mainly of C 3 COO molecules with only a relatively few 3 O and C 3 COO - ions: C 3 COO1aq2 2 O1l2 3 O 1aq2 C 3 COO - 1aq2 [16.11] The C 3 COO - ion is a stronger base than 2 O (Figure 16.4) and therefore the reverse reaction is favored more than the forward reaction. From these examples, we conclude that in every acid base reaction, equilibrium favors transfer of the proton from the stronger acid to the stronger base to form the weaker acid and the weaker base. Sample Exercise 16.3 predicting the Position of a Proton-Transfer Equilibrium For the following proton-transfer reaction use Figure 16.4 to predict whether the equilibrium lies to the left 1K c 6 12 or to the right 1K c 7 12: SO 4-1aq2 CO 3 2-1aq2 SO 4 2-1aq2 CO 3-1aq2 Solution Analyze We are asked to predict whether an equilibrium lies to the right, favoring products, or to the left, favoring reactants.

8 678 chapter 16 Acid Base Equilibria Plan This is a proton-transfer reaction, and the position of the equilibrium will favor the proton 2 going to the stronger of two bases. The two bases in the equation are CO - 3, the base in the 2 forward reaction, and SO - 4, the conjugate base of SO - 4. We can find the relative positions of these two bases in Figure 16.4 to determine which is the stronger base. 2 - Solve The CO 3 ion appears lower in the right-hand column in Figure 16.4 and is therefore a stronger base than SO Therefore, CO 3 will get the proton preferentially to become CO - 2-3, while SO 4 will remain mostly unprotonated. The resulting equilibrium lies to the right, favoring products (that is, K c 7 1): SO 4-1aq2 CO 3 2-1aq2 SO 4 2-1aq2 CO 3-1aq2 K c 7 1 Acid Base Conjugate base Conjugate acid Comment Of the two acids SO 4 - and CO 3 -, the stronger one 1SO 4-2 gives up a proton more readily, and the weaker one 1CO 3-2 tends to retain its proton. Thus, the equilibrium favors the direction in which the proton moves from the stronger acid and becomes bonded to the stronger base. Practice Exercise 1 Based on information in Figure 16.4, place the following equilibria in order from smallest to largest value of K c : (i) C 3 COO1aq2 S - 1aq2 C 3 COO - 1aq2 2 S1aq2 (ii) F - 1aq2 N 4 1aq2 F1aq2 N 3 1aq2 (iii) 2 CO 3 1aq2 Cl - 1aq2 CO - 3 1aq2 Cl1aq2 (a) i 6 ii 6 iii (b) ii 6 i 6 iii (c) iii 6 i 6 ii (d) ii 6 iii 6 i (e) iii 6 ii 6 i Practice Exercise 2 For each reaction, use Figure 16.4 to predict whether the equilibrium lies to the left or to the right: (a) PO 4 2-1aq2 2 O1l2 2 PO 4-1aq2 O - 1aq2 (b) N 4 1aq2 O - 1aq2 N 3 1aq2 2 O1l The Autoionization of Water One of the most important chemical properties of water is its ability to act as either a Brønsted Lowry acid or a Brønsted Lowry base. In the presence of an acid, it acts as a proton acceptor; in the presence of a base, it acts as a proton donor. In fact, one water molecule can donate a proton to another water molecule: 2 O(l) O 2 O(l) (aq) 3 O (aq) O O O O [16.12] Acid Base We call this process the autoionization of water. Because the forward and reverse reactions in Equation are extremely rapid, no water molecule remains ionized for long. At room temperature only about two out of every 10 9 water molecules are ionized at any given instant. Thus, pure water consists almost entirely of 2 O molecules and is an extremely poor conductor of electricity. Nevertheless, the autoionization of water is very important, as we will soon see.

9 The Ion Product of Water The equilibrium-constant expression for the autoionization of water is K c = 3 3 O 43O - 4 [16.13] The term [ 2 O] is excluded from the equilibrium-constant expression because we exclude the concentrations of pure solids and liquids. (Section 15.4) Because this expression refers specifically to the autoionization of water, we use the symbol K w to denote the equilibrium constant, which we call the ion-product constant for water. At 25 C, K w equals 1.0 * Thus, we have K w = 3 3 O 43O - 4 = 1.0 * at 25 C2 [16.14] Because we use 1aq2 and 3 O 1aq2 interchangeably to represent the hydrated proton, the autoionization reaction for water can also be written as 2 O1l2 1aq2 O - 1aq2 [16.15] Likewise, the expression for K w can be written in terms of either 3 O or, and K w has the same value in either case: K w = 3 3 O 43O - 4 = 3 43O - 4 = 1.0 * at 25 C2 [16.16] This equilibrium-constant expression and the value of K w at 25 C are extremely important, and you should commit them to memory. A solution in which 3 4 = 3O - 4 is said to be neutral. In most solutions, however, the and O - concentrations are not equal. As the concentration of one of these ions increases, the concentration of the other must decrease, so that the product of their concentrations always equals 1.0 * ( Figure 16.5). section 16.3 The Autoionization of Water 679 Go Figure Suppose that equal volumes of the middle and right samples in the figure were mixed. Would the resultant solution be acidic, neutral, or basic? ydrochloric acid Cl(aq) Water 2 O Sodium hydroxide NaO(aq) Acidic solution [ ] > [O ] [ ][O ] = Neutral solution [ ] = [O ] [ ][O ] = Basic solution [ ] < [O ] [ ][O ] = Figure 16.5 Relative concentrations of and O in aqueous solutions at 25 C. Sample Exercise 16.4 Calculating 3h 4 for Pure Water Calculate the values of 3 4 and 3O - 4 in a neutral aqueous solution at 25 C. Solution Analyze We are asked to determine the concentrations of and O - ions in a neutral solution at 25 C. Plan We will use Equation and the fact that, by definition, 3 4 = 3O - 4 in a neutral solution. Solve We will represent the concentration of and O - in neutral solution with x. This gives 3 43O - 4 = 1x21x2 = 1.0 * x 2 = 1.0 * x = 1.0 * 10-7 M = 3 4 = 3O - 4 In an acid solution 3 4 is greater than 1.0 * 10-7 M; in a basic solution 3 4 is less than 1.0 * 10-7 M.

10 680 chapter 16 Acid Base Equilibria Practice Exercise 1 In a certain acidic solution at 25 C, 3 4 is 100 times greater than 3O - 4. What is the value for 3O - 4 for the solution? (a) 1.0 * 10-8 M (b) 1.0 * 10-7 M (c) 1.0 * 10-6 M (d) 1.0 * 10-2 M (e) 1.0 * 10-9 M Practice Exercise 2 Indicate whether solutions with each of the following ion concentrations are neutral, acidic, or basic: (a) 3 4 = 4 * 10-9 M; (b) 3O - 4 = 1 * 10-7 M; (c) 3O - 4 = 1 * M. What makes Equation particularly useful is that it is applicable both to pure water and to any aqueous solution. Although the equilibrium between 1aq2 and O - 1aq2 as well as other ionic equilibria are affected somewhat by the presence of additional ions in solution, it is customary to ignore these ionic effects except in work requiring exceptional accuracy. Thus, Equation is taken to be valid for any dilute aqueous solution and can be used to calculate either 3 4 (if 3O - 4 is known) or 3O - 4 (if 3 4 is known). Sample Exercise 16.5 Calculating 3h 4 from 3oh - 4 Calculate the concentration of 1aq2 in (a) a solution in which 3O - 4 is M, (b) a solution in which 3O - 4 is 1.8 * 10-9 M. Note: In this problem and all that follow, we assume, unless stated otherwise, that the temperature is 25 C. Solution Analyze We are asked to calculate the 3 4 concentration in an aqueous solution where the hydroxide concentration is known. Solve (a) Using Equation 16.16, we have 3 43O - 4 = 1.0 * This solution is basic because 3O Plan We can use the equilibrium-constant expression for the autoionization of water and the value of K w to solve for each unknown concentration. 3 4 = 11.0 * O - 4 = 1.0 * = 1.0 * M (b) In this instance 3 4 = 11.0 * O - 4 This solution is acidic because O - 4 = 1.0 * * 10-9 = 5.6 * 10-6 M Practice Exercise 1 A solution has 3O - 4 = 4.0 * What is the value of 3 4 for the solution? (a) 2.5 * 10-8 M (b) 4.0 * 10-8 M (c) 2.5 * 10-7 M (d) 2.5 * 10-6 M (e) 4.0 * 10-6 M Practice Exercise 2 Calculate the concentration of O - 1aq2 in a solution in which (a) 3 4 = 2 * 10-6 M; (b) 3 4 = 3O - 4; (c) 3 4 = 200 * 3O The p Scale The molar concentration of 1aq2 in an aqueous solution is usually very small. For convenience, we therefore usually express 3 4 in terms of p, which is the negative logarithm in base 10 of 3 4:* p = -log[ ] [16.17] If you need to review the use of logarithms, see Appendix A. In Sample Exercise 16.4, we saw that 3 4 = 1.0 * 10-7 M for a neutral aqueous solution at 25 C. We can now use Equation to calculate the p of a neutral solution at 25 C: p = -log11.0 * = = 7.00 *Because 3 4 and 3 3 O 4 are used interchangeably, you might see p defined as -log 3 3 O 4.

11 section 16.4 The p Scale 681 Table 16.1 Relationships among 3h 4, 3oh 4, and p at 25 c Solution Type 3h 4 1M2 3oh 4 1M2 p Acidic * * Neutral 1.0 * * Basic * * Notice that the p is reported with two decimal places. We do so because only the numbers to the right of the decimal point are the significant figures in a logarithm. Because our original value for the concentration 11.0 * 10-7 M2 has two significant figures, the corresponding p has two decimal places (7.00). What happens to the p of a solution as we make the solution more acidic, so that 3 4 increases? Because of the negative sign in the logarithm term of Equation 16.17, the p decreases as 3 4 increases. For example, when we add sufficient acid to make 3 4 = 1.0 * 10-3 M the p is p = -log11.0 * = = 3.00 At 25 C the p of an acidic solution is less than We can also calculate the p of a basic solution, one in which 3O * 10-7 M. Suppose 3O - 4 = 2.0 * 10-3 M. We can use Equation to calculate 3 4 for this solution and Equation to calculate the p: 3 4 = K w 3O - 4 = 1.0 * * 10-3 = 5.0 * M p = -log15.0 * = At 25 C the p of a basic solution is greater than The relationships among 3 4, 3O - 4, and p are summarized in Table Give It Some Thought Is it possible for a solution to have a negative p? If so, would that p signify a basic or acidic solution? One might think that when 3 4 is very small, as is often the case, it would be unimportant. That reasoning is quite incorrect! Remember that many chemical processes depend on the ratio of changes in concentration. For example, if a kinetic rate law is first order in 3 4, doubling the concentration doubles the rate even if the change is merely from 1 * 10-7 M to 2 * 10-7 M. (Section 14.3) In biological systems, many reactions involve proton transfers and have rates that depend on 3 4. Because the speeds of these reactions are crucial, the p of biological fluids must be maintained within narrow limits. For example, human blood has a normal p range of 7.35 to Illness and even death can result if the p varies much from this narrow range. Sample Exercise 16.6 Calculating p from 3h 4 Calculate the p values for the two solutions of Sample Exercise Solution Analyze We are asked to determine the p of aqueous solutions for which we have already calculated 3 4. Plan We can calculate p using its defining equation, Equation Solve (a) In the first instance we found 3 4 to be 1.0 * M, so that p = -log11.0 * = = Because 1.0 * has two significant figures, the p has two decimal places, (b) For the second solution, 3 4 = 5.6 * 10-6 M. Before performing the calculation, it is helpful to estimate the p. To do so, we note that 3 4 lies between 1 * 10-6 and 1 * Thus, we expect the p to lie between 6.0 and 5.0. We use Equation to calculate the p: p = -log15.6 * = 5.25 Check After calculating a p, it is useful to compare it to your estimate. In this case the p, as we predicted, falls between 6 and 5. ad the calculated p and the estimate not agreed, we should have reconsidered our calculation or estimate or both.

12 682 chapter 16 Acid Base Equilibria Practice Exercise 1 A solution at 25 C has 3O - 4 = 6.7 * What is the p of the solution? (a) 0.83 (b) 2.2 (c) 2.17 (d) (e) 12 Practice Exercise 2 (a) In a sample of lemon juice, 3 4 = 3.8 * 10-4 M. What is the p? (b) A commonly available window-cleaning solution has 3O - 4 = 1.9 * 10-6 M. What is the p at 25 C? po and Other p Scales The negative logarithm is a convenient way of expressing the magnitudes of other small quantities. We use the convention that the negative logarithm of a quantity is labeled p (quantity). Thus, we can express the concentration of O - as po: po = -log 3O - 4 [16.18] Likewise, pk w equals -log K w. By taking the negative logarithm of both sides of the equilibrium-constant expression for water, K w = 3 43O - 4, we obtain -log3 4 1-log 3O - 42 = -log K w [16.19] from which we obtain the useful expression p po = at 25 C2 [16.20] The p and po values characteristic of a number of familiar solutions are shown in ( Figure 16.6). Notice that a change in 3 4 by a factor of 10 causes the p to change by 1. Thus, the concentration of 1aq2 in a solution of p 5 is 10 times the 1aq2 concentration in a solution of p 6. Go Figure Which is more acidic, black coffee or lemon juice? [ ] (M) p po [O ] (M) 1 ( ) Increasing acid strength Stomach acid Lemon juice Cola, vinegar Wine Tomatoes Black coffee Rain Saliva Milk uman blood Seawater Borax Lime water ousehold ammonia ousehold bleach Increasing base strength ( ) p po = 14 [ ][O ] = Figure 16.6 Concentrations of and O -, and p and po values of some common substances at 25 C.

13 section 16.4 The p Scale 683 Give It Some Thought If the po for a solution is 3.00, what is the p? Is the solution acidic or basic? Sample Exercise 16.7 Calculating 3h 4 from po A sample of freshly pressed apple juice has a po of Calculate 3 4. Solution Analyze We need to calculate 3 4 from po. Plan We will first use Equation 16.20, p po = 14.00, to calculate p from po. Then we will use Equation to determine the concentration of. Solve From Equation 16.20, we have p = po p = = 3.76 Next we use Equation 16.17: p = -log3 4 = 3.76 Thus, log3 4 = To find 3 4, we need to determine the antilogarithm of Your calculator will show this command as 10 x or INV log (these functions are usually above the log key). We use this function to perform the calculation: 3 4 = antilog = = 1.7 * 10-4 M Comment The number of significant figures in 3 4 is two because the number of decimal places in the p is two. Check Because the p is between 3.0 and 4.0, we know that 3 4 will be between 1.0 * 10-3 M and 1.0 * 10-4 M. Our calculated 3 4 falls within this estimated range. Practice Exercise 1 A solution at 25 C has po = Which of the following statements is or are true? (i) The solution is acidic. (ii) The p of the solution is (iii) For this solution, 3O - 4 = M. (a) Only one of the statements is true. (b) Statements (i) and (ii) are true. (c) Statements (i) and (iii) are true. (d) Statements (ii) and (iii) are true. (e) All three statements are true. Practice Exercise 2 A solution formed by dissolving an antacid tablet has a po of Calculate 3 4. Measuring p The p of a solution can be measured with a p meter ( Figure 16.7). A complete understanding of how this important device works requires a knowledge of electrochemistry, a subject we take up in Chapter 20. In brief, a p meter consists of a pair of electrodes connected to a meter capable of measuring small voltages, on the order of millivolts. A voltage, which varies with p, is generated when the electrodes are placed in a solution. This voltage is read by the meter, which is calibrated to give p. Although less precise, acid base indicators can be used to measure p. An acid base indicator is a colored substance that can exist in either an acid or a base form. The two forms have different colors. Thus, the indicator has one color at lower p and another at higher p. If you know the p at which the indicator turns from one form to the other, you can determine whether a solution has a higher or lower p than this value. Litmus, for example, changes color in the vicinity of p 7. The color change, however, is not very sharp. Red litmus indicates a p of about 5 or lower, and blue litmus indicates a p of about 8 or higher. Some common indicators are listed in Figure The chart tells us, for instance, that methyl red changes color over the p interval from about 4.5 to 6.0. Below p 4.5 it is in the acid form, which is red. In the interval between 4.5 and 6.0, it is gradually converted to its basic form, which is yellow. Once the p rises above 6 the conversion Figure 16.7 A digital p meter. The device is a millivoltmeter, and the electrodes immersed in a solution produce a voltage that depends on the p of the solution.

14 684 chapter 16 Acid Base Equilibria Go Figure Which of these indicators is best suited to distinguish between a solution that is slightly acidic and one that is slightly basic? Go Figure If a colorless solution turns pink when we add phenolphthalein, what can we conclude about the p of the solution? Methyl violet p range for color change Yellow Violet Thymol blue Red Yellow Yellow Blue Methyl orange Red Yellow Methyl red Red Yellow Bromthymol blue Yellow Blue Phenolphthalein Colorless Pink Alizarin yellow R Yellow Red Methyl red Figure 16.8 p ranges for common acid base indicators. Most indicators have a useful range of about 2 p units. is complete, and the solution is yellow. This color change, along with that of the indicators bromthymol blue and phenolphthalein, is shown in Figure Paper tape impregnated with several indicators is widely used for determining approximate p values. Bromthymol blue Phenolphthalein Figure 16.9 Solutions containing three common acid base indicators at various p values Strong Acids and Bases The chemistry of an aqueous solution often depends critically on p. It is therefore important to examine how p relates to acid and base concentrations. The simplest cases are those involving strong acids and strong bases. Strong acids and bases are strong electrolytes, existing in aqueous solution entirely as ions. There are relatively few common strong acids and bases (see Table 4.2). Strong Acids The seven most common strong acids include six monoprotic acids (Cl, Br, I, NO 3, ClO 3, and ClO 4 ), and one diprotic acid 1 2 SO 4 2. Nitric acid 1NO 3 2 exemplifies the behavior of the monoprotic strong acids. For all practical purposes, an aqueous solution of NO 3 consists entirely of 3 O and NO 3 - ions: NO 3 1aq2 2 O1l2 3 O 1aq2 NO 3-1aq2 1complete ionization2 [16.21] We have not used equilibrium arrows for this equation because the reaction lies entirely to the right. (Section 4.1) As noted in Section 16.3, we use 3 O 1aq2 and 1aq2 interchangeably to represent the hydrated proton in water. Thus, we can simplify this acid ionization equation to NO 3 1aq2 1aq2 NO 3-1aq2 In an aqueous solution of a strong acid, the acid is normally the only significant source of ions.* As a result, calculating the p of a solution of a strong monoprotic *If the concentration of the acid is 10-6 M or less, we also need to consider ions that result from 2 O autoionization. Normally, the concentration of from 2 O is so small that it can be neglected.

15 section 16.5 Strong Acids and Bases 685 acid is straightforward because 3 4 equals the original concentration of acid. In a 0.20 M solution of NO 3 1aq2, for example, 3 4 = 3NO 3-4 = 0.20 M. The situation with the diprotic acid 2 SO 4 is somewhat more complex, as we will see in Section Sample Exercise 16.8 Calculating the p of a Strong Acid What is the p of a M solution of ClO 4? Solution Analyze and Plan Because ClO 4 is a strong acid, it is completely ionized, giving 3 4 = 3ClO 4-4 = M. Solve p = -log = 1.40 Check Because 3 4 lies between 1 * 10-2 and 1 * 10-1, the p will be between 2.0 and 1.0. Our calculated p falls within the estimated range. Furthermore, because the concentration has two significant figures, the p has two decimal places. Practice Exercise 1 Order the following three solutions from smallest to largest p: (i) 0.20 M ClO 3 (ii) M NO 3 (iii) 1.50 M Cl (a) i 6 ii 6 iii (b) ii 6 i 6 iii (c) iii 6 i 6 ii (d) ii 6 iii 6 i (e) iii 6 ii 6 i Practice Exercise 2 An aqueous solution of NO 3 has a p of What is the concentration of the acid? Strong Bases The most common soluble strong bases are the ionic hydroxides of the alkali metals, such as NaO, KO, and the ionic hydroxides heavier alkaline earth metals, such as Sr1O2 2. These compounds completely dissociate into ions in aqueous solution. Thus, a solution labeled 0.30 M NaO consists of 0.30 M Na 1aq2 and 0.30 M O - 1aq2; there is essentially no undissociated NaO. Give It Some Thought Which solution has the higher p, a M solution of NaO or a M solution of Ba1O2 2? Sample Exercise 16.9 Calculating the p of a Strong Base What is the p of (a) a M solution of NaO, (b) a M solution of Ca1O2 2? Solution Analyze We are asked to calculate the p of two solutions of strong bases. Plan We can calculate each p by either of two equivalent methods. First, we could use Equation to calculate 3 4 and then use Equation to calculate the p. Alternatively, we could use 3O - 4 to calculate po and then use Equation to calculate the p. Solve (a) NaO dissociates in water to give one O - ion per formula unit. Therefore, the O - concentration for the solution in (a) equals the stated concentration of NaO, namely M. Method 1: Method 2: 3 4 = 1.0 * = 3.57 * M p = -log13.57 * = po = -log = 1.55 p = po = (b) Ca1O2 2 is a strong base that dissociates in water to give two O - ions per formula unit. Thus, the concentration of O - 1aq2 for the solution in part (b) is 2 * M2 = M. Method 1: Method 2: 3 4 = 1.0 * = 4.55 * M p = -log14.55 * = po = -log = 2.66 p = po = 11.34

16 686 chapter 16 Acid Base Equilibria Practice Exercise 1 Order the following three solutions from smallest to largest p: (i) M Ba1O2 2 (ii) M KO (iii) pure water (a) i 6 ii 6 iii (b) ii 6 i 6 iii (c) iii 6 i 6 ii (d) ii 6 iii 6 i (e) iii 6 ii 6 i Practice Exercise 2 What is the concentration of a solution of (a) KO for which the p is 11.89, (b) Ca1O2 2 for which the p is 11.68? Although all of the alkali metal hydroxides are strong electrolytes, LiO, RbO, and CsO are not commonly encountered in the laboratory. The hydroxides of the heavier alkaline earth metals Ca1O2 2, Sr1O2 2, and Ba1O2 2 :are also strong electrolytes. They have limited solubility, however, so they are used only when high solubility is not critical. Strongly basic solutions are also created by certain substances that react with water to form O - 1aq2. The most common of these contain the oxide ion. Ionic metal oxides, especially Na 2 O and CaO, are often used in industry when a strong base is needed. The O 2- reacts very exothermically with water to form O -, leaving virtually no O 2- in the solution: O 2-1aq2 2 O1l2 2 O - 1aq2 [16.22] Thus, a solution formed by dissolving mol of Na 2 O1s2 in enough water to form 1.0 L of solution has 3O - 4 = M and a p of Give It Some Thought The C 3 - ion is the conjugate base of C 4, and C 4 shows no evidence of being an acid in water. Write a balanced equation for the reaction of C 3 - and water Weak Acids Most acidic substances are weak acids and therefore only partially ionized in aqueous solution ( Figure 16.10). We can use the equilibrium constant for the ionization reaction to express the extent to which a weak acid ionizes. If we represent a general weak acid as A, we can write the equation for its ionization in either of the following ways, depending on whether the hydrated proton is represented as 3 O 1aq2 or 1aq2: or A1aq2 2 O1l2 3 O 1aq2 A - 1aq2 [16.23] A1aq2 1aq2 A - 1aq2 [16.24] These equilibria are in aqueous solution, so we will use equilibrium-constant expressions based on concentrations. Because 2 O is the solvent, it is omitted from the equilibrium-constant expression. (Section 15.4) Further, we add a subscript a on the equilibrium constant to indicate that it is an equilibrium constant for the ionization of an acid. Thus, we can write the equilibrium-constant expression as either: K a = 3 3O 43A - 4 3A4 or K a = 3 43A - 4 3A4 [16.25] K a is called the acid-dissociation constant for acid A. Table 16.2 shows the structural formulas, conjugate bases, and K a values for a number of weak acids. Appendix D provides a more complete list. Many weak acids are organic compounds composed entirely of carbon, hydrogen, and oxygen. These compounds usually contain some hydrogen atoms bonded to carbon atoms and some bonded to oxygen atoms. In almost all cases, the hydrogen atoms bonded to carbon do not ionize in water; instead, the acidic behavior of these compounds is due to the hydrogen atoms attached to oxygen atoms.

17 section 16.6 Weak Acids 687 A A A 3 O 3 O A(aq) 2 O(l) A (aq) 3 O (aq) A(aq) 2 O(l) A (aq) 3 O (aq) Strong acid A molecules completely dissociate Weak acid A molecules partially dissociate Figure Species present in a solution of a strong acid and a weak acid. Table 16.2 Some Weak Acids in Water at 25 c Acid Structural Formula* Conjugate Base K a Chlorous 1ClO 2 2 O Cl O ClO * 10-2 ydrofluoric (F) F F * 10-4 Nitrous 1NO 2 2 O N O NO * 10-4 O Benzoic 1C 6 5 COO2 O C C 6 5 COO * 10-5 O Acetic 1C 3 COO2 O C C C 3 COO * 10-5 ypochlorous (OCl) O Cl OCl * 10-8 ydrocyanic (CN) C N CN * Phenol 1OC O C 6 5 O * *The proton that ionizes is shown in red. The magnitude of K a indicates the tendency of the acid to ionize in water: The larger the value of K a, the stronger the acid. Chlorous acid 1ClO 2 2, for example, is the strongest acid in Table 16.2, and phenol 1OC is the weakest. For most weak acids K a values range from 10-2 to Give It Some Thought Based on the entries in Table 16.2, which element is most commonly bonded to the acidic hydrogen?

18 688 chapter 16 Acid Base Equilibria Calculating K a from p In order to calculate either the K a value for a weak acid or the p of its solutions, we will use many of the skills for solving equilibrium problems developed in Section In many cases the small magnitude of K a allows us to use approximations to simplify the problem. In doing these calculations, it is important to realize that proton-transfer reactions are generally very rapid. As a result, the measured or calculated p for a weak acid always represents an equilibrium condition. Sample Exercise Calculating K a from Measured p A student prepared a 0.10 M solution of formic acid (COO) and found its p at 25 C to be Calculate K a for formic acid at this temperature. Solution Analyze We are given the molar concentration of an aqueous solution of weak acid and the p of the solution, and we are asked to determine the value of K a for the acid. Plan Although we are dealing specifically with the ionization of a weak acid, this problem is very similar to the equilibrium problems we encountered in Chapter 15. We can solve this problem using the method first outlined in Sample Exercise 15.8, starting with the chemical reaction and a tabulation of initial and equilibrium concentrations. Solve The first step in solving any equilibrium problem is to write the equation for the equilibrium reaction. The ionization of formic acid can be written as The equilibrium-constant expression is From the measured p, we can calculate 3 4: COO1aq2 1aq2 COO - 1aq2 K a = 3 43COO - 4 3COO4 p = -log 3 4 = 2.38 log3 4 = = = 4.2 * 10-3 M To determine the concentrations of the species involved in the equilibrium, we imagine that the solution is initially 0.10 M in COO molecules. We then consider the ionization of the acid into and COO -. For each COO molecule that ionizes, one ion and one COO - ion are produced in solution. Because the p measurement indicates that 3 4 = 4.2 * 10-3 M at equilibrium, we can construct the following table: COO1aq2 1aq2 COO - 1aq2 Initial concentration (M) Change in concentration (M) -4.2 * * * 10-3 Equilibrium concentration (M) * * * 10-3 Notice that we have neglected the very small concentration of 1aq2 due to 2 O autoionization. Notice also that the amount of COO that ionizes is very small compared with the initial concentration of the acid. To the number of significant figures we are using, the subtraction yields 0.10 M: * M 0.10 M We can now insert the equilibrium concentrations into the expression for K a : K a = 14.2 * * = 1.8 * 10-4 Check The magnitude of our answer is reasonable because K a for a weak acid is usually between 10-2 and

19 section 16.6 Weak Acids 689 Practice Exercise 1 A 0.50 M solution of an acid A has p = What is the value of K a for the acid? (a) 1.7 * (b) 3.3 * 10-5 (c) 6.6 * 10-5 (d) 5.8 * 10-3 (e) 1.2 * 10-2 Practice Exercise 2 Niacin, one of the B vitamins, has the molecular structure O C O N A M solution of niacin has a p of What is the acid-dissociation constant for niacin? Percent Ionization We have seen that the magnitude of K a indicates the strength of a weak acid. Another measure of acid strength is percent ionization, defined as concentration of ionized A Percent ionization = * 100% [16.26] original concentration of A The stronger the acid, the greater the percent ionization. If we assume that the autoionization of 2 O is negligible, the concentration of acid that ionizes equals the concentration of 1aq2 that forms. Thus, the percent ionization for an acid A can be expressed as Percent ionization = 3 4 equilibrium * 100% [16.27] 3A4 initial For example, a M solution of NO 2 contains 3.7 * 10-3 M 1aq2 and its percent ionization is Percent ionization = 3 4 equilibrium 3NO 2 4 initial * 100% = 3.7 * 10-3 M M * 100% = 11% Sample Exercise Calculating Percent Ionization As calculated in Sample Exercise 16.10, a 0.10 M solution of formic acid (COO) contains 4.2 * 10-3 M 1aq2. Calculate the percentage of the acid that is ionized. Solution Analyze We are given the molar concentration of an aqueous solution of weak acid and the equilibrium concentration of 1aq2 and asked to determine the percent ionization of the acid. Plan The percent ionization is given by Equation Solve Percent ionization = 3 4 equilibrium 3COO4 initial * 100% = 4.2 * 10-3 M 0.10 M * 100% = 4.2 Practice Exercise 1 A M solution of an acid A has p = What is the percentage of the acid that is ionized? (a) 0.090% (b) 0.69% (c) 0.90% (d) 3.6% (e) 9.0% Practice Exercise 2 A M solution of niacin has a p of Calculate the percent ionization of the niacin.

20 690 chapter 16 Acid Base Equilibria Using K a to Calculate p Knowing the value of K a and the initial concentration of a weak acid, we can calculate the concentration of 1aq2 in a solution of the acid. Let s calculate the p at 25 C of a 0.30 M solution of acetic acid 1C 3 COO2, the weak acid responsible for the characteristic odor and acidity of vinegar. 1. Our first step is to write the ionization equilibrium: C 3 COO1aq2 1aq2 C 3 COO - 1aq2 [16.28] Notice that the hydrogen that ionizes is the one attached to an oxygen atom. 2. The second step is to write the equilibrium-constant expression and the value for the equilibrium constant. Taking K a = 1.8 * 10-5 from Table 16.2, we write K a = 3 43C 3 COO - 4 3C 3 COO4 = 1.8 * 10-5 [16.29] 3. The third step is to express the concentrations involved in the equilibrium reaction. This can be done with a little accounting, as described in Sample Exercise Because we want to find the equilibrium value for 3 4, let s call this quantity x. The concentration of acetic acid before any of it ionizes is 0.30 M. The chemical equation tells us that for each molecule of C 3 COO that ionizes, one 1aq2 and one C 3 COO - 1aq2 are formed. Consequently, if x moles per liter of 1aq2 form at equilibrium, x moles per liter of C 3 COO - 1aq2 must also form and x moles per liter of C 3 COO must be ionized: C 3 COO1aq2 1aq2 C 3 COO - 1aq2 Initial concentration (M) Change in concentration (M) Equilibrium concentration (M) x x x x2 x x 4. The fourth step is to substitute the equilibrium concentrations into the equilibriumconstant expression and solve for x: K a = 3 43C 3 COO - 4 3C 3 COO4 = 1x21x x = 1.8 * 10-5 [16.30] This expression leads to a quadratic equation in x, which we can solve by using either an equation-solving calculator or the quadratic formula. We can simplify the problem, however, by noting that the value of K a is quite small. As a result, we anticipate that the equilibrium lies far to the left and that x is much smaller than the initial concentration of acetic acid. Thus, we assume that x is negligible relative to 0.30, so that x is essentially equal to We can (and should!) check the validity of this assumption when we finish the problem. By using this assumption, Equation becomes Solving for x, we have K a = x = 1.8 * 10-5 x 2 = * = 5.4 * 10-6 x = 25.4 * 10-6 = 2.3 * = x = 2.3 * 10-3 M p = -log12.3 * = 2.64

21 section 16.6 Weak Acids 691 Now we check the validity of our simplifying assumption that x The value of x we determined is so small that, for this number of significant figures, the assumption is entirely valid. We are thus satisfied that the assumption was a reasonable one to make. Because x represents the moles per liter of acetic acid that ionize, we see that, in this particular case, less than 1% of the acetic acid molecules ionize: Percent ionization of C 3 COO = M 0.30 M * 100% = 0.77% As a general rule, if x is more than about 5% of the initial concentration value, it is better to use the quadratic formula. You should always check the validity of any simplifying assumptions after you have finished solving a problem. We have also made one other assumption, namely that all of the in the solution comes from ionization of C 3 COO. Are we justified in neglecting the autoionization of 2 O? The answer is yes the additional [ 4 due to water, which would be on the order of 10-7 M, is negligible compared to the [ 4 from the acid (which in this case is on the order of 10-3 M). In extremely precise work, or in cases involving very dilute solutions of acids, we would need to consider the autoionization of water more fully. Give It Some Thought Would a 1.0 * 10-8 M solution of Cl have p 6 7, p = 7, or p 7 7? Finally, we can compare the p value of this weak acid with the p of a solution of a strong acid of the same concentration. The p of the 0.30 M acetic acid is 2.64, but the p of a 0.30 M solution of a strong acid such as Cl is -log = As expected, the p of a solution of a weak acid is higher than that of a solution of a strong acid of the same molarity. (Remember, the higher the p value, the less acidic the solution.) Sample Exercise Using K a to Calculate p Calculate the p of a 0.20 M solution of CN. (Refer to Table 16.2 or Appendix D for the value of K a.) Solution Analyze We are given the molarity of a weak acid and are asked for the p. From Table 16.2, K a for CN is 4.9 * Solve Writing both the chemical equation for the ionization reaction that forms 1aq2 and the equilibrium-constant 1K a 2 expression for the reaction: Plan We proceed as in the example just worked in the text, writing the chemical equation and constructing a table of initial and equilibrium concentrations in which the equilibrium concentration of is our unknown. CN1aq2 1aq2 CN - 1aq2 K a = 3 43CN - 4 3CN4 = 4.9 * Next, we tabulate the concentrations of the species involved in the equilibrium reaction, letting x = 3 4 at equilibrium: Initial concentration (M) Change in concentration (M) Equilibrium concentration (M) CN1aq2 1aq2 CN - 1aq x x x x2 x x Substituting the equilibrium concentrations into the equilibriumconstant expression yields We next make the simplifying approximation that x, the amount of acid that dissociates, is small compared with the initial concentration of acid, x Thus, K a = 1x21x2 = 4.9 * x x = 4.9 * 10-10

22 692 chapter 16 Acid Base Equilibria Solving for x, we have x 2 = * = 0.98 * x = * = 9.9 * 10-6 M = 3 4 A concentration of 9.9 * 10-6 M is much smaller than 5% of 0.20, the initial CN concentration. Our simplifying approximation is therefore appropriate. We now calculate the p of the solution: p = -log3 4 = -log19.9 * = 5.00 Practice Exercise 1 What is the p of a 0.40 M solution of benzoic acid, C 6 5 COO? (The K a value for benzoic acid is given in Table 16.2.) (a) 2.30 (b) 2.10 (c) 1.90 (d) 4.20 (e) 4.60 Practice Exercise 2 The K a for niacin (Practice Exercise 16.10) is 1.5 * What is the p of a M solution of niacin? The properties of an acid solution that relate directly to the concentration of 1aq2, such as electrical conductivity and rate of reaction with an active metal, are less evident for a solution of a weak acid than for a solution of a strong acid of the same concentration. Figure presents an experiment that demonstrates this difference with 1 M C 3 COO and 1 M Cl. The concentration of 1aq2 in 1 M C 3 COO is only M, whereas the 1 M Cl solution contains 1 M 1aq2. As a result, the reaction rate with the metal is much faster in the Cl solution. As the concentration of a weak acid increases, the equilibrium concentration of 1aq2 increases, as expected. owever, as shown in Figure 16.12, the percent ionization decreases as the concentration increases. Thus, the concentration of 1aq2 is not directly proportional to the concentration of the weak acid. For example, doubling the concentration of a weak acid does not double the concentration of 1aq2. Reaction proceeds more rapidly in strong acid, leading to formation of larger 2 bubbles and rapid disappearance of metal Reaction complete in strong acid 2 bubbles show reaction still in progress in weak acid Reaction eventually goes to completion in both acids 1 M Cl(aq) [ ] = 1 M 1 M C 3 COO(aq) [ ] = M Figure Rates of the same reaction run in a weak acid and a strong acid. The bubbles are 2 gas, which along with metal cations, is produced when a metal is oxidized by an acid. (Section 4.4)

23 section 16.6 Weak Acids 693 Go Figure Is the trend observed in this graph consistent with Le Châtelier s principle? Explain. 6.0 Percent ionized As concentration increases, a smaller percentage of C 3 COO molecules dissociates Acid concentration (M) Figure Effect of concentration on percent ionization in an acetic acid solution. Sample Exercise Using the Quadratic Equation to Calculate p and Percent Ionization Calculate the p and percentage of F molecules ionized in a 0.10 M F solution. Solution Analyze We are asked to calculate the percent ionization of a solution of F. From Appendix D, we find K a = 6.8 * Plan We approach this problem as for previous equilibrium problems: We write the chemical equation for the equilibrium and tabulate the known and unknown concentrations of all species. We then substitute the equilibrium concentrations into the equilibrium-constant expression and solve for the unknown concentration of. Solve The equilibrium reaction and equilibrium concentrations are as follows: Initial concentration (M) Change in concentration (M) Equilibrium concentration (M) F1aq2 1aq2 F - 1aq x x x x2 x x The equilibrium-constant expression is When we try solving this equation using the approximation x 0.10 (that is, by neglecting the concentration of acid that ionizes), we obtain Because this approximation is greater than 5% of 0.10 M, however, we should work the problem in standard quadratic form. Rearranging, we have Substituting these values in the standard quadratic formula gives K a = 3 43F - 4 3F4 x = 8.2 * 10-3 M = 1x21x x x 2 = x216.8 * = 6.8 * * x = 6.8 * 10-4 x * x * 10-5 = 0 x = -6.8 * 10-4 { * * = -6.8 * 10-4 { 1.6 *

24 694 chapter 16 Acid Base Equilibria Of the two solutions, only the positive value for x is chemically reasonable. From that value, we can determine 3 4 and hence the p x = 3 4 = 3F - 4 = 7.9 * 10-3 M, so p = -log3 4 = 2.10 From our result, we can calculate the percent of molecules ionized: concentration ionized Percent ionization of F = original concentration * 100% Practice Exercise 1 What is the p of a M solution of F? (a) 1.58 (b) 2.10 (c) 2.30 (d) 2.58 (e) 2.64 = 7.9 * 10-3 M 0.10 M Practice Exercise 2 In Practice Exercise 2 for Sample Exercise 16.11, we found that the percent ionization of niacin 1K a = 1.5 * in a M solution is 2.7%. Calculate the percentage of niacin molecules ionized in a solution that is (a) M, (b) 1.0 * 10-3 M. * 100% = 7.9% Polyprotic Acids Acids that have more than one ionizable atom are known as polyprotic acids. Sulfurous acid 1 2 SO 3 2, for example, can undergo two successive ionizations: 2 SO 3 1aq2 1aq2 SO 3-1aq2 K a1 = 1.7 * 10-2 [16.31] SO 3-1aq2 1aq2 SO 3 2-1aq2 K a2 = 6.4 * 10-8 [16.32] Note that the acid-dissociation constants are labeled K a1 and K a2. The numbers on the constants refer to the particular proton of the acid that is ionizing. Thus, K a2 always refers to the equilibrium involving removal of the second proton of a polyprotic acid. We see that K a2 for sulfurous acid is much smaller than K a1. Because of electrostatic attractions, we would expect a positively charged proton to be lost more readily from the neutral 2 SO 3 molecule than from the negatively charged SO 3 - ion. This observation is general: It is always easier to remove the first proton from a polyprotic acid than to remove the second. Similarly, for an acid with three ionizable protons, it is easier to remove the second proton than the third. Thus, the K a values become successively smaller as successive protons are removed. Give It Some Thought What is the equilibrium associated with K a3 for 3 PO 4? Go Figure Citric acid has four hydrogen atoms bonded to oxygen. ow does the hydrogen atom that is not an acidic proton differ from the other three? O 2 C C 2 C O C O C C O Citric acid O O O Figure The structure of the polyprotic acid, citric acid. The acid-dissociation constants for common polyprotic acids are listed in Table 16.3, and Appendix D provides a more complete list. The structure of citric acid illustrates the presence of multiple ionizable protons Figure Table 16.3 Acid-Dissociation Constants of Some Common Polyprotic Acids Name Formula K a1 K a2 K a3 Ascorbic 2 C 6 6 O * * Carbonic 2 CO * * Citric 3 C 6 5 O * * * 10-7 Oxalic OOC COO 5.9 * * 10-5 Phosphoric 3 PO * * * Sulfurous 2 SO * * 10-8 Sulfuric 2 SO 4 Large 1.2 * 10-2 Tartaric C 2 2 O 2 1COO * * 10-5

25 section 16.6 Weak Acids 695 Notice in Table 16.3 that in most cases the K a values for successive losses of protons differ by a factor of at least Notice also that the value of K a1 for sulfuric acid is listed simply as large. Sulfuric acid is a strong acid with respect to the removal of the first proton. Thus, the reaction for the first ionization step lies completely to the right: 2 SO 4 1aq2 1aq2 SO 4-1aq2 1complete ionization2 owever, SO 4 - is a weak acid for which K a2 = 1.2 * For many polyprotic acids K a1 is much larger than subsequent dissociation constants, in which case the 1aq2 in the solution comes almost entirely from the first ionization reaction. As long as successive K a values differ by a factor of 10 3 or more, it is usually possible to obtain a satisfactory estimate of the p of polyprotic acid solutions by treating the acids as if they were monoprotic, considering only K a1. Sample Exercise Calculating the p of a Solution of a Polyprotic Acid The solubility of CO 2 in water at 25 C and 0.1 atm is M. The common practice is to assume that all the dissolved CO 2 is in the form of carbonic acid 1 2 CO 3 2, which is produced in the reaction What is the p of a M solution of 2 CO 3? CO 2 1aq2 2 O1l2 2 CO 3 1aq2 Solution Analyze We are asked to determine the p of a M solution of a polyprotic acid. Plan 2 CO 3 is a diprotic acid; the two acid-dissociation constants, K a1 and K a2 (Table 16.3), differ by more than a factor of Consequently, the p can be determined by considering only K a1, thereby treating the acid as if it were a monoprotic acid. Solve Proceeding as in Sample Exercises and 16.13, we can write the equilibrium reaction and equilibrium concentrations as Initial concentration (M) Change in concentration (M) Equilibrium concentration (M) 2 CO 3 1aq2 1aq2 CO - 3 1aq x x x x2 x x The equilibrium-constant expression is K a1 = 3 43CO CO 3 4 x = 4.0 * 10-5 M = 1x21x x = 4.3 * 10-7 Solving this quadratic equation, we get Alternatively, because K a1 is small, we can make the simplifying approximation that x is small, so that x Thus, 1x21x = 4.3 * 10-7 Solving for x, we have x 2 = * = 1.6 * 10-9 x = 3 4 = 3CO 3-4 = 21.6 * 10-9 = 4.0 * 10-5 M Because we get the same value (to 2 significant figures) our simplifying assumption was justified. The p is therefore p = -log3 4 = -log14.0 * = 4.40

26 696 chapter 16 Acid Base Equilibria Comment If we were asked for 3CO we would need to use K a2. Let s illustrate that calculation. Using our calculated values of 3CO 3-4 and 3 4 and setting 3CO = y, we have Initial concentration (M) Change in concentration (M) Equilibrium concentration (M) CO - 3 1aq2 1aq2 CO 2-3 1aq2 4.0 * * y y y 14.0 * y * 10-5 y2 y Assuming that y is small relative to 4.0 * 10-5, we have K a2 = 3 43CO 3 3CO y = 5.6 * M = 3CO = 14.0 * y2 4.0 * 10-5 = 5.6 * We see that the value for y is indeed very small compared with 4.0 * 10-5, showing that our assumption was justified. It also shows that the ionization of CO 3 - is negligible relative to that of 2 CO 3, as far as production of is concerned. owever, it is the only source of CO 3 2-, which has a very low concentration in the solution. Our calculations thus tell us that in a solution of carbon dioxide in water, most of the CO 2 is in the form of CO 2 or 2 CO 3, only a small fraction ionizes to form and CO 3 -, and an even smaller fraction ionizes to give CO Notice also that 3CO is numerically equal to K a2. Practice Exercise 1 What is the p of a 0.28 M solution of ascorbic acid (Vitamin C)? (See Table 16.3 for K a1 and K a2.) (a) 2.04 (b) 2.32 (c) 2.82 (d) 4.65 (e) 6.17 Practice Exercise 2 (a) Calculate the p of a M solution of oxalic acid 1 2 C 2 O 4 2. (See Table 16.3 for K a1 and K a2.) (b) Calculate the concentration of oxalate ion, 3C 2 O 4 2-4, in this solution Weak Bases Many substances behave as weak bases in water. Weak bases react with water, abstracting protons from 2 O, thereby forming the conjugate acid of the base and O - ions: B1aq2 2 O1l2 B 1aq2 O - 1aq2 [16.33] The equilibrium-constant expression for this reaction can be written as K b = 3B 43O - 4 3B4 [16.34] Water is the solvent, so it is omitted from the equilibrium-constant expression. One of the most commonly encountered weak bases is ammonia, N 3 : N 3 1aq2 2 O1l2 N 4 1aq2 O - 1aq2 K b = 3N 4 43O - 4 3N 3 4 [16.35] As with K w and K a, the subscript b in K b denotes that the equilibrium constant refers to a particular type of reaction, namely the ionization of a weak base in water. The constant K b, the base-dissociation constant, always refers to the equilibrium in which a base reacts with 2 O to form the corresponding conjugate acid and O -. Table 16.4 lists the Lewis structures, conjugate acids, and K b values for a number of weak bases in water. Appendix D includes a more extensive list. These bases contain one or more lone pairs of electrons because a lone pair is necessary to form the bond with. Notice that in the neutral molecules in Table 16.4, the lone pairs are on nitrogen atoms. The other bases listed are anions derived from weak acids.

27 section 16.7 Weak Bases 697 Table 16.4 Some Weak Bases in Water at 25 c Base Structural Formula* Conjugate Acid K b Ammonia 1N 3 2 N N * 10-5 Pyridine 1C 5 5 N2 N C 5 5 N 1.7 * 10-9 ydroxylamine 1ON 2 2 N O ON * 10-8 Methylamine 1C 3 N 2 2 N C 3 C 3 N * 10-4 ydrosulfide ion 1S - 2 S 2 S 1.8 * 10-7 Carbonate ion 1CO C CO 3 O O O * 10-4 ypochlorite ion 1ClO - 2 Cl O ClO 3.3 * 10-7 *The atom that accepts the proton is shown in blue. Sample Exercise Using K b to Calculate oh - Calculate the concentration of O - in a 0.15 M solution of N 3. Solution Analyze We are given the concentration of a weak base and asked to determine the concentration of O -. Plan We will use essentially the same procedure here as used in solving problems involving the ionization of weak acids that is, write the chemical equation and tabulate initial and equilibrium concentrations. Solve The ionization reaction and equilibriumconstant expression are Ignoring the concentration of 2 O because it is not involved in the equilibrium-constant expression, the equilibrium concentrations are N 3 1aq2 2 O1l2 N 4 1aq2 O - 1aq2 K b = 3N 4 43O - 4 3N 3 4 Initial concentration (M) Change in concentration (M) Equilibrium concentration (M) = 1.8 * 10-5 N 3 1aq2 2 O1l2 N 4 1aq2 O x x x x2 x x Inserting these quantities into the equilibriumconstant expression gives K b = 3N 4 43O - 4 = 1x21x2 3N x = 1.8 * 10-5 Because K b is small, the amount of N 3 that reacts with water is much smaller than the N 3 concentration, and so we can neglect x relative to 0.15 M. Then we have x = 1.8 * 10-5 x 2 = * = 2.7 * 10-6 x = 3N 4 4 = 3O - 4 = 22.7 * 10-6 = 1.6 * 10-3 M

28 698 chapter 16 Acid Base Equilibria Check The value obtained for x is only about 1% of the N 3 concentration, 0.15 M. Therefore, neglecting x relative to 0.15 was justified. Comment You may be asked to find the p of a solution of a weak base. Once you have found 3O - 4, you can proceed as in Sample Exercise 16.9, where we calculated the p of a strong base. In the present sample exercise, we have seen that the 0.15 M solution of N 3 contains 3O - 4 = 1.6 * 10-3 M. Thus, po = -log11.6 * = 2.80, and p = = The p of the solution is above 7 because we are dealing with a solution of a base. Practice Exercise 1 What is the p of a 0.65 M solution of pyridine, C 5 5 N? (See Table 16.4 for K b.) (a) 4.48 (b) 8.96 (c) 9.52 (d) 9.62 (e) 9.71 Practice Exercise 2 Which of the following compounds should produce the highest p as a 0.05 M solution: pyridine, methylamine, or nitrous acid? Go Figure When hydroxylamine acts as a base, which atom accepts the proton? Ammonia N 3 Types of Weak Bases Weak bases fall into two general categories. The first category is neutral substances that have an atom with a nonbonding pair of electrons that can accept a proton. Most of these bases, including all uncharged bases in Table 16.4, contain a nitrogen atom. These substances include ammonia and a related class of compounds called amines ( Figure 16.14). In organic amines, at least one N bond in N 3 is replaced with an N C bond. Like N 3, amines can abstract a proton from a water molecule by forming an N bond, as shown here for methylamine: N C 3 (aq) 2 O(l) N C 3 (aq) O (aq) [16.36] Methylamine C 3 N 2 Anions of weak acids make up the second general category of weak bases. In an aqueous solution of sodium hypochlorite (NaClO), for example, NaClO dissociates to Na and ClO - ions. The Na ion is always a spectator ion in acid base reactions. (Section 4.3) The ClO - ion, however, is the conjugate base of a weak acid, hypochlorous acid. Consequently, the ClO - ion acts as a weak base in water: ClO - 1aq2 2 O1l2 ClO1aq2 O - 1aq2 K b = 3.3 * 10-7 [16.37] In Figure 16.6 we saw that bleach is quite basic (p values of 12 13). Common chlorine bleach is typically a 5% NaOCl solution. ydroxylamine N 2 O Figure Structures of ammonia and two simple amines. Sample Exercise Using p to Determine the Concentration of a Salt A solution made by adding solid sodium hypochlorite (NaClO) to enough water to make 2.00 L of solution has a p of Using the information in Equation 16.37, calculate the number of moles of NaClO added to the water. Solution Analyze NaClO is an ionic compound consisting of Na and ClO - ion. As such, it is a strong electrolyte that completely dissociates in solution into Na, a spectator ion, and ClO - ion, a weak base with K b = 3.3 * 10-7 (Equation 16.37). Given this information we must calculate the number of moles of NaClO needed to increase the p of 2.00-L of water to Plan From the p, we can determine the equilibrium concentration of O -. We can then construct a table of initial and equilibrium concentrations in which the initial concentration of ClO - is our unknown. We can calculate 3ClO - 4 using the expression for K b.

29 section 16.8 Relationship Between k a and k b 699 Solve We can calculate 3O - 4 by using either Equation or Equation 16.20; we will use the latter method here: po = p = = O - 4 = = 3.2 * 10-4 M This concentration is high enough that we can assume that Equation is the only source of O - ; that is, we can neglect any O - produced by the autoionization of 2 O. We now assume a value of x for the initial concentration of ClO - and solve the equilibrium problem in the usual way. Initial concentration (M) Change in concentration (M) Equilibrium concentration (M) ClO - 1aq2 2 O1l2 ClO1aq2 O - 1aq2 x * * * x * * * 10-4 We now use the expression for the basedissociation constant to solve for x: K b = 3ClO43O- 4 3ClO - 4 = 13.2 * = 3.3 * 10-7 x * 10 x = 13.2 * * * = 0.31 M We say that the solution is 0.31 M in NaClO even though some of the ClO - ions have reacted with water. Because the solution is 0.31 M in NaClO and the total volume of solution is 2.00 L, 0.62 mol of NaClO is the amount of the salt that was added to the water. Practice Exercise 1 The benzoate ion, C 6 5 COO -, is a weak base with K b = 1.6 * ow many moles of sodium benzoate are present in 0.50 L of a solution of NaC 6 5 COO if the p is 9.04? (a) 0.38 (b) 0.66 (c) 0.76 (d) 1.5 (e) 2.9 Practice Exercise 2 What is the molarity of an aqueous N 3 solution that has a p of 11.17? 16.8 Relationship Between K a and K b We have seen in a qualitative way that the stronger an acid, the weaker its conjugate base. To see if we can find a corresponding quantitative relationship, let s consider the N 4 and N 3 conjugate acid base pair. Each species reacts with water. For the acid, N 4, the equilibrium is N 4 1aq2 2 O1l2 N 3 1aq2 3 O 1aq2 or written in its simpler form: N 4 1aq2 N 3 1aq2 1aq2 [16.38] For the base, N 3, the equilibrium is N 3 1aq2 2 O1l2 N 4 1aq2 O - 1aq2 [16.39] Each equilibrium is expressed by a dissociation constant: K a = 3N N 4 4 K b = 3N 4 43O - 4 3N 3 4 When we add Equations and 16.39, the N 4 and N 3 species cancel and we are left with the autoionization of water: N 4 1aq2 N 3 1aq2 1aq2 N 3 1aq2 2 O1l2 N 4 1aq2 O - 1aq2 2 O1l2 1aq2 O - 1aq2 Recall that when two equations are added to give a third, the equilibrium constant associated with the third equation equals the product of the equilibrium constants of the first two equations. (Section 15.3)

30 700 chapter 16 Acid Base Equilibria Applying this rule to our present example, we see that when we multiply K a and K b, we obtain K a * K b = a 3N N 4 4 = 3 43O - 4 = K w b a 3N 4 43O - 4 b 3N 3 4 Thus, the product of K a and K b is the ion-product constant for water, K w (Equation 16.16). We expect this result because adding Equations and gave us the autoionization equilibrium for water, for which the equilibrium constant is K w. The above result holds for any conjugate acid base pair. In general, the product of the acid-dissociation constant for an acid and the base-dissociation constant for its conjugate base equals the ion-product constant for water: K a * K b = K w 1for a conjugate acid9base pair2 [16.40] As the strength of an acid increases (K a gets larger), the strength of its conjugate base must decrease (K b gets smaller) so that the product K a * K b remains 1.0 * at 25 C. Table 16.5 demonstrates this relationship. Remember, this important relationship applies only to conjugate acid base pairs. By using Equation 16.40, we can calculate K b for any weak base if we know K a for its conjugate acid. Similarly, we can calculate K a for a weak acid if we know K b for its conjugate base. As a practical consequence, ionization constants are often listed for only one member of a conjugate acid base pair. For example, Appendix D does not contain K b values for the anions of weak acids because they can be readily calculated from the tabulated K a values for their conjugate acids. Recall that we often express 3 4 as p: p = -log 3 4. (Section 16.4) This p nomenclature is often used for other very small numbers. For example, if you look up the values for acid- or base-dissociation constants in a chemistry handbook, you may find them expressed as pk a or pk b : pk a = -log K a and pk b = -log K b [16.41] Using this nomenclature, Equation can be written in terms of pk a and pk b by taking the negative logarithm of both sides: pk a pk b = pk w = at 25 C 1conjugate acid9base pair2 [16.42] Give It Some Thought K a for acetic acid is 1.8 * What is the first digit of the pk a value for acetic acid? Table 16.5 Some Conjugate Acid Base Pairs Acid K a Base K b NO 3 (Strong acid) - NO 3 (Negligible basicity) F 6.8 * 10-4 F * C 3 COO 1.8 * 10-5 C 3 COO * CO * CO * 10-8 N 4 CO * N * * CO * 10-4 O - (Negligible acidity) O 2 - (Strong base)

31 section 16.8 Relationship Between Ka and Kb 701 Chemistry Put to Work Amines and Amine ydrochlorides Many low-molecular-weight amines have a fishy odor. Amines and N3 are produced by the anaerobic (absence of O2) decomposition of dead animal or plant matter. Two such amines with very disagreeable aromas are 2N1C224N2, putrescine, and 2N1C225N2, cadaverine. The names of these substances are testaments to their repugnant odors! Many drugs, including quinine, codeine, caffeine, and amphetamine, are amines. Like other amines, these substances are weak bases; the amine nitrogen is readily protonated upon treatment with an acid. The resulting products are called acid salts. If we use A as the abbreviation for an amine, the acid salt formed by reaction with hydrochloric acid can be written ACl-. It can also be written as A # Cl and referred to as a hydrochloride. Amphetamine hydrochloride, for example, is the acid salt formed by treating amphetamine with Cl: C2 examples of over-the-counter medications that contain amine hydrochlorides as active ingredients are shown in Figure Related Exercises: 16.9, 16.73, 16.74, , , N2(aq) Cl(aq) C C3 Amphetamine C2 C N3Cl (aq) C3 Amphetamine hydrochloride Acid salts are much less volatile, more stable, and generally more water soluble than the corresponding amines. For this reason, many drugs that are amines are sold and administered as acid salts. Some Figure Some over-the-counter medications in which an amine hydrochloride is a major active ingredient. Sa m p le Exercise Calculating Ka or Kb for a Conjugate Acid Base Pair Calculate (a) Kb for the fluoride ion, (b) Ka for the ammonium ion. Solution Analyze We are asked to determine dissociation constants for F -, the conjugate base of F, and N4, the conjugate acid of N3. Plan We can use the tabulated K values for F and N3 and the relationship between Ka and Kb to calculate the dissociation constants for their conjugates, F - and N4. Solve (a) For the weak acid F, Table 16.2 and Appendix D give Ka = 6.8 * We can use Equation to calculate Kb for the conjugate base, F -: Kw 1.0 * Kb = = = 1.5 * Ka 6.8 * 10-4 (b) For N3, Table 16.4 and in Appendix D give Kb = 1.8 * 10-5, and this value in Equation gives us Ka for the conjugate acid, N4: Kw 1.0 * Ka = = = 5.6 * Kb 1.8 * 10-5 Check The respective K values for F - and N4 are listed in Table 16.5, where we see that the values calculated here agree with those in Table Practice Exercise 1 By using information from Appendix D, put the following three substances in order of weakest to strongest base: (i) 1C323N, (ii) COO-, (iii) BrO-. (a) i 6 ii 6 iii (b) ii 6 i 6 iii (c) iii 6 i 6 ii (d) ii 6 iii 6 i (e) iii 6 ii 6 i. Practice Exercise 2 (a) Based on information in Appendix D, which of these anions has the largest base-dissociation constant: NO2-, PO43 -, or N3-? (b) The base quinoline has the structure N Its conjugate acid is listed in handbooks as having a pka of What is the base-dissociation constant for quinoline?

32 702 chapter 16 Acid Base Equilibria 16.9 Acid Base Properties of Salt Solutions Even before you began this chapter, you were undoubtedly aware of many substances that are acidic, such as NO 3, Cl, and 2 SO 4, and others that are basic, such as NaO and N 3. owever, our discussion up to this point in the chapter has indicated that ions can also exhibit acidic or basic properties. For example, we calculated K a for N 4 and K b for F - in Sample Exercise Such behavior implies that salt solutions can be acidic or basic. Before proceeding with further discussions of acids and bases, let s examine the way dissolved salts can affect p. Because nearly all salts are strong electrolytes, we can assume that any salt dissolved in water is completely dissociated. Consequently, the acid base properties of salt solutions are due to the behavior of the cations and anions. Many ions react with water to generate 1aq2 or O - 1aq2 ions. This type of reaction is often called hydrolysis. The p of an aqueous salt solution can be predicted qualitatively by considering the salt s cations and anions. An Anion s Ability to React with Water In general, an anion A - in solution can be considered the conjugate base of an acid. For example, Cl - is the conjugate base of Cl, and C 3 COO - is the conjugate base of C 3 COO. Whether an anion reacts with water to produce hydroxide ions depends on the strength of the anion s conjugate acid. To identify the acid and assess its strength, we add a proton to the anion s formula. If the acid A determined in this way is one of the seven strong acids listed at the beginning of Section 16.5, the anion has a negligible tendency to produce O - ions from water and does not affect the p of the solution. The presence of Cl - in an aqueous solution, for example, does not result in the production of any O - and does not affect the p. Thus, Cl - is always a spectator ion in acid base chemistry. If A is not one of the seven common strong acids, it is a weak acid. In this case, the conjugate base A - is a weak base and it reacts to a small extent with water to produce the weak acid and hydroxide ions: A - 1aq2 2 O1l2 A1aq2 O - 1aq2 [16.43] The O - ion generated in this way increases the p of the solution, making it basic. Acetate ion, for example, being the conjugate base of a weak acid, reacts with water to produce acetic acid and hydroxide ions, thereby increasing the p of the solution: C 3 COO - 1aq2 2 O1l2 C 3 COO1aq2 O - 1aq2 [16.44] Give It Some Thought Will NO 3 - ions affect the p of a solution? What about CO 3 2- ions? The situation is more complicated for salts containing anions that have ionizable protons, such as SO 3 -. These salts are amphiprotic (Section 16.2), and how they behave in water is determined by the relative magnitudes of K a and K b for the ion, as shown in Sample Exercise If K a 7 K b, the ion causes the solution to be acidic. If K b 7 K a, the solution is made basic by the ion. A Cation s Ability to React with Water Polyatomic cations containing one or more protons can be considered the conjugate acids of weak bases. The N 4 ion, for example, is the conjugate acid of the weak base N 3. Thus, N 4 is a weak acid and will donate a proton to water, producing hydronium ions and thereby lowering the p: N 4 1aq2 2 O1l2 N 3 1aq2 3 O 1aq2 [16.45] Many metal ions react with water to decrease the p of an aqueous solution. This effect is most pronounced for small, highly charged cations like Fe 3 and Al 3, as illustrated

33 section 16.9 Acid Base Properties of Salt Solutions 703 by the K a values for metal cations in Table A comparison of Fe 2 and Fe 3 values in the table illustrates how acidity increases as ionic charge increases. Notice that K a values for the 3 ions in Table 16.6 are comparable to the values for familiar weak acids, such as acetic acid 1K a = 1.8 * In contrast, the ions of alkali and alkaline earth metals, being relatively large and not highly charged, do not react with water and therefore do not affect p. Note that these are the same cations found in the strong bases (Section 16.5). The different tendencies of four cations to lower the p of a solution are illustrated in Figure The mechanism by which metal ions produce acidic solutions is shown in Figure Because metal ions are positively charged, they attract the unshared Go Figure Why do we need to use two different acid-base indicators in this figure? Table 16.6 Acid-Dissociation Constants for Metal Cations in Aqueous Solution at 25 c Cation K a Fe * Zn * Ni * Fe * 10-3 Cr * 10-4 Al * 10-5 NaNO 3 Bromothymol blue p = 7.0 Ca(NO 3 ) 2 Bromothymol blue p = 6.9 Zn(NO 3 ) 2 Methyl red p = 5.5 Al(NO 3 ) 3 Methyl orange p = 3.5 Figure Effect of cations on solution p. The p values of 1.0 M solutions of four nitrate salts are estimated using acid base indicators. Interaction between Fe 3 and oxygen of bound 2 O molecule weakens O bonds lost, charge of complex ion changes from 3 to 2 3 O created, solution becomes acidic 3 2 [Fe( 2 O) 6 ] 3 (aq) 2 O(l) [Fe( 2 O) 5 (O)] 2 (aq) 3 O (aq) Figure A hydrated Fe 3 ion acts as an acid by donating an to a free 2 O molecule, forming 3 O.

34 704 chapter 16 Acid Base Equilibria electron pairs of water molecules and become hydrated. (Section 13.1) The larger the charge on the metal ion, the stronger the interaction between the ion and the oxygen of its hydrating water molecules. As the strength of this interaction increases, the O bonds in the hydrating water molecules become weaker. This facilitates transfer of protons from the hydration water molecules to solvent water molecules. Combined Effect of Cation and Anion in Solution To determine whether a salt forms an acidic, a basic, or a neutral solution when dissolved in water, we must consider the action of both cation and anion. There are four possible combinations. 1. If the salt contains an anion that does not react with water and a cation that does not react with water, we expect the p to be neutral. Such is the case when the anion is a conjugate base of a strong acid and the cation is either from group 1A or one of the heavier members of group 2A 1Ca 2, Sr 2, and Ba 2 2. Examples: NaCl, Ba1NO 3 2 2, RbClO If the salt contains an anion that reacts with water to produce hydroxide ions and a cation that does not react with water, we expect the p to be basic. Such is the case when the anion is the conjugate base of a weak acid and the cation is either from group 1A or one of the heavier members of group 2A 1Ca 2, Sr 2, and Ba 2 2. Examples: NaClO, RbF, BaSO If the salt contains a cation that reacts with water to produce hydronium ions and an anion that does not react with water, we expect the p to be acidic. Such is the case when the cation is a conjugate acid of a weak base or a small cation with a charge of 2 or greater. Examples: N 4 NO 3, AlCl 3, Fe1NO If the salt contains an anion and a cation both capable of reacting with water, both hydroxide ions and hydronium ions are produced. Whether the solution is basic, neutral, or acidic depends on the relative abilities of the ions to react with water. Examples: N 4 ClO, Al1C 3 COO2 3, CrF 3. Sample Exercise Determining Whether Salt Solutions Are Acidic, Basic, or Neutral Determine whether aqueous solutions of each of these salts are acidic, basic, or neutral: (a) Ba1C 3 COO2 2, (b) N 4 Cl, (c) C 3 N 3 Br, (d) KNO 3, (e) Al1ClO Solution Analyze We are given the chemical formulas of five ionic compounds (salts) and asked whether their aqueous solutions will be acidic, basic, or neutral. Plan We can determine whether a solution of a salt is acidic, basic, or neutral by identifying the ions in solution and by assessing how each ion will affect the p. Solve (a) This solution contains barium ions and acetate ions. The cation is an ion of a heavy alkaline earth metal and will therefore not affect the p. The anion, C 3 COO -, is the conjugate base of the weak acid C 3 COO and will hydrolyze to produce O - ions, thereby making the solution basic (combination 2). (b) In this solution, N 4 is the conjugate acid of a weak base 1N 3 2 and is therefore acidic. Cl - is the conjugate base of a strong acid (Cl) and therefore has no influence on the p of the solution. Because the solution contains an ion that is acidic 1N 4 2 and one that has no influence on p 1Cl - 2, the solution of N 4 Cl will be acidic (combination 3). (c) ere C 3 N 3 is the conjugate acid of a weak base 1C 3 N 2, an amine2 and is therefore acidic, and Br - is the conjugate base of a strong acid (Br) and therefore p neutral. Because the solution contains one ion that is acidic and one that has no influence on p, the solution of C 3 N 3 Br will be acidic (combination 3). (d) This solution contains the K ion, which is a cation of group 1A, - and the NO 3 ion, which is the conjugate base of the strong acid NO 3. Neither of the ions will react with water to any appreciable extent, making the solution neutral (combination 1). (e) This solution contains Al 3 - and ClO 4 ions. Cations, such as Al 3, - that have a charge of 3 or higher are acidic. The ClO 4 ion is the conjugate base of a strong acid 1ClO 4 2 and therefore does not affect p. Thus, the solution of Al1ClO will be acidic (combination 3). Practice Exercise 1 Order the following solutions from lowest to highest p: (i) 0.10 M NaClO, (ii) 0.10 M KBr, (iii) 0.10 M N 4 ClO 4. (a) i 6 ii 6 iii (b) ii 6 i 6 iii (c) iii 6 i 6 ii (d) ii 6 iii 6 i (e) iii 6 ii 6 i Practice Exercise 2 Indicate which salt in each of the following pairs forms the more acidic (or less basic) M solution: (a) NaNO 3 or Fe1NO 3 2 3, (b) KBr or KBrO, (c) C 3 N 3 Cl or BaCl 2, (d) N 4 NO 2 or N 4 NO 3.

35 section Acid Base Behavior and Chemical Structure 705 Sample Exercise predicting Whether the Solution of an Amphiprotic Anion Is Acidic or Basic Predict whether the salt Na 2 PO 4 forms an acidic solution or a basic solution when dissolved in water. Solution Analyze We are asked to predict whether a solution of Na 2 PO 4 is acidic or basic. This substance is an ionic compound composed of Na 2- and PO 4 ions. Plan We need to evaluate each ion, predicting whether it is acidic or basic. Because Na is a cation of group 1A, it has no influence on p. Thus, our analysis of whether the solution is acidic or 2-2- basic must focus on the behavior of the PO 4 ion. We need to consider that PO 4 can act as either an acid or a base: As acid PO 2-4 1aq2 1aq2 PO 3-4 1aq2 [16.46] As base PO 2-4 1aq2 2 O 2 PO - 4 1aq2 O - 1aq2 [16.47] Of these two reactions, the one with the larger equilibrium constant determines whether the solution is acidic or basic. Solve The value of K a for Equation is K a3 for 3 PO 4 : 4.2 * (Table 16.3). For Equation 16.47, we must calculate K b for the base PO 4 from the value of K a for its conjugate acid, 2-2 PO - 4, and the relationship K a * K b = K w 1Equation The relevant value of K a for - 2 PO 4 is K a2 for 3 PO 4 : 6.2 * 10-8 (from Table 16.3). We therefore have K b 1PO * K a 1 2 PO 4-2 = K w = 1.0 * K b 1PO = 1.0 * * 10-8 = 1.6 * This K b value is more than 10 5 times larger than K a for PO 4 2- ; thus, the reaction in Equation predominates over that in Equation 16.46, and the solution is basic. Practice Exercise 1 ow many of the following salts are expected to produce acidic solutions (see Table 16.3 for data): NaSO 4, NaC 2 O 4, Na 2 PO 4, and NaCO 3? (a) 0 (b) 1 (c) 2 (d) 3 (e) 4 Practice Exercise 2 Predict whether the dipotassium salt of citric acid 1K 2 C 6 5 O 7 2 forms an acidic or basic solution in water (see Table 16.3 for data) Acid Base Behavior and Chemical Structure When a substance is dissolved in water, it may behave as an acid, behave as a base, or exhibit no acid base properties. ow does the chemical structure of a substance determine which of these behaviors is exhibited by the substance? For example, why do some substances that contain O groups behave as bases, releasing O - ions into solution, whereas others behave as acids, ionizing to release ions? In this section we discuss briefly the effects of chemical structure on acid base behavior. Factors That Affect Acid Strength A molecule containing will act as a proton donor (an acid) only if the A bond is polarized such that the atom has a partial positive charge. (Section 8.4) Recall that we indicate such polarization in this way: A In ionic hydrides, such as Na, the bond is polarized in the opposite way: the atom possesses a negative charge and behaves as a proton acceptor (a base). Nonpolar A bonds, such as the C bond in C 4, produce neither acidic nor basic aqueous solutions.

36 706 chapter 16 Acid Base Equilibria A second factor that helps determine whether a molecule containing an A bond will donate a proton is the strength of the bond. (Section 8.8) Very strong bonds are less easily broken than weaker ones. This factor is important, for example, in the hydrogen halides. The F bond is the most polar A bond. You therefore might expect F to be a very strong acid if bond polarity were all that mattered. owever, the A bond strength increases as you move up the group: 299 kj>mol in I, 366 kj>mol in Br, 431 kj>mol in Cl, and 567 kj>mol in F. Because F has the highest bond strength among the hydrogen halides, it is a weak acid, whereas all the other hydrogen halides are strong acids in water. A third factor that affects the ease with which a hydrogen atom ionizes from A is the stability of the conjugate base, A -. In general, the greater the stability of the conjugate base, the stronger the acid. The strength of an acid is often a combination of all three factors. Binary Acids For a series of binary acids A in which A represents members of the same group in the periodic table, the strength of the A bond is generally the most important factor determining acid strength. The strength of an A bond tends to decrease as the element A increases in size. As a result, the bond strength decreases and acidity increases down a group. Thus, Cl is a stronger acid than F, and 2 S is a stronger acid than 2 O. Bond polarity is the major factor determining acidity for binary acids A when A represents members of the same period. Thus, acidity increases as the electronegativity of the element A increases, as it generally does moving from left to right across a period. (Section 8.4) For example, the difference in acidity of the period 2 elements is C 4 6 N 3 V 2 O 6 F. Because the C bond is essentially nonpolar, C 4 shows no tendency to form and C - 3 ions. Although the N bond is polar, N 3 has a nonbonding pair of electrons on the nitrogen atom that dominates its chemistry, so N 3 acts as a base rather than an acid. The periodic trends in the acid strengths of binary compounds of hydrogen and the nonmetals of periods 2 and 3 are summarized in Figure Go Figure Are the acid properties of I what you would expect from this figure? 4A 5A 6A 7A C 4 Neither acid nor base Si 4 Neither acid nor base N 3 Weak base K b = P 3 Very weak base K b = O 2 S Weak acid K a = Se Weak acid K a = F Weak acid K a = Cl Strong acid Br Strong acid Increasing acid strength Increasing acid strength Figure Trends in acid strength for the binary hydrides of periods 2 4.

37 section Acid Base Behavior and Chemical Structure 707 Oxyacids Many common acids, such as sulfuric acid, contain one or more O bonds: O O S O O Acids in which O groups and possibly additional oxygen atoms are bound to a central atom are called oxyacids. At first it may seem confusing that the O group, which we know behaves as a base, is also present in some acids. Let s take a closer look at what factors determine whether a given O group behaves as a base or as an acid. Consider an O group bound to some atom Y, which might in turn have other groups attached to it: Y O At one extreme, Y might be a metal, such as Na or Mg. Because of the low electronegativity of metals, the pair of electrons shared between Y and O is completely transferred to oxygen, and an ionic compound containing O - is formed. Such compounds are therefore sources of O - ions and behave as bases, as in NaO and Mg1O2 2. When Y is a nonmetal, the bond to O is covalent and the substance does not readily lose O -. Instead, these compounds are either acidic or neutral. Generally, as the electronegativity of Y increases, so does the acidity of the substance. This happens for two reasons: First, as electron density is drawn toward Y, the O bond becomes weaker and more polar, thereby favoring loss of. Second, because the conjugate base of any acid YO is usually an anion, its stability generally increases as the electronegativity of Y increases. This trend is illustrated by the K a values of the hypohalous acids (YO acids where Y is a halide ion), which decrease as the electronegativity of the halogen atom decreases ( Figure 16.19). Go Figure At equilibrium, which of the two species with a halogen atom (green) is present in greater concentration? 1 As the electronegativity of Y increases, electron density shifts toward Y 2 The O bond becomes more polar 3 Protons are more readily transferred to 2 O, leading to increased acid strength Substance ypochlorous acid ypobromous acid ypoiodous acid Water Y O Cl O Br O I O O Electronegativity of Y Dissociation constant K a = K a = K a = K w = Figure Acidity of the hypohalous oxyacids (YO) as a function of electronegativity of Y.

38 708 chapter 16 Acid Base Equilibria Many oxyacids contain additional oxygen atoms bonded to the central atom Y. These atoms pull electron density from the O bond, further increasing its polarity. Increasing the number of oxygen atoms also helps stabilize the conjugate base by increasing its ability to spread out its negative charge. Thus, the strength of an acid increases as additional electronegative atoms bond to the central atom Y. For example, the strength of the chlorine oxyacids 1Y = Cl2 steadily increases as O atoms are added: ypochlorous Chlorous Chloric Perchloric O O O Cl O Cl O O Cl O O Cl O O K a = K a = Strong acid Strong acid Increasing acid strength Because the oxidation number of Y increases as the number of attached O atoms increases, this correlation can be stated in an equivalent way: In a series of oxyacids, the acidity increases as the oxidation number of the central atom increases. Give It Some Thought Which acid has the larger acid-dissociation constant, IO 2 or BrO 3? Sample Exercise Predicting Relative Acidities from Composition and Structure Arrange the compounds in each series in order of increasing acid strength: (a) As 3, Br, K, 2 Se; (b) 2 SO 4, 2 SeO 3, 2 SeO 4. Solution Analyze We are asked to arrange two sets of compounds in order from weakest acid to strongest acid. In (a), the substances are binary compounds containing, and in (b) the substances are oxyacids. Plan For the binary compounds, we will consider the electronegativities of As, Br, K, and Se relative to the electronegativity of. The higher the electronegativity of these atoms, the higher the partial positive charge on and so the more acidic the compound. For the oxyacids, we will consider both the electronegativities of the central atom and the number of oxygen atoms bonded to the central atom. Solve (a) Because K is on the left side of the periodic table, it has a very low electronegativity (0.8, from Figure 8.7, p. 310). As a result, the hydrogen in K carries a negative charge. Thus, K should be the least acidic (most basic) compound in the series. Arsenic and hydrogen have similar electronegativities, 2.0 and 2.1, respectively. This means that the As bond is nonpolar, and so As 3 has little tendency to donate a proton in aqueous solution. The electronegativity of Se is 2.4, and that of Br is 2.8. Consequently, the Br bond is more polar than the Se bond, giving Br the greater tendency to donate a proton. (This expectation is confirmed by Figure 16.18, where we see that 2 Se is a weak acid and Br a strong acid.) Thus, the order of increasing acidity is K 6 As Se 6 Br. (b) The acids 2 SO 4 and 2 SeO 4 have the same number of O atoms and the same number of O groups. In such cases, the acid strength increases with increasing electronegativity of the central atom. Because S is slightly more electronegative than Se (2.5 vs 2.4), we predict that 2 SO 4 is more acidic than 2 SeO 4. For acids with the same central atom, the acidity increases as the number of oxygen atoms bonded to the central atom increases. Thus, 2 SeO 4 should be a stronger acid than 2 SeO 3. We predict the order of increasing acidity to be 2 SeO SeO SO 4. Practice Exercise 1 Arrange the following substances in order from weakest to strongest acid: ClO 3, OI, BrO 2, ClO 2, IO 2 (a) IO 2 6 OI 6 ClO 3 6 BrO 2 6 ClO 2 (b) OI 6 IO 2 6 BrO 2 6 ClO 2 6 ClO 3 (c) BrO 2 6 IO 2 6 ClO 2 6 OI 6 ClO 3 (d) ClO 3 6 ClO 2 6 BrO 2 6 IO 2 6 OI (e) OI 6 ClO 2 6 BrO 2 6 IO 2 6 ClO 3 Practice Exercise In each pair, choose the compound that gives the more acidic (or less basic) solution: (a) Br, F; (b) P 3, 2 S; (c) NO 2, NO 3 ; (d) 2 SO 3, 2 SeO 3.

39 Carboxylic Acids Another large group of acids is illustrated by acetic acid, a weak acid 1K a = 1.8 * : O C C O The portion of the structure shown in red is called the carboxyl group, which is often written COO. Thus, the chemical formula of acetic acid is written as C 3 COO, where only the hydrogen atom in the carboxyl group can be ionized. Acids that contain a carboxyl group are called carboxylic acids, and they form the largest category of organic acids. Formic acid and benzoic acid are further examples of this large and important category of acids: O O C O C O Formic acid Benzoic acid Two factors contribute to the acidic behavior of carboxylic acids. First, the additional oxygen atom attached to the carbon of the carboxyl group draws electron density from the O bond, increasing its polarity and helping to stabilize the conjugate base. Second, the conjugate base of a carboxylic acid (a carboxylate anion) can exhibit resonance (Section 8.6), which contributes to the stability of the anion by spreading the negative charge over several atoms: O O section Acid Base Behavior and Chemical Structure 709 C C O resonance C C O Give It Some Thought What group of atoms is present in all carboxylic acids? Chemistry and Life The Amphiprotic Behavior of Amino Acids As we will discuss in greater detail in Chapter 24, amino acids are the building blocks of proteins. The general structure of amino acids is N R C Amine group (basic) O C O Carboxyl group (acidic) where different amino acids have different R groups attached to the central carbon atom. For example, in glycine, the simplest amino acid, R is a hydrogen atom, and in alanine R is a C 3 group: C 3 2 N C COO 2 N Glycine Alanine Amino acids contain a carboxyl group and can therefore serve as acids. They also contain an N 2 group, characteristic of amines C COO (Section 16.7), and thus they can also act as bases. Amino acids, therefore, are amphiprotic. For glycine, we might expect the acid and base reactions with water to be Acid: 2 N C 2 COO1aq2 2 O1l2 2 N C 2 COO - 1aq2 3 O 1aq2 [16.48] Base: 2 N C 2 COO1aq2 2 O1l2 3 N C 2 COO1aq2 O - 1aq2 [16.49] The p of a solution of glycine in water is about 6.0, indicating that it is a slightly stronger acid than base. The acid base chemistry of amino acids is more complicated than shown in Equations and 16.49, however. Because the COO group can act as an acid and the N 2 group can act as a base, amino acids undergo a self-contained Brønsted Lowry acid base reaction in which the proton of the carboxyl group is transferred to the basic nitrogen atom: O O N C C O N C C proton transfer Neutral molecule Zwitterion O

40 710 chapter 16 Acid Base Equilibria Although the form of the amino acid on the right in this equation is electrically neutral overall, it has a positively charged end and a negatively charged end. A molecule of this type is called a zwitterion (German for hybrid ion ). Do amino acids exhibit any properties indicating that they behave as zwitterions? If so, their behavior should be similar to that of ionic substances. (Section 8.2) Crystalline amino acids have relatively high melting points, usually above 200 C, which is characteristic of ionic solids. Amino acids are far more soluble in water than in nonpolar solvents. In addition, the dipole moments of amino acids are large, consistent with a large separation of charge in the molecule. Thus, the ability of amino acids to act simultaneously as acids and bases has important effects on their properties. Related Exercise: , Lewis Acids and Bases For a substance to be a proton acceptor (a Brønsted Lowry base), it must have an unshared pair of electrons for binding the proton, as, for example, in N 3. Using Lewis structures, we can write the reaction between and N 3 as N N G. N. Lewis was the first to notice this aspect of acid base reactions. e proposed a more general definition of acids and bases that emphasizes the shared electron pair: A Lewis acid is an electron-pair acceptor, and a Lewis base is an electron-pair donor. Every base that we have discussed thus far whether O -, 2 O, an amine, or an anion is an electron-pair donor. Everything that is a base in the Brønsted Lowry sense (a proton acceptor) is also a base in the Lewis sense (an electron-pair donor). In the Lewis theory, however, a base can donate its electron pair to something other than. The Lewis definition therefore greatly increases the number of species that can be considered acids; in other words, is a Lewis acid but not the only one. For example, the reaction between N 3 and BF 3 occurs because BF 3 has a vacant orbital in its valence shell. (Section 8.7) It therefore acts as an electron-pair acceptor (a Lewis acid) toward N 3, which donates the electron pair: F F N B F N B F F F Lewis base Lewis acid Give It Some Thought What feature must any molecule or ion have in order to act as a Lewis acid? Our emphasis throughout this chapter has been on water as the solvent and on the proton as the source of acidic properties. In such cases we find the Brønsted Lowry definition of acids and bases to be the most useful. In fact, when we speak of a substance as being acidic or basic, we are usually thinking of aqueous solutions and using these terms in the Arrhenius or Brønsted Lowry sense. The advantage of the Lewis definitions of acid and base is that they allow us to treat a wider variety of reactions, including

41 section Lewis Acids and Bases 711 those that do not involve proton transfer, as acid base reactions. To avoid confusion, a substance such as BF 3 is rarely called an acid unless it is clear from the context that we are using the term in the sense of the Lewis definition. Instead, substances that function as electron-pair acceptors are referred to explicitly as Lewis acids. Lewis acids include molecules that, like BF 3, have an incomplete octet of electrons. In addition, many simple cations can function as Lewis acids. For example, Fe 3 interacts strongly with cyanide ions to form the ferricyanide ion: Fe 3 63:C N:4-3Fe1C N: The Fe 3 ion has vacant orbitals that accept the electron pairs donated by the cyanide ions. (We will learn more in Chapter 23 about just which orbitals are used by the Fe 3 ion.) The metal ion is highly charged, too, which contributes to the interaction with CN - ions. Some compounds containing multiple bonds can behave as Lewis acids. For example, the reaction of carbon dioxide with water to form carbonic acid 1 2 CO 3 2 can be pictured as an attack by a water molecule on CO 2, in which the water acts as an electron-pair donor and the CO 2 as an electron-pair acceptor: O O O O C O O C O O C O One electron pair of one of the carbon oxygen double bonds is moved onto the oxygen, leaving a vacant orbital on the carbon, which means the carbon can accept an electron pair donated by 2 O. The initial acid base product rearranges by transferring a proton from the water oxygen to a carbon dioxide oxygen, forming carbonic acid. The hydrated cations we encountered in Section 16.9, such as 3Fe1 2 O in Figure 16.17, form through the reaction between the cation acting as a Lewis acid and the water molecules acting as Lewis bases. When a water molecule interacts with the positively charged metal ion, electron density is drawn from the oxygen ( Figure 16.20). This flow of electron density causes the O bond to become more polarized; as a result, water molecules bound to the metal ion are more acidic than Weak electrostatic interaction means small electron density shift to cation Cation has little effect on O bond strength, solution remains neutral 1 Cation weakens O bond strength, solvating 2 O can readily donate, solution becomes acidic Strong electrostatic interaction means significant electron density shift to cation 3 Figure The acidity of a hydrated cation depends on cation charge.

42 712 chapter 16 Acid Base Equilibria those in the bulk solvent. This effect becomes more pronounced as the charge of the cation increases, which explains why 3 cations are much more acidic than cations with smaller charges. The Lewis acid base concept allows many ideas developed in this chapter to be used more broadly in chemistry, including rections in solvents other than water. If you take a course in organic chemistry, you will see a number of important rections that require the presence of a Lewis acid in order to proceed. The interaction of lone pairs on one molecule or ion with vacant orbitals on another molecule or ion is one of the most important concepts in chemistry, as you will see throughout your studies. Sample Integrative Exercise Phosphorous acid 1 3 PO 3 2 has the Lewis structure Putting Concepts Together O P O O (a) Explain why 3 PO 3 is diprotic and not triprotic. (b) A 25.0-mL sample of an 3 PO 3 solution titrated with M NaO requires 23.3 ml of NaO to neutralize both acidic protons. What is the molarity of the 3 PO 3 solution? (c) The original solution from part (b) has a p of Calculate the percent ionization and K a1 for 3 PO 3, assuming that K a1 W K a2. (d) ow does the osmotic pressure of a M solution of Cl compare qualitatively with that of a M solution of 3 PO 3? Explain. Solution We will use what we have learned about molecular structure and its impact on acidic behavior to answer part (a). We will then use stoichiometry and the relationship between p and 3 4 to answer parts (b) and (c). Finally, we will consider percent ionization in order to compare the osmotic pressure of the two solutions in part (d). (a) Acids have polar X bonds. From Figure 8.7 (p. 310) we see that the electronegativity of is 2.1 and that of P is also 2.1. Because the two elements have the same electronegativity, the P bond is nonpolar. (Section 8.4) Thus, this cannot be acidic. The other two atoms, however, are bonded to O, which has an electronegativity of 3.5. The O bonds are, therefore, polar with having a partial positive charge. These two atoms are consequently acidic. (b) The chemical equation for the neutralization reaction is 3 PO 3 1aq2 2NaO1aq2 Na 2 PO 3 1aq2 2 2 O1l2 From the definition of molarity, M = mol>l, we see that moles = M * L. (Section 4.5) Thus, the number of moles of NaO added to the solution is L mol>l2 = 2.38 * 10-3 mol NaO The balanced equation indicates that 2 mol of NaO is consumed for each mole of 3 PO 3. Thus, the number of moles of 3 PO 3 in the sample is * 10-3 mol NaO2a 1 mol 3PO 3 2 mol NaO b = 1.19 * 10-3 mol 3 PO 3 The concentration of the 3 PO 3 solution, therefore, equals * 10-3 mol2> L2 = M. (c) From the p of the solution, 1.59, we can calculate 3 4 at equilibrium: 3 4 = antilog = = M 1two significant figures2 Because K a1 W K a2, the vast majority of the ions in solution are from the first ionization step of the acid. 3 PO 3 1aq2 1aq2 2 PO 3-1aq2 Because one 2 PO 3 - ion forms for each ion formed, the equilibrium concentrations of and 2 PO 3 - are equal: 3 4 = 3 2 PO 3-4 = M. The equilibrium concentration of 3 PO 3 equals the initial concentration minus the amount that ionizes to form

43 chapter Summary and Key Terms 713 and 2 PO 3 - : 3 3 PO 3 4 = M M = M (two significant figures). These results can be tabulated as follows: 3 PO 3 1aq2 1aq2 2 PO 3-1aq2 Initial concentration (M) Change in concentration (M) Equilibrium concentration (M) The percent ionization is percent ionization = 3 4 equilibrium 3 3 PO 3 4 initial The first acid-dissociation constant is K a1 = PO PO 3 4 * 100% = M * 100% = 55% M = = (d) Osmotic pressure is a colligative property and depends on the total concentration of particles in solution. (Section 13.5) Because Cl is a strong acid, a M solution will contain M 1aq2 and M Cl - 1aq2, or a total of mol>l of particles. Because 3 PO 3 is a weak acid, it ionizes to a lesser extent than Cl and, hence, there are fewer particles in the 3 PO 3 solution. As a result, the 3 PO 3 solution will have the lower osmotic pressure. Chapter Summary and Key Terms Introduction to Acids and Bases (Section 16.1) Acids and bases were first recognized by the properties of their aqueous solutions. For example, acids turn litmus red, whereas bases turn litmus blue. Arrhenius recognized that the properties of acidic solutions are due to 1aq2 ions and those of basic solutions are due to O - 1aq2 ions. Brønsted Lowry Acids and Bases (Section 16.2) The Brønsted Lowry concept of acids and bases is more general than the Arrhenius concept and emphasizes the transfer of a proton 1 2 from an acid to a base. The ion, which is merely a proton with no surrounding valence electrons, is strongly bound to water. For this reason, the hydronium ion, 3 O 1aq2, is often used to represent the predominant form of in water instead of the simpler 1aq2. A Brønsted Lowry acid is a substance that donates a proton to another substance; a Brønsted Lowry base is a substance that accepts a proton from another substance. Water is an amphiprotic substance, one that can function as either a Brønsted Lowry acid or base, depending on the substance with which it reacts. The conjugate base of a Brønsted Lowry acid is the species that remains when a proton is removed from the acid. The conjugate acid of a Brønsted Lowry base is the species formed by adding a proton to the base. Together, an acid and its conjugate base (or a base and its conjugate acid) are called a conjugate acid base pair. The acid base strengths of conjugate acid base pairs are related: The stronger an acid, the weaker is its conjugate base; the weaker an acid, the stronger is its conjugate base. In every acid base reaction, the position of the equilibrium favors the transfer of the proton from the stronger acid to the stronger base. Autoionization of Water (Section 16.3) Water ionizes to a slight degree, forming 1aq2 and O - 1aq2. The extent of this autoionization is expressed by the ion-product constant for water: K w = 3 43O - 4 = 1.0 * C2. This relationship holds for both pure water and aqueous solutions. The K w expression indicates that the product of 3 4 and 3O - 4 is a constant. Thus, as 3 4 increases, 3O - 4 decreases. Acidic solutions are those that contain more 1aq2 than O - 1aq2, whereas basic solutions contain more O - 1aq2 than 1aq2. When 3 4 = 3O - 4, the solution is neutral. The p Scale (Section 16.4) The concentration of 1aq2 can be expressed in terms of p: p = -log3 4. At 25 C the p of a neutral solution is 7.00, whereas the p of an acidic solution is below 7.00, and the p of a basic solution is above This p notation is also used to represent the negative logarithm of other small quantities, as in po and pk w. The p of a solution can be measured using a p meter, or it can be estimated using acid base indicators. Strong Acids and Bases (Section 16.5) Strong acids are strong electrolytes, ionizing completely in aqueous solution. The common strong acids are Cl, Br, I, NO 3, ClO 3, ClO 4, and 2 SO 4. The conjugate bases of strong acids have negligible basicity. Common strong bases are the ionic hydroxides of the alkali metals and the heavy alkaline earth metals. Weak Acids and Bases (Sections 16.6 and 16.7) Weak acids are weak electrolytes; only a small fraction of the molecules exist in solution in ionized form. The extent of ionization is expressed by the acid-dissociation constant, K a, which is the equilibrium constant for the reaction A1aq2 1aq2 A - 1aq2, which can also be written as A1aq2 2 O1l2 3 O 1aq2 A - 1aq2. The larger the value of K a, the stronger is the acid. For solutions of the same concentration, a stronger acid also has a larger percent ionization. The concentration of a weak acid and its K a value can be used to calculate the p of a solution. Polyprotic acids, such as 3 PO 4, have more than one ionizable proton. These acids have acid-dissociation constants that decrease in magnitude in the order K a1 7 K a2 7 K a3. Because nearly all the

44 714 chapter 16 Acid Base Equilibria 1aq2 in a polyprotic acid solution comes from the first dissociation step, the p can usually be estimated satisfactorily by considering only K a1. Weak bases include N 3, amines, and the anions of weak acids. The extent to which a weak base reacts with water to generate the corresponding conjugate acid and O - is measured by the base-dissociation constant, K b. K b is the equilibrium constant for the reaction B1aq2 2 O1l2 B 1aq2 O - 1aq2, where B is the base. Relationship Between K a and K b (Section 16.8) The relationship between the strength of an acid and the strength of its conjugate base is expressed quantitatively by the equation K a * K b = K w, where K a and K b are dissociation constants for conjugate acid base pairs. This equation explains the inverse relationship between the strength of an acid and the strength of its conjugate base. Acid Base Properties of Salts (Section 16.9) The acid base properties of salts can be ascribed to the behavior of their respective cations and anions. The reaction of ions with water, with a resultant change in p, is called hydrolysis. The cations of the alkali metals and the alkaline earth metals as well as the anions of strong acids, such as Cl -, Br -, I -, and NO 3 -, do not undergo hydrolysis. They are always spectator ions in acid base chemistry. A cation that is the conjugate acid of a weak base produces upon hydrolysis. An anion that is the conjugate base of a weak acid produces O - upon hydrolysis. Acid Base Behavior and Chemical Structure (Section 16.10) The tendency of a substance to show acidic or basic characteristics in water can be correlated with its chemical structure. Acid character requires the presence of a highly polar X bond. Acidity is also favored when the X bond is weak and when the X - ion is very stable. For oxyacids with the same number of O groups and the same number of O atoms, acid strength increases with increasing electronegativity of the central atom. For oxyacids with the same central atom, acid strength increases as the number of oxygen atoms attached to the central atom increases. Carboxylic acids, which are organic acids containing the COO group, are the most important class of organic acids. The presence of delocalized p bonding in the conjugate base is a major factor responsible for the acidity of these compounds. Lewis Acids and Bases (Section 16.11) The Lewis concept of acids and bases emphasizes the shared electron pair rather than the proton. A Lewis acid is an electron-pair acceptor, and a Lewis base is an electron-pair donor. The Lewis concept is more general than the Brønsted Lowry concept because it can apply to cases in which the acid is some substance other than, and to solvents other than water. Learning Outcomes After studying this chapter, you should be able to: Define and identify Arrhenius acids and bases. (Section 16.1) Describe the nature of the hydrated proton, represented as either (aq) or 3 O (aq). (Section 16.2) Define and identify Brønsted Lowry acids and bases and identify conjugate acid base pairs. (Section 16.2) Correlate the strength of an acid to the strength of its conjugate base. (Section 16.2) Explain how the equilibrium position of a proton-transfer reaction relates to the strengths of the acids and bases involved. (Section 16.3) Describe the autoionization of water and explain how [ 3 O ] and [O ] are related via K w. (Section 16.3) Calculate the p of a solution given [ 3 O ] or [O ]. (Section 16.4) Calculate the p of a strong acid or strong base given its concentration. (Section 16.5) Calculate K a or K b for a weak acid or weak base given its concentration and the p of the solution, and vice versa. (Sections 16.6 and 16.7) Calculate the p of a weak acid or weak base or its percent ionization given its concentration and K a or K b. (Sections 16.6 and 16.7) Calculate K b for a weak base given K a of its conjugate acid, and similarly calculate K a from K b. (Section 16.8) Predict whether an aqueous solution of a salt will be acidic, basic, or neutral. (Section 16.9) Predict the relative strength of a series of acids from their molecular structures. (Section 16.10) Define and identify Lewis acids and bases. (Section 16.11) Key Equations K w = 3 3 O 43O - 4 = 3 43O - 4 = 1.0 * [16.16] Ion product of water at 25 C p = -log3 4 [16.17] Definition of p po = -log[o - ] [16.18] Definition of po p po = [16.20] Relationship between p and po K a = 3 3O 43A - 4 or K 3A4 a = 3 43A - 4 3A4 [16.25] Acid-dissociation constant for a weak acid, A Percent ionization = 3 4 equilibrium 3A4 initial * 100% [16.27] Percent ionization of a weak acid

45 Exercises 715 K b = 3B 43O - 4 3B4 [16.34] Base-dissociation constant for a weak base, B K a * K b = K w [16.40] Relationship between acid- and base-dissociation constants of a conjugate acid base pair pk a = -log K a and pk b = -log K b [16.41] Definitions of pk a and pk b Exercises Visualizing Concepts 16.1 (a) Identify the Brønsted Lowry acid and base in the reaction (b) The p of solution B is definitely greater than (c) The p of solution B is greater than that of solution A. [Section 16.4] = = N = Cl (b) Identify the Lewis acid and base in the reaction. [Sections 16.2 and 16.11] 16.2 The following diagrams represent aqueous solutions of two monoprotic acids, A 1A = X or Y2. The water molecules have been omitted for clarity. (a) Which is the stronger acid, X or Y? (b) Which is the stronger base, X - or Y -? (c) If you mix equal concentrations of X and NaY, will the equilibrium X1aq2 Y - 1aq2 Y1aq2 X - 1aq2 lie mostly to the right 1K c 7 12 or to the left 1K c 6 12? [Section 16.2] Solution A Solution B 16.4 The probe of the p meter shown here is sitting in a beaker that contains a clear liquid. (a) You are told the liquid is pure water, a solution of Cl(aq), or a solution of KO(aq). Which one is it? (b) If the liquid is one of the solutions, what is its molarity? (c) Why is the temperature given on the p meter? [Sections 16.4 and 16.5] = A = = A 3 O X Y 16.3 The indicator methyl orange has been added to both of the following solutions. Based on the colors, classify each statement as true or false: (a) The p of solution A is definitely less than The following diagrams represent aqueous solutions of three acids, X, Y, and Z. The water molecules have been omitted for clarity, and the hydrated proton is represented as rather than 3 O. (a) Which of the acids is a strong acid? Explain. (b) Which acid would have the smallest acid-dissociation constant, K a? (c) Which solution would have the highest p? [Sections 16.5 and 16.6]

46 716 chapter 16 Acid Base Equilibria X Y Z Molecule A Molecule B 16.6 The graph given below shows 3 4 vs. concentration for an aqueous solution of an unknown substance. (a) Is the substance a strong acid, a weak acid, a strong base, or a weak base? (b) Based on your answer to (a), can you determine the value of the p of the solution when the concentration is 0.18 M? (c) Would the line go exactly through the origin of the plot? [Sections 16.5 and 16.6] Molecule C [ ] 16.9 Phenylephrine, an organic substance with molecular formula C 9 13 NO 2, is used as a nasal decongenstant in over-thecounter medications. The molecular structure of phenylephrine is shown below using the usual shortcut organic nomenclature. (a) Would you expect a solution of phenylephrine to be acidic, neutral, or basic? (b) One of the active ingredients in Alka-Seltzer PLUS cold medication is phenylephrine hydrochloride. ow does this ingredient differ from the structure shown below? (c) Would you expect a solution of phenylephrine hydrochloride to be acidic, neutral, or basic? [Sections 16.8 and 16.9] Concentration (M) O O N C (a) Which of these three lines represents the effect of concentration on the percent ionization of a weak acid? (b) Explain in qualitative terms why the curve you chose has the shape it does. [Section 16.6] Percent ionized 0 0 Acid concentration 16.8 Each of the three molecules shown here contains an O group, but one molecule acts as a base, one as an acid, and the third is neither acid nor base. (a) Which one acts as a base? Why does only this molecule act as a base? (b) Which one acts as an acid? (c) Why is the remaining molecule neither acidic nor basic? [Sections 16.6 and 16.7] A B C Which of the following diagrams best represents an aqueous solution of NaF? (For clarity, the water molecules are not shown.) Will this solution be acidic, neutral, or basic? [Section 16.9] Solution A Solution B Na F O F Solution C Consider the molecular models shown here, where X represents a halogen atom. (a) If X is the same atom in both molecules, which molecule will be more acidic? (b) Does the acidity of each molecule increase or decrease as the electronegativity of the atom X increases? [Section 16.10]

47 Exercises 717 X O (a) (a) The following diagram represents the reaction of PCl 4 with Cl -. Draw the Lewis structures for the reactants and products, and identify the Lewis acid and the Lewis base in the reaction. (b) The following reaction represents a hydrated cation losing a proton. ow does the equilibrium constant for the reaction change as the charge of the cation increases? [Sections 16.9 and 16.11] n X O C (b) O (n 1) Arrhenius and Brønsted Lowry Acids and Bases (Sections 16.1 and 16.2) (a) What is the difference between the Arrhenius and the Brønsted Lowry definitions of an acid? (b) N 3 1g2 and Cl(g) react to form the ionic solid N 4 Cl1s2. Which substance is the Brønsted Lowry acid in this reaction? Which is the Brønsted Lowry base? (a) What is the difference between the Arrhenius and the Brønsted Lowry definitions of a base? (b) Can a substance behave as an Arrhenius base if it does not contain an O group? Explain (a) Give the conjugate base of the following Brønsted Lowry acids: (i) IO 3, (ii) N 4. (b) Give the conjugate acid of the following Brønsted Lowry bases: (i) O 2-, (ii) 2 PO (a) Give the conjugate base of the following Brønsted Lowry acids: (i) COO, (ii) PO 2-4. (b) Give the conjugate acid of the following Brønsted Lowry bases: (i) SO 2-4, (ii) C 3 N Designate the Brønsted Lowry acid and the Brønsted Lowry base on the left side of each of the following equations, and also designate the conjugate acid and conjugate base of each on the right side: (a) N 4 1aq2 CN - 1aq2 CN1aq2 N 3 1aq2 (b) 1C N1aq2 2 O1l2 1C N 1aq2 O - 1aq2 (c) COO1aq2 PO 3-4 1aq2 COO - 1aq2 PO 2-4 1aq Designate the Brønsted Lowry acid and the Brønsted Lowry base on the left side of each equation, and also designate the conjugate acid and conjugate base of each on the right side. (a) BrO1aq2 2 O1l2 3 O 1aq2 BrO - 1aq2 (b) SO - 4 1aq2 CO - 3 1aq2 2 SO - 4 1aq2 2 CO 3 1aq2 (c) SO - 3 1aq2 3 O 1aq2 2 SO 3 1aq2 2 O1l (a) The hydrogen sulfite ion 1SO is amphiprotic. Write a balanced chemical equation showing how it acts as an acid toward water and another equation showing how it acts as a base toward water. (b) What is the conjugate acid of SO - 3? What is its conjugate base? (a) Write an equation for the reaction in which 2 C 6 7 O - 5 1aq2 acts as a base in 2 O1l2. (b) Write an equation for the reaction in which 2 C 6 7 O - 5 1aq2 acts as an acid in 2 O1l2. (c) What is the conjugate acid of 2 C 6 7 O - 5 1aq2? What is its conjugate base? Label each of the following as being a strong base, a weak base, or a species with negligible basicity. In each case write the formula of its conjugate acid, and indicate whether the conjugate acid is a strong acid, a weak acid, or a species with negligible acidity: (a) C 3 COO -, (b) CO - 3, (c) O 2-, (d) Cl -, (e) N Label each of the following as being a strong acid, a weak acid, or a species with negligible acidity. In each case write the formula of its conjugate base, and indicate whether the conjugate base is a strong base, a weak base, or a species with negligible basicity: (a) COO, (b) 2, (c) C 4, (d) F, (e) N (a) Which of the following is the stronger Brønsted Lowry acid, BrO or Br? (b) Which is the stronger Brønsted Lowry base, F - or Cl -? (a) Which of the following is the stronger Brønsted Lowry acid, ClO 3 or ClO 2? (b) Which is the stronger Brønsted Lowry base, S - or SO - 4? Predict the products of the following acid base reactions, and predict whether the equilibrium lies to the left or to the right of the equation: (a) O 2-1aq2 2 O1l2 (b) C 3 COO1aq2 S - 1aq2 (c) NO - 2 1aq2 2 O1l Predict the products of the following acid base reactions, and predict whether the equilibrium lies to the left or to the right of the equation: (a) N 4 1aq2 O - 1aq2 (b) C 3 COO - 1aq2 3 O 1aq2 (c) CO - 3 1aq2 F - 1aq2 Autoionization of Water (Section 16.3) If a neutral solution of water, with p = 7.00, is cooled to 10 C, the p rises to Which of the following three statements is correct for the cooled water: (i) O - 4, (ii) 3 4 = 3O - 4, or (iii) O - 4? (a) Write a chemical equation that illustrates the autoionization of water. (b) Write the expression for the ion-product constant for water, K w. (c) If a solution is described as basic, which of the following is true: (i) O - 4, (ii) 3 4 = 3O - 4, or (iii) O - 4?

48 718 chapter 16 Acid Base Equilibria Calculate 3 4 for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) 3O - 4 = M; (b) 3O - 4 = 8.8 * 10-9 M; (c) a solution in which 3O - 4 is 100 times greater than Calculate 3O - 4 for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) 3 4 = M; (b) 3 4 = 2.5 * M; (c) a solution in which 3 4 is 1000 times greater than 3O At the freezing point of water 10 C2, K w = 1.2 * Calculate 3 4 and 3O - 4 for a neutral solution at this temperature Deuterium oxide 1D 2 O, where D is deuterium, the hydrogen-2 isotope) has an ion-product constant, K w, of 8.9 * at 20 C. Calculate 3D 4 and 3OD - 4 for pure (neutral) D 2 O at this temperature. The p Scale (Section 16.4) By what factor does 3 4 change for a p change of (a) 2.00 units, (b) 0.50 units? Consider two solutions, solution A and solution B. 3 4 in solution A is 250 times greater than that in solution B. What is the difference in the p values of the two solutions? Complete the following table by calculating the missing entries and indicating whether the solution is acidic or basic. 3h 4 oh 1aq2 ph poh Acidic or Basic? 7.5 * 10-3 M 3.6 * M Complete the following table by calculating the missing entries. In each case indicate whether the solution is acidic or basic. ph poh 3h 4 3oh 4 Acidic or Basic? * M 8.5 * 10-2 M The average p of normal arterial blood is At normal body temperature 137 C2, K w = 2.4 * Calculate 3 4, 3O - 4, and po for blood at this temperature Carbon dioxide in the atmosphere dissolves in raindrops to produce carbonic acid 1 2 CO 3 2, causing the p of clean, unpolluted rain to range from about 5.2 to 5.6. What are the ranges of 3 4 and 3O - 4 in the raindrops? Addition of the indicator methyl orange to an unknown solution leads to a yellow color. The addition of bromthymol blue to the same solution also leads to a yellow color. (a) Is the solution acidic, neutral, or basic? (b) What is the range (in whole numbers) of possible p values for the solution? (c) Is there another indicator you could use to narrow the range of possible p values for the solution? Addition of phenolphthalein to an unknown colorless solution does not cause a color change. The addition of bromthymol blue to the same solution leads to a yellow color. (a) Is the solution acidic, neutral, or basic? (b) Which of the following can you establish about the solution: (i) A minimum p, (ii) A maximum p, or (iii) A specific range of p values? (c) What other indicator or indicators would you want to use to determine the p of the solution more precisely? Strong Acids and Bases (Section 16.5) Is each of the following statements true or false? (a) All strong acids contain one or more atoms. (b) A strong acid is a strong electrolyte. (c) A 1.0-M solution of a strong acid will have p = Determine whether each of the following is true or false: (a) All strong bases are salts of the hydroxide ion. (b) The addition of a strong base to water produces a solution of p (c) Because Mg1O2 2 is not very soluble, it cannot be a strong base Calculate the p of each of the following strong acid solutions: (a) 8.5 * 10-3 M Br, (b) 1.52 g of NO 3 in 575 ml of solution, (c) 5.00 ml of M ClO 4 diluted to 50.0 ml, (d) a solution formed by mixing 10.0 ml of M Br with 20.0 ml of M Cl Calculate the p of each of the following strong acid solutions: (a) M NO 3, (b) g of ClO 3 in 2.00 L of solution, (c) ml of 1.00 M Cl diluted to L, (d) a mixture formed by adding 50.0 ml of M Cl to 125 ml of M I Calculate 3O - 4 and p for (a) 1.5 * 10-3 M Sr1O2 2, (b) g of LiO in ml of solution, (c) 1.00 ml of M NaO diluted to 2.00 L, (d) a solution formed by adding 5.00 ml of M KO to 15.0 ml of 9.5 * 10-2 M Ca1O Calculate 3O - 4 and p for each of the following strong base solutions: (a) M KO, (b) g of KO in ml of solution, (c) 10.0 ml of M Ca1O2 2 diluted to ml, (d) a solution formed by mixing 20.0 ml of M Ba1O2 2 with 40.0 ml of 8.2 * 10-3 M NaO Calculate the concentration of an aqueous solution of NaO that has a p of Calculate the concentration of an aqueous solution of Ca1O2 2 that has a p of Weak Acids (Section 16.6) Write the chemical equation and the K a expression for the ionization of each of the following acids in aqueous solution. First show the reaction with 1aq2 as a product and then with the hydronium ion: (a) BrO 2, (b) C 2 5 COO Write the chemical equation and the K a expression for the acid dissociation of each of the following acids in aqueous solution. First show the reaction with 1aq2 as a product and then with the hydronium ion: (a) C 6 5 COO, (b) CO Lactic acid 1C 3 C(O2COO) has one acidic hydrogen. A 0.10 M solution of lactic acid has a p of Calculate K a Phenylacetic acid 1C 6 5 C 2 COO2 is one of the substances that accumulates in the blood of people with phenylketonuria, an inherited disorder that can cause mental retardation or even death. A M solution of C 6 5 C 2 COO has a p of Calculate the K a value for this acid.

49 Exercises A M solution of chloroacetic acid 1ClC 2 COO2 is 11.0% ionized. Using this information, calculate 3ClC 2 COO - 4, 3 4, 3ClC 2 COO4, and K a for chloroacetic acid A M solution of bromoacetic acid 1BrC 2 COO2 is 13.2% ionized. Calculate 3 4, 3BrC 2 COO - 4, 3BrC 2 COO4 and K a for bromoacetic acid A particular sample of vinegar has a p of If acetic acid is the only acid that vinegar contains 1K a = 1.8 * , calculate the concentration of acetic acid in the vinegar If a solution of F 1K a = 6.8 * has a p of 3.65, calculate the concentration of hydrofluoric acid The acid-dissociation constant for benzoic acid 1C 6 5 COO2 is 6.3 * Calculate the equilibrium concentrations of 3 O, C 6 5 COO -, and C 6 5 COO in the solution if the initial concentration of C 6 5 COO is M The acid-dissociation constant for chlorous acid 1ClO 2 2 is 1.1 * Calculate the concentrations of 3 O, ClO 2 -, and ClO 2 at equilibrium if the initial concentration of ClO 2 is M Calculate the p of each of the following solutions (K a and K b values are given in Appendix D): (a) M propionic acid 1C 2 5 COO2, (b) M hydrogen chromate ion 1CrO 4-2, (c) M pyridine 1C 5 5 N Determine the p of each of the following solutions (K a and K b values are given in Appendix D): (a) M hypochlorous acid, (b) M hydrazine, (c) M hydroxylamine Saccharin, a sugar substitute, is a weak acid with pk a = 2.32 at 25 C. It ionizes in aqueous solution as follows: NC 7 4 SO 3 1aq2 1aq2 NC 7 4 SO 3-1aq2 What is the p of a 0.10 M solution of this substance? The active ingredient in aspirin is acetylsalicylic acid 1C 9 7 O 4 2, a monoprotic acid with K a = 3.3 * 10-4 at 25 C. What is the p of a solution obtained by dissolving two extra-strength aspirin tablets, containing 500 mg of acetylsalicylic acid each, in 250 ml of water? Calculate the percent ionization of hydrazoic acid 1N 3 2 in solutions of each of the following concentrations (K a is given in Appendix D): (a) M, (b) M, (c) M Calculate the percent ionization of propionic acid 1C 2 5 COO2 in solutions of each of the following concentrations 1K a is given in Appendix D): (a) M, (b) M, (c) M Citric acid, which is present in citrus fruits, is a triprotic acid (Table 16.3). (a) Calculate the p of a M solution of citric acid. (b) Did you have to make any approximations or assumptions in completing your calculations? (c) Is the concentration of citrate ion 1C 6 5 O equal to, less than, or greater than the ion concentration? Tartaric acid is found in many fruits, including grapes, and is partially responsible for the dry texture of certain wines. Calculate the p and the tartrate ion 1C 4 4 O concentration for a M solution of tartaric acid, for which the acid-dissociation constants are listed in Table Did you have to make any approximations or assumptions in your calculation? Weak Bases (Section 16.7) Consider the base hydroxylamine, N 2 O. (a) What is the conjugate acid of hydroxylamine? (b) When it acts as a base, which atom in hydroxylamine accepts a proton? (c) There are two atoms in hydroxylamine that have nonbonding electron pairs that could act as proton acceptors. Use Lewis structures and formal charges (Section 8.5) to rationalize why one of these two atoms is a much better proton acceptor than the other The hypochlorite ion, ClO -, acts as a weak base. (a) Is ClO - a stronger or weaker base than hydroxylamine? (b) When ClO - acts as a base, which atom, Cl or O, acts as the proton acceptor? (c) Can you use formal charges to rationalize your answer to part (b)? Write the chemical equation and the K b expression for the reaction of each of the following bases with water: (a) dimethylamine, 1C N; (b) carbonate ion, CO 2-3 ; (c) formate ion, CO Write the chemical equation and the K b expression for the reaction of each of the following bases with water: (a) propylamine, C 3 7 N 2 ; (b) monohydrogen phosphate ion, PO 2-4 ; (c) benzoate ion, C 6 5 CO Calculate the molar concentration of O - in a M solution of ethylamine 1C 2 5 N 2 ; K b = 6.4 * Calculate the p of this solution Calculate the molar concentration of O - in a M solution of hypobromite ion 1BrO - ; K b = 4.0 * What is the p of this solution? Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base: C ON1aq2 2 O1l2 C ON 1aq2 O - 1aq2 A M solution of ephedrine has a p of (a) What are the equilibrium concentrations of C ON, C ON, and O -? (b) Calculate K b for ephedrine Codeine 1C NO 3 2 is a weak organic base. A 5.0 * 10-3 M solution of codeine has a p of Calculate the value of K b for this substance. What is the pk b for this base? The K a K b Relationship; Acid Base Properties of Salts (Sections 16.8 and 16.9) Although the acid-dissociation constant for phenol 1C 6 5 O2 is listed in Appendix D, the base-dissociation constant for the phenolate ion 1C 6 5 O - 2 is not. (a) Explain why it is not necessary to list both K a for phenol and K b for the phenolate ion. (b) Calculate K b for the phenolate ion. (c) Is the phenolate ion a weaker or stronger base than ammonia? Use the acid-dissociation constants in Table 16.3 to arrange these oxyanions from strongest base to weakest: SO 2-4, CO 2-3, SO 2-3, and PO (a) Given that K a for acetic acid is 1.8 * 10-5 and that for hypochlorous acid is 3.0 * 10-8, which is the stronger acid? (b) Which is the stronger base, the acetate ion or the hypochlorite ion? (c) Calculate K b values for C 3 COO - and ClO (a) Given that K b for ammonia is 1.8 * 10-5 and that for hydroxylamine is 1.1 * 10-8, which is the stronger base?

50 720 chapter 16 Acid Base Equilibria (b) Which is the stronger acid, the ammonium ion or the hydroxylammonium ion? (c) Calculate K a values for N 4 and 3 NO Using data from Appendix D, calculate 3O - 4 and p for each of the following solutions: (a) 0.10 M NaBrO, (b) M NaS, (c) a mixture that is 0.10 M in NaNO 2 and 0.20 M in Ca1NO Using data from Appendix D, calculate 3O - 4 and p for each of the following solutions: (a) M NaF, (b) M Na 2 S, (c) a mixture that is M in NaC 3 COO and M in Ba1C 3 COO A solution of sodium acetate 1NaC 3 COO2 has a p of What is the molarity of the solution? Pyridinium bromide 1C 5 5 NBr2 is a strong electrolyte that dissociates completely into C 5 5 N and Br -. A solution of pyridinium bromide has a p of What is the molarity of the solution? Predict whether aqueous solutions of the following compounds are acidic, basic, or neutral: (a) N 4 Br, (b) FeCl 3, (c) Na 2 CO 3, (d) KClO 4, (e) NaC 2 O Predict whether aqueous solutions of the following substances are acidic, basic, or neutral: (a) AlCl 3, (b) NaBr, (c) NaClO, (d) 3C 3 N 3 4NO 3, (e) Na 2 SO An unknown salt is either NaF, NaCl, or NaOCl. When mol of the salt is dissolved in water to form L of solution, the p of the solution is What is the identity of the salt? An unknown salt is either KBr, N 4 Cl, KCN, or K 2 CO 3. If a M solution of the salt is neutral, what is the identity of the salt? Acid Base Character and Chemical Structure (Section 16.10) Explain the following observations: (a) NO 3 is a stronger acid than NO 2 ; (b) 2 S is a stronger acid than 2 O; (c) 2 SO 4 is a stronger acid than SO - 4 ; (d) 2 SO 4 is a stronger acid than 2 SeO 4 ; (e) CCl 3 COO is a stronger acid than CCl 3 COO Explain the following observations: (a) Cl is a stronger acid than 2 S; (b) 3 PO 4 is a stronger acid than 3 AsO 4 ; (c) BrO 3 is a stronger acid than BrO 2 ; (d) 2 C 2 O 4 is a stronger acid than C 2 O - 4 ; (e) benzoic acid 1C 6 5 COO2 is a stronger acid than phenol 1C 6 5 O Based on their compositions and structures and on conjugate acid base relationships, select the stronger base in each of the following pairs: (a) BrO - or ClO -, (b) BrO - or BrO - 2, 2- (c) PO 4 or 2 PO Based on their compositions and structures and on conjugate acid base relationships, select the stronger base in each of - the following pairs: (a) NO 3 or NO - 3-2, (b) PO 4 or AsO 3-4, - (c) CO 3 or CO Indicate whether each of the following statements is true or false. For each statement that is false, correct the statement to make it true. (a) In general, the acidity of binary acids increases from left to right in a given row of the periodic table. (b) In a series of acids that have the same central atom, acid strength increases with the number of hydrogen atoms bonded to the central atom. (c) ydrotelluric acid 1 2 Te2 is a stronger acid than 2 S because Te is more electronegative than S Indicate whether each of the following statements is true or false. For each statement that is false, correct the statement to make it true. (a) Acid strength in a series of A molecules increases with increasing size of A. (b) For acids of the same general structure but differing electronegativities of the central atoms, acid strength decreases with increasing electronegativity of the central atom. (c) The strongest acid known is F because fluorine is the most electronegative element. Lewis Acids and Bases (Section 16.11) If a substance is an Arrhenius base, is it necessarily a Brønsted Lowry base? Is it necessarily a Lewis base? If a substance is a Lewis acid, is it necessarily a Brønsted Lowry acid? Is it necessarily an Arrhenius acid? Identify the Lewis acid and Lewis base among the reactants in each of the following reactions: (a) Fe1ClO s2 6 2 O1l2 Fe1 2 O aq2 3 ClO - 4 1aq2 (b) CN - 1aq2 2 O1l2 CN1aq2 O - 1aq2 (c) 1C N1g2 BF 3 1g2 1C NBF 3 1s2 (d) IO1lq2 N - 2 1lq2 N 3 1lq2 IO - 1lq2 (lq denotes liquid ammonia as solvent) Identify the Lewis acid and Lewis base in each of the following reactions: (a) NO 2 1aq2 O - 1aq2 NO - 2 1aq2 2 O1l2 (b) FeBr 3 1s2 Br - 1aq2 FeBr - 4 1aq2 (c) Zn 2 1aq2 4 N 3 1aq2 Zn1N aq2 (d) SO 2 1g2 2 O1l2 2 SO 3 1aq Predict which member of each pair produces the more acidic aqueous solution: (a) K or Cu 2, (b) Fe 2 or Fe 3, (c) Al 3 or Ga Which member of each pair produces the more acidic aqueous solution: (a) ZnBr 2 or CdCl 2, (b) CuCl or Cu1NO 3 2 2, (c) Ca1NO or NiBr 2? Additional Exercises Indicate whether each of the following statements is correct or incorrect. (a) Every Brønsted Lowry acid is also a Lewis acid. (b) Every Lewis acid is also a Brønsted Lowry acid. (c) Conjugate acids of weak bases produce more acidic solutions than conjugate acids of strong bases. (d) K ion is acidic in water because it causes hydrating water molecules to become more acidic. (e) The percent ionization of a weak acid in water increases as the concentration of acid decreases A solution is made by adding g Ca1O2 2 1s2, 50.0 ml of 1.40 M NO 3, and enough water to make a final volume

51 Additional Exercises 721 of 75.0 ml Assuming that all of the solid dissolves, what is the p of the final solution? The odor of fish is due primarily to amines, especially methylamine 1C 3 N 2 2. Fish is often served with a wedge of lemon, which contains citric acid. The amine and the acid react forming a product with no odor, thereby making the less-than-fresh fish more appetizing. Using data from Appendix D, calculate the equilibrium constant for the reaction of citric acid with methylamine, if only the first proton of the citric acid 1K a1 2 is important in the neutralization reaction. [16.102] For solutions of a weak acid, a graph of p versus the logarithm of the initial acid concentration should be a straight line. What is the magnitude of the slope of that line? emoglobin plays a part in a series of equilibria involving protonation-deprotonation and oxygenation-deoxygenation. The overall reaction is approximately as follows: Oxalic acid 1 2 C 2 O 4 2 is a diprotic acid. By using data in Appendix D as needed, determine whether each of the following statements is true: (a) 2 C 2 O 4 can serve as both a Brønsted 2- Lowry acid and a Brønsted Lowry base. (b) C 2 O 4 is the conjugate base of C 2 O - 4. (c) An aqueous solution of the strong electrolyte KC 2 O 4 will have p Succinic acid 1 2 C 4 6 O 4 2, which we will denote 2 Suc, is a biologically relevant diprotic acid with the structure shown below. It is closely related to tartaric acid and malic acid (Figure 16.1). At 25 C, the acid-dissociation constants for succinic acid are K a1 = 6.9 * 10-5 and K a2 = 2.5 * (a) Determine the p of a 0.32 M solution of 2 Suc at 25 C, assuming that only the first dissociation is relevant. (b) Determine the molar concentration of Suc 2 - in the solution in part (a). (c) Is the assumption you made in part (a) justified by the result from part (b)? (d) Will a solution of the salt NaSuc be acidic, neutral, or basic? b 1aq2 O 2 1aq2 bo 2 1aq2 1aq2 O where b stands for hemoglobin and bo 2 for oxyhemoglobin. (a) The concentration of O 2 is higher in the lungs and lower in the tissues. What effect does high 3O 2 4 have on the position of this equilibrium? (b) The normal p of blood is 7.4. Is the blood acidic, basic, or neutral? (c) If the blood p is lowered by the presence of large amounts of acidic metabolism products, a condition known as acidosis results. What effect does lowering blood p have on the ability of hemoglobin to transport O 2? [16.104] Calculate the p of a solution made by adding 2.50 g of lithium oxide 1Li 2 O2 to enough water to make L of solution Benzoic acid 1C 6 5 COO2 and aniline 1C 6 5 N 2 2 are both derivatives of benzene. Benzoic acid is an acid with K a = 6.3 * 10-5 and aniline is a base with K a = 4.3 * O C O Benzoic acid N Aniline (a) What are the conjugate base of benzoic acid and the conjugate acid of aniline? (b) Anilinium chloride 1C 6 5 N 3 Cl2 is a strong electrolyte that dissociates into anilinium ions 1C 6 5 N 3 2 and chloride ions. Which will be more acidic, a 0.10 M solution of benzoic acid or a 0.10 M solution of anilinium chloride? (c) What is the value of the equilibrium constant for the following equilibrium? C 6 5 COO1aq2 C 6 5 N 2 1aq2 C 6 5 COO - 1aq2 C 6 5 N 3 1aq What is the p of a solution that is 2.5 * 10-9 M in NaO? Does your answer make sense? What assumption do we normally make that is not valid in this case? O C O C C C O Butyric acid is responsible for the foul smell of rancid butter. The pk a of butyric acid is (a) Calculate the pk b for the butyrate ion. (b) Calculate the p of a M solution of butyric acid. (c) Calculate the p of a M solution of sodium butyrate Arrange the following 0.10 M solutions in order of increasing acidity (decreasing p): (i) N 4 NO 3, (ii) NaNO 3, (iii) C 3 COON 4, (iv) NaF, (v) C 3 COONa A 0.25 M solution of a salt NaA has p = What is the value of K a for the parent acid A? [16.112] The following observations are made about a diprotic acid 2 A: (i) A 0.10 M solution of 2 A has p = (ii) A 0.10 M solution of the salt NaA is acidic. Which of the following could be the value of pk a2 for 2 A: (i) 3.22, (ii) 5.30, (iii) 7.47, or (iv) 9.82? Many moderately large organic molecules containing basic nitrogen atoms are not very soluble in water as neutral molecules, but they are frequently much more soluble as their acid salts. Assuming that p in the stomach is 2.5, indicate whether each of the following compounds would be present in the stomach as the neutral base or in the protonated form: nicotine, K b = 7 * 10-7 ; caffeine, K b = 4 * ; strychnine, K b = 1 * 10-6 ; quinine, K b = 1.1 * The amino acid glycine 1 2 N C 2 COO2 can participate in the following equilibria in water: 2 N C 2 COO 2 O 2 N C 2 COO - 3 O K a = 4.3 * N C 2 COO 2 O 3 N C 2 COO O - K b = 6.0 * 10-5 (a) Use the values of K a and K b to estimate the equilibrium constant for the intramolecular proton transfer to form a zwitterion: 2 N C 2 COO 3 N C 2 COO -

52 722 chapter 16 Acid Base Equilibria (b) What is the p of a M aqueous solution of glycine? (c) What would be the predominant form of glycine in a solution with p 13? With p 1? The structural formula for acetic acid is shown in Table Replacing hydrogen atoms on the carbon with chlorine atoms causes an increase in acidity, as follows: Using Lewis structures as the basis of your discussion, explain the observed trend in acidities in the series. Calculate the p of a M solution of each acid. Acid Formula K a 125 C2 Acetic C 3 COO 1.8 * 10-5 Chloroacetic C 2 ClCOO 1.4 * 10-3 Dichloroacetic CCl 2 COO 3.3 * 10-2 Trichloroacetic CCl 3 COO 2 * 10-1 Integrative Exercises Calculate the number of 1aq2 ions in 1.0 ml of pure water at 25 C ow many milliliters of concentrated hydrochloric acid solution (36.0% Cl by mass, density = 1.18 g>ml) are required to produce 10.0 L of a solution that has a p of 2.05? The volume of an adult s stomach ranges from about 50 ml when empty to 1 L when full. If the stomach volume is 400 ml and its contents have a p of 2, how many moles of does the stomach contain? Assuming that all the comes from Cl, how many grams of sodium hydrogen carbonate will totally neutralize the stomach acid? Atmospheric CO 2 levels have risen by nearly 20% over the past 40 years from 315 ppm to 380 ppm. (a) Given that the average p of clean, unpolluted rain today is 5.4, determine the p of unpolluted rain 40 years ago. Assume that carbonic acid 1 2 CO 3 2 formed by the reaction of CO 2 and water is the only factor influencing p. CO 2 1g2 2 O1l2 2 CO 3 1aq2 (b) What volume of CO 2 at 25 C and 1.0 atm is dissolved in a 20.0-L bucket of today s rainwater? At 50 C, the ion-product constant for 2 O has the value K w = 5.48 * (a) What is the p of pure water at 50 C? (b) Based on the change in K w with temperature, predict whether is positive, negative, or zero for the autoionization reaction of water: 2 2 O1l2 3 O 1aq2 O - 1aq In many reactions the addition of AlCl 3 produces the same effect as the addition of. (a) Draw a Lewis structure for AlCl 3 in which no atoms carry formal charges, and determine its structure using the VSEPR method. (b) What characteristic is notable about the structure in part (a) that helps us understand the acidic character of AlCl 3? (c) Predict the result of the reaction between AlCl 3 and N 3 in a solvent that does not participate as a reactant. (d) Which acid base theory is most suitable for discussing the similarities between AlCl 3 and? What is the boiling point of a 0.10 M solution of NaSO 4 if the solution has a density of g>ml? Use average bond enthalpies from Table 8.4 to estimate the enthalpies of the following gas-phase reactions: Reaction 1: F1g2 2 O1g2 F - 1g2 3 O 1g2 Reaction 2: Cl1g2 2 O1g2 Cl - 1g2 3 O 1g2 Are both reactions exothermic? ow do these values relate to the different strengths of hydrofluoric and hydrochloric acid? Cocaine is a weak organic base whose molecular formula is C NO 4. An aqueous solution of cocaine was found to have a p of 8.53 and an osmotic pressure of 52.7 torr at 15 C. Calculate K b for cocaine. [16.125] The iodate ion is reduced by sulfite according to the following reaction: IO 3-1aq2 3SO 3 2-1aq2 I - 1aq2 3 SO 4 2-1aq2 The rate of this reaction is found to be first order in IO - 3, first order in SO 2-3, and first order in. (a) Write the rate law for the reaction. (b) By what factor will the rate of the reaction change if the p is lowered from 5.00 to 3.50? Does the reaction proceed more quickly or more slowly at the lower p? (c) By using the concepts discussed in Section 14.6, explain how the reaction can be p-dependent even though does not appear in the overall reaction (a) Using dissociation constants from Appendix D, determine the value for the equilibrium constant for each of the following reactions. (i) CO - 3 1aq2 O - 1aq2 CO 2-3 1aq2 2 O1l2 (ii) N 4 1aq2 CO 2-3 1aq2 N 3 1aq2 CO - 3 1aq2 (b) We usually use single arrows for reactions when the forward reaction is appreciable (K much greater than 1) or when products escape from the system, so that equilibrium is never established. If we follow this convention, which of these equilibria might be written with a single arrow?

53 Design an Experiment 723 Design an Experiment Your professor gives you a bottle that contains a clear liquid. You are told that the liquid is a pure substance that is volatile, soluble in water, and might be an acid or a base. Design experiments to elucidate the following about this unknown sample. (a) Determine whether the substance in the sample is an acid or a base. (b) Suppose the substance is an acid. ow would you determine whether it is a strong acid or a weak acid? (c) If the substance is a weak acid, how would you determine the value of K a for the substance? (d) Suppose the substance were a weak acid and that you are also given a solution of NaO(aq) of known molarity. What procedure would you use to isolate a pure sample of the sodium salt of the substance? (e) Now suppose that the substance were a base rather than an acid. ow would you adjust the procedures in parts (b) and (c) to determine if the substance were a strong or weak base, and, if weak, the value of K b?

54 17 Additional Aspects of Aqueous Equilibria Water, the most common and most important solvent on Earth, occupies its position of importance because of its abundance and its exceptional ability to dissolve a wide variety of substances. Coral reefs are a striking example of aqueous chemistry at work in nature. Coral reefs are built by tiny animals called stony corals, which secrete a hard calcium carbonate exoskeleton. Over time, the stony corals build up large networks of calcium carbonate upon which a reef is built. The size of such structures can be immense, as illustrated by the Great Barrier Reef. Stony corals make their exoskeletons from dissolved Ca 2 and CO ions. This process is aided by the fact that the CO concentration is supersaturated in most parts of the ocean. owever, well documented increases in the amount of CO 2 in the atmosphere threaten to upset the aqueous chemistry that stony corals depend on. As atmospheric CO 2 levels increase, the amount of CO 2 dissolved in the ocean also increases. This lowers the p of the ocean and leads to a decrease in the CO concentration. As a result it becomes more difficult for stony corals and other important ocean creatures to maintain their exoskeletons. We will take a closer look at the consequences of ocean acidification later in the chapter. To understand the chemistry that underlies coral reef formation and other processes in the ocean and in aqueous systems such as living cells, we must understand the concepts of aqueous equilibria. In this chapter, we take a step toward understanding such complex solutions by looking first at further applications of acid base equilibria. The idea is to consider not only solutions in which there is a single solute but A microscopic view of coral and sand. This image shows coral and sand particles, magnified by 100x. What s Ahead 17.1 The Common-Ion Effect We begin by considering a specific example of Le Châtelier s principle known as the common-ion effect Buffers We consider the composition of buffered solutions and learn how they resist p change when small amounts of a strong acid or strong base are added to them Acid Base Titrations We examine acid base titrations and explore how to determine p at any point in an acid base titration Solubility Equilibria We learn how to use solubilityproduct constants to determine to what extent a sparingly soluble salt dissolves in water Factors That Affect Solubility We investigate some of the factors that affect solubility, including the common-ion effect and the effect of acids.

55 17.6 Precipitation and Separation of Ions We learn how differences in solubility can be used to separate ions through selective precipitation Qualitative Analysis for Metallic Elements We explain how the principles of solubility and complexation equilibria can be used to identify ions in solution.

56 726 chapter 17 Additional Aspects of Aqueous Equilibria also those containing a mixture of solutes. We then broaden our discussion to include two additional types of aqueous equilibria: those involving slightly soluble salts and those involving the formation of metal complexes in solution. For the most part, the discussions and calculations in this chapter are extensions of those in Chapters 15 and The Common-Ion Effect In Chapter 16, we examined the equilibrium concentrations of ions in solutions containing a weak acid or a weak base. We now consider solutions that contain a weak acid, such as acetic acid 1C 3 COO2, and a soluble salt of that acid, such as sodium acetate 1C 3 COONa2. Notice that these solutions contain two substances that share a common ion, C 3 COO -. It is instructive to view these solutions from the perspective of Le Châtelier s principle. (Section 15.7) Sodium acetate is a soluble ionic compound and therefore a strong electrolyte. (Section 4.1) Consequently, it dissociates completely in aqueous solution to form Na and C 3 COO - ions: C 3 COONa1aq2 Na 1aq2 C 3 COO - 1aq2 In contrast, C 3 COO is a weak electrolyte that ionizes only partially, represented by the dynamic equilibrium C 3 COO1aq2 1aq2 C 3 COO - 1aq2 [17.1] The equilibrium constant for Equation 17.1 is K a = 1.8 * 10-5 at 25 C (Table 16.2). If we add sodium acetate to a solution of acetic acid in water, the C 3 COO - from C 3 COONa causes the equilibrium concentrations of the substances in Equation 17.1 to shift to the left, thereby decreasing the equilibrium concentration of 1aq2: C 3 COO1aq2 1aq2 C 3 COO - 1aq2 Addition of C 3 COO shifts equilibrium concentrations, lowering [ ] In other words, the presence of the added acetate ion causes the acetic acid to ionize less than it normally would. The equilibrium constant itself does not change; it is the relative concentrations of products and reactants in the equilibrium expression that change. Whenever a weak electrolyte and a strong electrolyte containing a common ion are together in solution, the weak electrolyte ionizes less than it would if it were alone in solution. We call this observation the common-ion effect. Sample Exercise 17.1 Calculating the p When a Common Ion Is Involved What is the p of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution? Solution Analyze We are asked to determine the p of a solution of a weak electrolyte 1C 3 COO2 and a strong electrolyte 1C 3 COONa2 that share a common ion, C 3 COO -. Plan In any problem in which we must determine the p of a solution containing a mixture of solutes, it is helpful to proceed by a series of logical steps: (1) Consider which solutes are strong electrolytes and which are weak electrolytes, and identify the major species in solution. (2) Identify the important equilibrium reaction that is the source of and therefore determines p. (3) Tabulate the concentrations of ions involved in the equilibrium. (4) Use the equilibrium-constant expression to calculate 3 4 and then p.

57 section 17.1 The Common-Ion Effect 727 Solve First, because C 3 COO is a weak electrolyte and C 3 COONa is a strong electrolyte, the major species in the solution are C 3 COO (a weak acid), Na (which is neither acidic nor basic and is therefore a spectator in the acid base chemistry), and C 3 COO - (which is the conjugate base of C 3 COO). Second, 3 4 and, therefore, the p are controlled by the dissociation equilibrium of C 3 COO: (We have written the equilibrium using 1aq2 rather than 3 O 1aq2, but both representations of the hydrated hydrogen ion are equally valid.) Third, we tabulate the initial and equilibrium concentrations as we did in solving other equilibrium problems in Chapters 15 and 16: C 3 COO1aq2 1aq2 C 3 COO - 1aq2 C 3 COO1aq2 1aq2 C 3 COO - 1aq2 Initial (M) Change (M) -x x x Equilibrium (M) ( x) x x2 The equilibrium concentration of C 3 COO - (the common ion) is the initial concentration that is due to C 3 COONa (0.30 M) plus the change in concentration (x) that is due to the ionization of C 3 COO. Now we can use the equilibrium-constant expression: K a = 1.8 * 10-5 = 3 43C 3 COO - 4 3C 3 COO4 The dissociation constant for C 3 COO at 25 C is from Table 16.2, or Appendix D; addition of C 3 COONa does not change the value of this constant. Substituting the equilibriumconstant concentrations from our table into the equilibrium expression gives: K a = 1.8 * 10-5 = x10.30 x x Because K a is small, we assume that x is small compared to the original concentrations of C 3 COO and C 3 COO - (0.30 M each). Thus, we can ignore the very small x relative to 0.30 M, giving K a = 1.8 * 10-5 = x The resulting value of x is indeed small relative to 0.30, justifying the approximation made in simplifying the problem. x = 1.8 * 10-5 M = 3 4 Finally, we calculate the p from the equilibrium concentration of 1aq2: p = -log11.8 * = 4.74 Comment In Section 16.6, we calculated that a 0.30 M solution of C 3 COO has a p of 2.64, corresponding to 3 4 = 2.3 * 10-3 M. Thus, the addition of C 3 COONa has substantially decreased 3 4, as we expect from Le Châtelier s principle. Practice Exercise 1 For the generic equilibrium A 1aq2 1aq2 A - 1aq2, which of these statements is true? (a) The equilibrium constant for this reaction changes as the p changes. (b) If you add the soluble salt KA to a solution of A that is at equilibrium, the concentration of A would decrease. (c) If you add the soluble salt KA to a solution of A that is at equilibrium, the concentration of A - would decrease. (d) If you add the soluble salt KA to a solution of A that is at equilibrium, the p would increase. Practice Exercise 2 Calculate the p of a solution containing M nitrous acid 1NO 2, K a = 4.5 * and 0.10 M potassium nitrite 1KNO 2 2.

58 728 chapter 17 Additional Aspects of Aqueous Equilibria Sample Exercise 17.2 Calculating Ion Concentrations When a Common Ion Is Involved Calculate the fluoride ion concentration and p of a solution that is 0.20 M in F and 0.10 M in Cl. Solution Analyze We are asked to determine the concentration of F - and the p in a solution containing the weak acid F and the strong acid Cl. In this case the common ion is. Plan We can again use the four steps outlined in Sample Exercise Solve Because F is a weak acid and Cl is a strong acid, the major species in solution are F,, and Cl -. The Cl -, which is the conjugate base of a strong acid, is merely a spectator ion in any acid base chemistry. The problem asks for 3F - 4, which is formed by ionization of F. Thus, the important equilibrium is F1aq2 1aq2 F - 1aq2 The common ion in this problem is the hydrogen (or hydronium) ion. Now we can tabulate the initial and equilibrium concentrations of each species involved in this equilibrium: F1aq2 1aq2 F - 1aq2 Initial (M) Change (M) -x x x Equilibrium (M) x x2 x The equilibrium constant for the ionization of F, from Appendix D, is 6.8 * Substituting the equilibrium-constant concentrations into the equilibrium expression gives K a = 6.8 * 10-4 = 3 43F - 4 = 3F4 Comment Notice that for all practical purposes, the hydrogen ion concentration is due entirely to the Cl; the F makes a negligible contribution by comparison. Practice Exercise 1 Calculate the concentration of the lactate ion in a solution that is M in lactic acid 1C 3 C(O2COO, pk a = 3.86) and M in Cl. (a) 4.83 M, (b) M, (c) 7.3 * 10-3 M, (d) 3.65 * 10-3 M, (e) 1.73 * 10-4 M. Practice Exercise 2 Calculate the formate ion concentration and p of a solution that is M in formic acid 1COO, K a = 1.8 * and 0.10 M in NO x21x x If we assume that x is small relative to 0.10 or x M, this expression simplifies to = 6.8 * x = * = 1.4 * 10-3 M = 3F - 4 This F - concentration is substantially smaller than it would be in a 0.20 M solution of F with no added Cl. The common ion,, suppresses the ionization of F. The concentration of 1aq2 is 3 4 = x2 M 0.10 M Thus, p = 1.00 Sample Exercises 17.1 and 17.2 both involve weak acids. The ionization of a weak base is also decreased by the addition of a common ion. For example, the addition of N 4 (as from the strong electrolyte N 4 Cl) causes the equilibrium concentrations of the reagents in Equation 17.2 to shift to the left, decreasing the equilibrium concentration of O - and lowering the p: N 3 1aq2 2 O1l2 N 4 1aq2 O - 1aq2 [17.2] Addition of N 4 shifts equilibrium concentrations, lowering [O ]

59 section 17.2 Buffers 729 Give It Some Thought If solutions of N 4 Cl1aq2 and N 3 1aq2 are mixed, which ions in the resulting solution are spectator ions in any acid base chemistry occurring in the solution? What equilibrium reaction determines 3O - 4 and, therefore, the p of the solution? 17.2 Buffers Solutions that contain high concentrations (10-3 M or more) of a weak conjugate acid base pair and that resist drastic changes in p when small amounts of strong acid or strong base are added to them are called buffered solutions (or merely buffers). uman blood, for example, is a complex buffered solution that maintains the blood p at about 7.4. (Section 17.2, Blood as a Buffered Solution ) Much of the chemical behavior of seawater is determined by its p, buffered at about 8.1 to 8.3 near the surface. (Section 17.5, Ocean Acidification ) Buffers find many important applications in the laboratory and in medicine ( Figure 17.1). Many biological reactions occur at the optimal rates only when properly buffered. If you ever work in a biochemistry lab, you will very likely to have prepare specific buffers in which to run your biochemical reactions Figure 17.1 Standard buffers. For laboratory work, prepackaged buffers at specific p values can be purchased. Composition and Action of Buffers A buffer resists changes in p because it contains both an acid to neutralize added O - ions and a base to neutralize added ions. The acid and base that make up the buffer, however, must not consume each other through a neutralization reaction. (Section 4.3) These requirements are fulfilled by a weak acid base conjugate pair, such as C 3 COO>C 3 COO - or N 4 >N 3. The key is to have roughly equal concentrations of both the weak acid and its conjugate base. There are two ways to make a buffer: 1. Mix a weak acid or a weak base with a salt of that acid or base. For example, the C 3 COO>C 3 COO - buffer can be prepared by adding C 3 COONa to a solution of C 3 COO. Similarly, the N 4 >N 3 buffer can be prepared by adding N 4 Cl to a solution of N Make the conjugate acid or base from a solution of weak base or acid by the addition of strong acid or base. For example, to make the C 3 COO>C 3 COO - buffer, you could start with a solution of C 3 COO and add some NaO to the solution enough to neutralize about half of C 3 COO according to the reaction. C 3 COO O - C 3 COO - 2 O (Section 4.3) Neutralization reactions have very large equilibrium constants, and so the amount of acetate formed will only be limited by the relative amounts of the acid and strong base that are mixed. The resulting solution is the same as if you added sodium acetate to the acetic acid solution: You will have comparable quantities of both acetic acid and its conjugate base in solution. By choosing appropriate components and adjusting their relative concentrations, we can buffer a solution at virtually any p. Give It Some Thought Which of these conjugate acid base pairs will not function as a buffer: C 2 5 COO and C 2 5 COO -, CO 3 - and CO 3 2 -, or NO 3 and NO 3 -? Explain. To understand how a buffer works, let s consider one composed of a weak acid A and one of its salts MA, where M could be Na, K, or any other cation that does not react with water. The acid-dissociation equilibrium in this buffered solution involves both the acid and its conjugate base: A1aq2 1aq2 A - 1aq2 [17.3]

60 730 chapter 17 Additional Aspects of Aqueous Equilibria The corresponding acid-dissociation-constant expression is K a = 3 43A - 4 3A4 [17.4] Solving this expression for 3 4, we have 3 4 = K a 3A4 3A - 4 [17.5] We see from this expression that 3 4 and, thus, the p are determined by two factors: the value of K a for the weak-acid component of the buffer and the ratio of the concentrations of the conjugate acid base pair, 3A4>3A - 4. If O - ions are added to this buffered solution, they react with the buffer acid component to produce water and A - : O - 1aq2 A1aq2 2 O1l2 A - 1aq2 [17.6] added base This neutralization reaction causes [A] to decrease and 3A - 4 to increase. As long as the amounts of A and A - in the buffer are large relative to the amount of O - added, the ratio 3A4>3A - 4 does not change much and, thus, the change in p is small. If ions are added, they react with the base component of the buffer: 1aq2 A - 1aq2 A1aq2 [17.7] added acid This reaction can also be represented using 3 O : 3 O 1aq2 A - 1aq2 A1aq2 2 O1l2 Using either equation, we see that this reaction causes 3A - 4 to decrease and [A] to increase. As long as the change in the ratio 3A4>3A - 4 is small, the change in p will be small. Figure 17.2 shows an A/A - buffer consisting of equal concentrations of hydrofluoric acid and fluoride ion (center). The addition of O - reduces [F] and increases 3F - 4, whereas the addition of 3 4 reduces 3F - 4 and increases 3F4. Presence of F counteracts addition of base; p increase is small Presence of F counteracts addition of acid; p decrease is small After addition of O Initial buffered solution [F] = [F ] After addition of Add O Add F F F F F F 2 O F O F F F Figure 17.2 Buffer action. The p of an F>F - buffered solution changes by only a small amount in response to addition of an acid or base.

61 section 17.2 Buffers 731 It is possible to overwhelm a buffer by adding too much strong acid or strong base. We will examine this in more detail a little later in this chapter. Give It Some Thought (a) What happens when NaO is added to a buffer composed of C 3 COO and C 3 COO -? (b) What happens when Cl is added to this buffer? Calculating the p of a Buffer Because conjugate acid base pairs share a common ion, we can use the same procedures to calculate the p of a buffer that we used to treat the common-ion effect in Sample Exercise Alternatively, we can take an approach based on an equation derived from Equation Taking the negative logarithm of both sides of Equation 17.5, we have -log[ 3A4 4 = -logak a 3A - 4 b = -logk a - log 3A4 3A - 4 Because -log3 4 = p and -logk a = pk a, we have p = pk a - log 3A4 3A - 4 = pk a log 3A- 4 3A4 [17.8] (Remember the logarithm rules in Appendix A.2, if you are not sure how this calculation works.) In general, p = pk a log 3base4 [17.9] 3acid4 where [acid] and [base] refer to the equilibrium concentrations of the conjugate acid base pair. Note that when 3base4 = 3acid4, we have p = pk a. Equation 17.9 is known as the enderson asselbalch equation. Biologists, biochemists, and others who work frequently with buffers often use this equation to calculate the p of buffers. In doing equilibrium calculations, we have seen that we can normally neglect the amounts of the acid and base of the buffer that ionize. Therefore, we can usually use the initial concentrations of the acid and base components of the buffer directly in Equation 17.9, as seen in Sample Exercise owever, the assumption that the initial concentrations of the acid and base components in the buffer are equal to the equilibrium concentrations is just that: an assumption. There may be times when you will need to be more careful, as seen in Sample Exercise Sample Exercise 17.3 Calculating the p of a Buffer What is the p of a buffer that is 0.12 M in lactic acid 3C 3 C1O2COO, or C 3 5 O 3 4 and 0.10 M in sodium lactate 3C 3 C1O2COONa or NaC 3 5 O 3 4? For lactic acid, K a = 1.4 * Solution Analyze We are asked to calculate the p of a buffer containing lactic acid 1C 3 5 O 3 2 and its conjugate base, the lactate ion 1C 3 5 O 3-2. Plan We will first determine the p using the method described in Section Because C 3 5 O 3 is a weak electrolyte and NaC 3 5 O 3 is a strong electrolyte, the major species in solution are C 3 5 O 3, Na, and C 3 5 O 3 -. The Na ion is a spectator ion. The C 3 5 O 3 >C 3 5 O 3 - conjugate acid base pair determines 3 4 and, thus, p; 3 4 can be determined using the acid-dissociation equilibrium of lactic acid.

62 732 chapter 17 Additional Aspects of Aqueous Equilibria Solve The initial and equilibrium concentrations of the species involved in this equilibrium are C 3 C1O2COO 1aq2 1aq2 C 3 C1O2COO - 1aq2 Initial (M) Change (M) -x x x Equilibrium (M) ( x) x x2 The equilibrium concentrations are governed by the equilibrium expression: K a = 1.4 * 10-4 = 3 43C 3 5 O 3 3C 3 5 O 3 4 Because K a is small and a common ion is present, we expect x to be small relative to either 0.12 or 0.10 M. Thus, our equation can be simplified to give K a = 1.4 * 10-4 = x Solving for x gives a value that justifies our approximation: 3 4 = x = a b11.4 * = 1.7 * 10-4 M Then, we can solve for p: p = -log11.7 * = 3.77 Alternatively, we can use the enderson asselbalch equation with the initial concentrations of acid and base to calculate p directly: p = pk a log 3base4 3acid4 = = 3.77 = 3.85 loga b Practice Exercise 1 If the p of a buffer solution is equal to the pk a of the acid in the buffer, what does this tell you about the relative concentrations of the acid and conjugate base forms of the buffer components? (a) The acid concentration must be zero. (b) The base concentration must be zero. (c) The acid and base concentrations must be equal. (d) The acid and base concentrations must be equal to the K a. (e) The base concentration must be 2.3 times as large as the acid concentration. Practice Exercise 2 Calculate the p of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. (Refer to Appendix D.) - 4 = x10.10 x x Sample Exercise 17.4 Calculating p When the enderson asselbalch Equation May Not Be Accurate Calculate the p of a buffer that initially contains 1.00 * 10-3 M C 3 COO and 1.00 * 10-4 M C 3 COONa in the following two ways: (i) using the enderson asselbalch equation and (ii) making no assumptions about quantities (which means you will need to use the quadratic equation). The K a of C 3 COO is 1.80 * Solution Analyze We are asked to calculate the p of a buffer two different ways. We know the initial concentrations of the weak acid and its conjugate base, and the K a of the weak acid. Plan We will first use the enderson asselbalch equation, which relates pk a and ratio of acid base concentrations to the p. This will be straightforward. Then, we will redo the calculation making no assumptions about any quantities, which means we will need to write out the initial/change/ equilibrium concentrations, as we have done before. In addition, we will need to solve for quantities using the quadratic equation (since we cannot make assumptions about unknowns being small). Solve (i) The enderson asselbalch equation is We know the K a of the acid 11.8 * , so we know pk a 1pK a = -log K a = We know the initial concentrations of the base, sodium acetate, and the acid, acetic acid, which we will assume are the same as the equilibrium concentrations. p = pk a log 3base4 3acid4

63 section 17.2 Buffers 733 Therefore, we have p = 4.74 log * * Therefore, p = = 3.74 (ii) Now we will redo the calculation, without making any assumptions at all. We will solve for x, which represents the concentration at equilibrium, in order to calculate p. C 3 COO 1aq2 C 3 COO - 1aq2 1aq2 Initial (M) 1.00 * * Change (M) -x x x Equilibrium (M) * x * 10-4 x2 x 3C 3 COO - 4[ 4 3 C 3 COO] * 10-4 x21x * x2 = K a = 1.8 * * 10-4 x x 2 = 1.8 * * x2 x * 10-4 x = 1.8 * * 10-5 x x * 10-4 x * 10-8 = 0 x = * 10-4 { * * = * 10-4 { * = 8.76 * 10-5 = 3 4 p = 4.06 Comment In Sample Exercise 17.3, the calculated p is the same whether we solve exactly using the quadratic equation or make the simplifying assumption that the equilibrium concentrations of acid and base are equal to their initial concentrations. The simplifying assumption works because the concentrations of the acid base conjugate pair are both a thousand times larger than K a. In this Sample Exercise, the acid base conjugate pair concentrations are only as large as K a. Therefore, we cannot assume that x is small compared to the initial concentrations (that is, that the initial concentrations are essentially equal to the equilibrium concentrations). The best answer to this Sample Exercise is p = 4.06, obtained without assuming x is small. Thus we see that the assumptions behind the enderson asselbalch equation become less accurate as the acid/base becomes stronger and/or its concentration becomes smaller. Practice Exercise 1 A buffer is made with sodium acetate 1C 3 COONa2 and acetic acid 1C 3 COO2; the K a for acetic acid is 1.80 * The p of the buffer is What is the ratio of the equilibrium concentration of sodium acetate to that of acetic acid? (a) , (b) 0.174, (c) 0.840, (d) 5.75, (e) Not enough information is given to answer this question. Practice Exercise 2 Calculate the final, equilibrium p of a buffer that initially contains 6.50 * 10-4 M OCl and 7.50 * 10-4 M NaOCl. The K a of OCl is 3.0 * In Sample Exercise 17.3 we calculated the p of a buffered solution. Often we will need to work in the opposite direction by calculating the amounts of the acid and its conjugate base needed to achieve a specific p. This calculation is illustrated in Sample Exercise Sample Exercise 17.5 Preparing a Buffer ow many moles of N 4 Cl must be added to 2.0 L of 0.10 M N 3 to form a buffer whose p is 9.00? (Assume that the addition of N 4 Cl does not change the volume of the solution.) Solution Analyze We are asked to determine the amount of N 4 ion required to prepare a buffer of a specific p.

64 734 chapter 17 Additional Aspects of Aqueous Equilibria Plan The major species in the solution will be N 4, Cl -, and N 3. Of these, the Cl - ion is a spectator (it is the conjugate base of a strong acid). Thus, the N 4 >N 3 conjugate acid base pair will determine the p of the buffer. The equilibrium relationship between N 4 and N 3 is given by the base-dissociation reaction for N 3 : N 3 1aq2 2 O1l2 N 4 1aq2 O - 1aq2 K b = 3N 4 43O - 4 3N 3 4 The key to this exercise is to use this K b expression to calculate 3N 4 4. Solve We obtain 3O - 4 from the given p: and so po = p = = O - 4 = 1.0 * 10-5 M = 1.8 * 10-5 Because K b is small and the common ion 3N 4 4 is present, the equilibrium concentration of N 3 essentially equals its initial concentration: 3N 3 4 = 0.10 M We now use the expression for K b to calculate 3N 4 4 : 3N 3N = K b 3O - 4 = 11.8 * * = 0.18 M Thus, for the solution to have p = 9.00, 3N 4 4 must equal 0.18 M. The number of moles of N 4 Cl needed to produce this concentration is given by the product of the volume of the solution and its molarity: 12.0 L mol N 4 Cl>L2 = 0.36 mol N 4 Cl Comment Because N 4 and N 3 are a conjugate acid base pair, we could use the enderson asselbalch equation (Equation 17.9) to solve this problem. To do so requires first using Equation to calculate pk a for N 4 from the value of pk b for N 3. We suggest you try this approach to convince yourself that you can use the enderson asselbalch equation for buffers for which you are given K b for the conjugate base rather than K a for the conjugate acid. Practice Exercise 1 Calculate the number of grams of ammonium chloride that must be added to 2.00 L of a M ammonia solution to obtain a buffer of p = Assume the volume of the solution does not change as the solid is added. K b for ammonia is 1.80 * (a) 60.7 g, (b) 30.4 g, (c) 1.52 g, (d) g, (e) 1.59 * 10-5 g. Practice Exercise 2 Calculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid 1C 6 5 COO2 to produce a p of Refer to Appendix D. Buffer Capacity and p Range Two important characteristics of a buffer are its capacity and its effective p range. Buffer capacity is the amount of acid or base the buffer can neutralize before the p begins to change to an appreciable degree. The buffer capacity depends on the amount of acid and base used to prepare the buffer. According to Equation 17.5, for example, the p of a 1-L solution that is 1 M in C 3 COO and 1 M in C 3 COONa is the same as the p of a 1-L solution that is 0.1 M in C 3 COO and 0.1 M in C 3 COONa. The first solution has a greater buffering capacity, however, because it contains more C 3 COO and C 3 COO -. The p range of any buffer is the p range over which the buffer acts effectively. Buffers most effectively resist a change in p in either direction when the concentrations of weak acid and conjugate base are about the same. From Equation 17.9 we see that when the concentrations of weak acid and conjugate base are equal, p = pk a. This relationship gives the optimal p of any buffer. Thus, we usually try to select a buffer whose acid form has a pk a close to the desired p. In practice, we find that if the concentration of one component of the buffer is more than 10 times the concentration of the other component, the buffering action is poor. Because log 10 = 1, buffers usually have a usable range within {1 p unit of pk a (that is, a range of p = pk a {1).

65 section 17.2 Buffers 735 Give It Some Thought The K a values for nitrous acid 1NO 2 2 and hypochlorous (ClO) acid are 4.5 * 10-4 and 3.0 * 10-8, respectively. Which one would be more suitable for use in a solution buffered at p = 7.0? What other substances would be needed to make the buffer? Addition of Strong Acids or Bases to Buffers Let s now consider in a more quantitative way how a buffered solution responds to addition of a strong acid or base. In this discussion, it is important to understand that neutralization reactions between strong acids and weak bases proceed essentially to completion, as do those between strong bases and weak acids. This is because water is a produce of the reaction, and you have an equilibrium constant of 1>K w = in your favor when making water. (Section 16.3) Thus, as long as we do not exceed the buffering capacity of the buffer, we can assume that the strong acid or strong base is completely consumed by reaction with the buffer. Consider a buffer that contains a weak acid A and its conjugate base A -. When a strong acid is added to this buffer, the added is consumed by A - to produce A; thus, [A] increases and 3A - 4 decreases. (See Equation 17.7.) Upon addition of a strong base, the added O - is consumed by A to produce A - ; in this case [A] decreases and 3A - 4 increases. (See Equation 17.6.) These two situations are summarized in Figure To calculate how the p of the buffer responds to the addition of a strong acid or a strong base, we follow the strategy outlined in Figure 17.3: 1. Consider the acid base neutralization reaction and determine its effect on [A] and 3A - 4. This step is a limiting reactant stoichiometry calculation. (Section 3.6 and 3.7) 2. Use the calculated values of [A] and 3A - 4 along with K a to calculate 3 4. This step is an equilibrium calculation and is most easily done using the enderson asselbalch equation (if the concentrations of the weak acid base pair are very large compared to K a for the acid). Strong acid reacts with conjugate base component of buffer Add strong acid A A Buffer containing A and A Calculate new values of [A] and [A ] Use K a, [A], and [A ] to calculate [ ] p Add strong base A O A 2 O Strong base reacts with weak acid component of buffer 1 Stoichiometry 2 Equilibrium calculation calculation Figure 17.3 Calculating the p of a buffer after addition of a strong acid or strong base.

66 736 chapter 17 Additional Aspects of Aqueous Equilibria Sample Exercise 17.6 Calculating p Changes in Buffers L buffer M C 3 COO M C 3 COO p = 4.74 A buffer is made by adding mol C 3 COO and mol C 3 COONa to enough water to make L of solution. The p of the buffer is 4.74 (Sample Exercise 17.1). (a) Calculate the p of this solution after 5.0 ml of 4.0 M NaO1aq2 solution is added. (b) For comparison, calculate the p of a solution made by adding 5.0 ml of 4.0 M NaO1aq2 solution to L of pure water. Solution Analyze We are asked to determine the p of a buffer after addition of a small amount of strong base and to compare the p change with the p that would result if we were to add the same amount of strong base to pure water. Plan Solving this problem involves the two steps outlined in Figure First we do a stoichiometry calculation to determine how the added O - affects the buffer composition. Then we use the resultant buffer composition and either the enderson asselbalch equation or the equilibrium-constant expression for the buffer to determine the p. Solve (a) Stoichiometry Calculation: The O - provided by NaO reacts with C 3 COO, the weak acid component of the buffer. Since volumes are changing, it is prudent to figure out how many moles of reactants and products would be produced, then divide by the final volume later to obtain concentrations. Prior to this neutralization reaction, there are mol each of C 3 COO and C 3 COO -. The amount of base added is L * 4.0 mol>l = mol. Neutralizing the mol O - requires mol of C 3 COO. Consequently, the amount of C 3 COO decreases by mol, and the amount of the product of the neutralization, C 3 COO -, increases by mol. We can create a table to see how the composition of the buffer changes as a result of its reaction with O - : Add 5.0 ml of 4.0 M NaO(aq) L 2 O Add 5.0 ml of 4.0 M NaO(aq) p = 4.80 p increases by 0.06 p units p = 7.00 p = p increases by 5.30 p units Figure 17.4 Effect of adding a strong base to a buffered solution and to water. C 3 COO1aq2 O - 1aq2 2 O1l2 C 3 COO - 1aq2 Before reaction (mol) Change (limiting reactant) (mol) After reaction (mol) Equilibrium Calculation: We now turn our attention to the equilibrium for the ionization of acetic acid, the relationship that determines the buffer p: C 3 COO1aq2 1aq2 C 3 COO - 1aq2 Using the quantities of C 3 COO and C 3 COO - remaining in the buffer after the reaction with strong base, we determine the p using the enderson asselbalch equation. The volume of the solution is now L L = L due to addition of the NaO solution: mol>1.005 L p = 4.74 log mol>1.005 L = 4.80 (b) To determine the p of a solution made by adding mol of NaO to L of pure water, we first determine the concentration of O - ions in solution, 3O - 4 = mol>1.005 L = M We use this value in Equation to calculate po and then use our calculated po value in Equation to obtain p: po = -log3o - 4 = -log = 1.70 p = = Comment Note that the small amount of added NaO changes the p of water significantly. In contrast, the p of the buffer changes very little when the NaO is added, as summarized in Figure Practice Exercise 1 Which of these statements is true? (a) If you add strong acid or base to a buffer, the p will never change. (b) In order to do calculations in which strong acid or base is added to a buffer, you only need to use the enderson asselbalch equation. (c) Strong bases react with strong acids, but not weak acids. (d) If you add a strong acid or base to a buffer, the buffer s pk a or pk b will change. (e) In order to do calculations in which strong acid or base is added to a buffer, you need to calculate the amounts of substances from the neutralization reaction and then equilibrate.

67 section 17.2 Buffers 737 Practice Exercise 2 Determine (a) the p of the original buffer described in Sample Exercise 17.6 after the addition of mol Cl and (b) the p of the solution that would result from the addition of mol Cl to L of pure water. Chemistry and Life Blood as a Buffered Solution Chemical reactions that occur in living systems are often extremely sensitive to p. Many of the enzymes that catalyze important biochemical reactions are effective only within a narrow p range. For this reason, the human body maintains a remarkably intricate system of buffers, both within cells and in the fluids that transport cells. Blood, the fluid that transports oxygen to all parts of the body, is one of the most prominent examples of the importance of buffers in living beings. uman blood has a normal p of 7.35 to Any deviation from this range can have extremely disruptive effects on the stability of cell membranes, the structures of proteins, and the activities of enzymes. Death may result if the blood p falls below 6.8 or rises above 7.8. When the p falls below 7.35, the condition is called acidosis; when it rises above 7.45, the condition is called alkalosis. Acidosis is the more common tendency because metabolism generates several acids in the body. The major buffer system used to control blood p is the carbonic acid bicarbonate buffer system. Carbonic acid 1 2 CO 3 2 and bicarbonate ion 1CO 3-2 are a conjugate acid base pair. In addition, carbonic acid decomposes into carbon dioxide gas and water. The important equilibria in this buffer system are 1aq2 CO 3-1aq2 2 CO 3 1aq2 2 O1l2 CO 2 1g2 [17.10] Several aspects of these equilibria are notable. First, although carbonic acid is diprotic, the carbonate ion 1CO is unimportant in this system. Second, one component of this equilibrium, CO 2, is a gas, which provides a mechanism for the body to adjust the equilibria. Removal of CO 2 via exhalation shifts the equilibria to the right, consuming ions. Third, the buffer system in blood operates at p 7.4, which is fairly far removed from the pk a1 value of 2 CO 3 (6.1 at physiological temperatures). For the buffer to have a p of 7.4, the ratio [base]/[acid] must be about 20. In normal blood plasma, the concentrations of CO 3 - and 2 CO 3 are about M and M, respectively. Consequently, the buffer has a high capacity to neutralize additional acid but only a low capacity to neutralize additional base. The principal organs that regulate the p of the carbonic acid bicarbonate buffer system are the lungs and kidneys. When the concentration of CO 2 rises, the equilibrium concentrations in Equation shift to the left, which leads to the formation of more and a drop in p. This change is detected by receptors in the brain that trigger a reflex to breathe faster and deeper, increasing the rate at which CO 2 is expelled from the lungs and thereby shifting the equilibrium concentrations back to the right. When the blood p becomes too high, the kidneys remove CO 3 - from the blood. This shifts the equilibrium concentrations to the left, increasing the concentration of. As a result, the p decreases. Regulation of blood p relates directly to the effective transport of O 2 throughout the body. The protein hemoglobin, found in red blood cells ( Figure 17.5), carries oxygen. emoglobin (b) reversibly binds both O 2 and. These two substances compete for the b, which can be represented approximately by the equilibrium b O 2 bo 2 [17.11] Oxygen enters the blood through the lungs, where it passes into the red blood cells and binds to b. When the blood reaches tissue in which the concentration of O 2 is low, the equilibrium concentrations in Equation shift to the left and O 2 is released. During periods of strenuous exertion, three factors work together to ensure delivery of O 2 to active tissues. The role of each factor can be understood by applying Le Châtelier s principle to Equation 17.11: 1. O 2 is consumed, causing the equilibrium concentrations to shift to the left, releasing more O Large amounts of CO 2 are produced by metabolism, which increases 3 4 and causes the equilibrium concentrations to shift to the left, releasing O Body temperature rises. Because Equation is exothermic, the increase in temperature shifts the equilibrium concentrations to the left, releasing O 2. In addition to the factors causing release of O 2 to tissues, the decrease in p stimulates an increase in breathing rate, which furnishes more O 2 and eliminates CO 2. Without this elaborate series of equilibrium shifts and p changes, the O 2 in tissues would be rapidly depleted, making further activity impossible. Under such conditions, the buffering capacity of the blood and the exhalation of CO 2 through the lungs are essential to keep the p from dropping too low, thereby triggering acidosis. Related Exercises: 17.29, Figure 17.5 Red blood cells. A scanning electron micrograph of red blood cells traveling through a small branch of an artery. The red blood cells are approximately mm in diameter.

68 738 chapter 17 Additional Aspects of Aqueous Equilibria Go Figure In which direction do you expect the p to change as NaO is added to the Cl solution? Burette containing NaO(aq) of known concentration p meter Beaker containing Cl(aq) of unknown concentration Figure 17.6 Measuring p during a titration Acid Base Titrations Titrations are procedures in which one reactant is slowly added into a solution of another reactant, while equilibrium concentrations along the way are monitored. (Section 4.6) There are two main reasons to do titrations: (1) you want to know the concentration of one of the reactants or (2) you want to know the equilibrium constant for the reaction. In an acid base titration, a solution containing a known concentration of base is slowly added to an acid (or the acid is added to the base). (Section 4.6) Acid base indicators can be used to signal the equivalence point of a titration (the point at which stoichiometrically equivalent quantities of acid and base have been brought together). Alternatively, a p meter can be used to monitor the progress of the reaction ( Figure 17.6), producing a p titration curve, a graph of the p as a function of the volume of titrant added. The shape of the titration curve makes it possible to determine the equivalence point. The curve can also be used to select suitable indicators and to determine the K a of the weak acid or the K b of the weak base being titrated. To understand why titration curves have certain characteristic shapes, we will examine the curves for three kinds of titrations: (1) strong acid strong base, (2) weak acid strong base, and (3) polyprotic acid strong base. We will also briefly consider how these curves relate to those involving weak bases. Strong Acid Strong Base Titrations The titration curve produced when a strong base is added to a strong acid has the general shape shown in Figure 17.7, which depicts the p change that occurs as M Go Figure What volume of NaO(aq) would be needed to reach the equivalence point if the concentration of the added base were M? 14 p Equivalence point p = 7.0 at equivalence point, NaCl(aq) salt solution 4 2 Equivalence point occurs when moles base = moles acid ml NaO Cl Na O 1 Only Cl(aq) 2 consumed as 3 completely 4 No left to present before O added, forming neutralized by react with excess titration 2 O (p < 7.0) O (p = 7.0) O (p > 7.0) Figure 17.7 Titration of a strong acid with a strong base. The p curve for titration of 50.0 ml of a M solution of hydrochloric acid with a M solution of NaO1aq2. For clarity, water molecules have been omitted from the molecular art.

69 section 17.3 Acid Base Titrations 739 NaO is added to 50.0 ml of M Cl. The p can be calculated at various stages of the titration. To help understand these calculations, we can divide the curve into four regions: 1. Initial p: The p of the solution before the addition of any base is determined by the initial concentration of the strong acid. For a solution of M Cl, 3 4 = M and p = -log = Thus, the initial p is low. 2. Between initial p and equivalence point: As NaO is added, the p increases slowly at first and then rapidly in the vicinity of the equivalence point. The p before the equivalence point is determined by the concentration of acid not yet neutralized. This calculation is illustrated in Sample Exercise 17.7(a). 3. Equivalence point: At the equivalence point an equal number of moles of NaO and Cl have reacted, leaving only a solution of their salt, NaCl. The p of the solution is 7.00 because the cation of a strong base (in this case Na ) and the anion of a strong acid (in this case Cl - ) are neither acids nor bases and, therefore, have no appreciable effect on p. (Section 16.9) 4. After equivalence point: The p of the solution after the equivalence point is determined by the concentration of excess NaO in the solution. This calculation is illustrated in Sample Exercise 17.7(b). Sample Exercise 17.7 Calculations for a Strong Acid Strong Base Titration Calculate the p when (a) 49.0 ml and (b) 51.0 ml of M NaO solution have been added to 50.0 ml of M Cl solution. Solution Analyze We are asked to calculate the p at two points in the titration of a strong acid with a strong base. The first point is just before the equivalence point, so we expect the p to be determined by the small amount of strong acid that has not yet been neutralized. The second point is just after the equivalence point, so we expect this p to be determined by the small amount of excess strong base. Plan (a) As the NaO solution is added to the Cl solution, 1aq2 reacts with O - 1aq2 to form 2 O. Both Na and Cl - are spectator ions, having negligible effect on the p. To determine the p of the solution, we must first determine how many moles of were originally present and how many moles of O - were added. We can then calculate how many moles of each ion remain after the neutralization reaction. To calculate 3 4, and hence p, we must also remember that the volume of the solution increases as we add titrant, thus diluting the concentration of all solutes present. Therefore, it is best to deal with moles first, and then convert to molarities using total solution volumes (volume of acid plus volume of base). Solve The number of moles of in the original Cl solution is given by the product of the volume of the solution and its molarity: L soln2a mol b = 5.00 * 10-3 mol 1 L soln Likewise, the number of moles of O -, in mol O ml of M NaO is L soln2a b = 4.90 * 10-3 mol O - 1 L soln Because we have not reached the equivalence point, there are more moles of present than O -. Therefore, O - is the limiting reactant. Each mole of O - reacts with 1 mol of. Using the convention introduced in Sample Exercise 17.6, we have 1aq2 O - 1aq2 2 O1l2 Before reaction (mol) 5.00 * * 10-3 Change (limiting reactant) (mol) * * 10-3 After reaction (mol) 0.10 * The volume of the reaction mixture increases as the NaO solution is added to the Cl solution. Thus, at this point in the titration, the volume in the titration flask is 50.0 ml 49.0 ml = 99.0 ml = L Thus, the concentration of 1aq2 in the flask is 3 4 = moles 1aq2 liters soln = 0.10 * 10-3 mol L = 1.0 * 10-3 M The corresponding p is -log11.0 * = 3.00

70 740 chapter 17 Additional Aspects of Aqueous Equilibria Plan (b) We proceed in the same way as we did in part (a) except we are now past the equivalence point and have more O - in the solution than. As before, the initial number of moles of each reactant Solve is determined from their volumes and concentrations. The reactant present in smaller stoichiometric amount (the limiting reactant) is consumed completely, leaving an excess of hydroxide ion. 1aq2 O - 1aq2 2 O1l2 Before reaction (mol) 5.00 * * 10-3 Change (limiting * * 10-3 reactant) (mol) After reaction (mol) * 10-3 In this case, the volume in the titration flask is ence, the concentration of O - 1aq2 in the flask is and we have 50.0 ml 51.0 ml = ml = L 3O - 4 = moles O- 1aq2 liters soln po = -log11.0 * = 3.00 = 0.10 * 10-3 mol L p = po = = = 1.0 * 10-3 M Comment Note that the p increased by only two p units, from 1.00 (Figure 17.7) to 3.00, after the first 49.0 ml of NaO solution was added, but jumped by eight p units, from 3.00 to 11.00, as 2.0 ml of base solution was added near the equivalence point. Such a rapid rise in p near the equivalence point is a characteristic of titrations involving strong acids and strong bases. Practice Exercise 1 An acid base titration is performed: ml of an unknown concentration of Cl (aq) is titrated to the equivalence point with 36.7 ml of a M aqueous solution of NaO. Which of the following statements is not true of this titration? (a) The Cl solution is less concentrated than the NaO solution. (b) The p is less than 7 after adding 25 ml of NaO solution. (c) The p at the equivalence point is (d) If an additional 1.00 ml of NaO solution is added beyond the equivalence point, the p of the solution is more than (e) At the equivalence point, the O - concentration in the solution is 3.67 * 10-3 M. Practice Exercise 2 Calculate the p when (a) 24.9 ml and (b) 25.1 ml of M NO 3 have been added to 25.0 ml of M KO solution Titration of a solution of a strong base with a solution of a strong acid yields an analogous curve of p versus added acid. In this case, however, the p is high at the outset of the titration and low at its completion ( Figure 17.8). The p at the equivalence point is still 7.0 (at 25 C), just like the strong acid strong base titration. p Equivalence point Give It Some Thought What is the p at the equivalence point when 0.10 M NO 3 is used to titrate a volume of solution containing 0.30 g of KO? ml acid Figure 17.8 Titration of a strong base with a strong acid. The p curve for titration of 50.0 ml of a M solution of a strong base with a M solution of a strong acid. Weak Acid Strong Base Titrations The curve for titration of a weak acid by a strong base is similar to the curve in Figure Consider, for example, the curve for titration of 50.0 ml of M acetic acid with M NaO shown in Figure We can calculate the p at points along this curve, using principles we discussed earlier, which means again dividing the curve into four regions: 1. Initial p: We use K a to calculate this p, as shown in Section The calculated p of M C 3 COO is 2.89.

71 section 17.3 Acid Base Titrations 741 Go Figure If the acetic acid being titrated here were replaced by hydrochloric acid, would the amount of base needed to reach the equivalence point change? Would the p at the equivalence point change? 14 p change near the equivalence point is smaller than in strong acid strong base titration 12 p Equivalence point p > 7.0 at equivalence point because C 3 COO is a weak base that reacts with water to make O 4 2 Equivalence point occurs when moles base = moles acid ml NaO C 3 COO C 3 COO Na O 1 C 3 COO(aq) 2 Added O converts 3 Acid completely 4 solution before titration C 3 COO(aq) into C 3 COO (aq), forming buffer solution neutralized by added base, C 3 COONa(aq) salt solution results No acid left to react with excess O Figure 17.9 Titration of a weak acid with a strong base. The p curve for titration of 50.0 ml of a M solution of acetic acid with a M solution of NaO1aq2. For clarity, water molecules have been omitted from the molecular art. 2. Between initial p and equivalence point: Prior to reaching the equivalence point, the acid is being neutralized, and its conjugate base is being formed: C 3 COO1aq2 O - 1aq2 C 3 COO - 1aq2 2 O1l2 [17.12] Thus, the solution contains a mixture of C 3 COO and C 3 COO -. Calculating the p in this region involves two steps. First, we consider the neutralization reaction between C 3 COO and O - to determine 3C 3 COO] and 3C 3 COO - 4. Next, we calculate the p of this buffer pair using procedures developed in Sections 17.1 and The general procedure is diagrammed in Figure and illustrated in Sample Exercise Equivalence point: The equivalence point is reached when 50.0 ml of M NaO has been added to the 50.0 ml of M C 3 COO. At this point, the 5.00 * 10-3 mol of NaO completely reacts with the 5.00 * 10-3 mol

72 742 chapter 17 Additional Aspects of Aqueous Equilibria Neutralization reaction: [A] decreases [A ] increases Solution containing weak acid Add strong base A O A 2 O Calculate new values of [A] and [A ] Use K a, [A], and [A ] to calculate [ ] p 1 Stoichiometry 2 Equilibrium calculation calculation Figure Procedure for calculating p when a weak acid is partially neutralized by a strong base. of C 3 COO to form 5.00 * 10-3 mol of C 3 COONa. The Na ion of this salt has no significant effect on the p. The C 3 COO - ion, however, is a weak base whose reaction with water cannot be neglected, and the p at the equivalence point is therefore greater than 7. In general, the p at the equivalence point is always above 7 in a weak acid strong base titration because the anion of the salt formed is a weak base. The procedure for calculating the p of the solution of a weak base is described in Section 16.7 and is shown in Sample Exercise After equivalence point (excess base): In this region, 3O - 4 from the reaction of C 3 COO - with water is negligible relative to 3O - 4 from the excess NaO. Thus, the p is determined by the concentration of O - from the excess NaO. The method for calculating p in this region is therefore like that illustrated in Sample Exercise 17.7(b). Thus, the addition of 51.0 ml of M NaO to 50.0 ml of either M Cl or M C 3 COO yields the same p, Notice by comparing Figures 17.7 and 17.9 that the titration curves for a strong acid and a weak acid are the same after the equivalence point. Sample Exercise 17.8 Calculations for a Weak Acid Strong Base Titration Calculate the p of the solution formed when 45.0 ml of M NaO is added to 50.0 ml of M C 3 COO 1K a = 1.8 * Solution Analyze We are asked to calculate the p before the equivalence point of the titration of a weak acid with a strong base. Plan We first must determine the number of moles of C 3 COO and C 3 COO - present after the neutralization reaction (the stoichiometry calculation). We then calculate p using K a, 3C 3 COO], and 3C 3 COO - 4 (the equilibrium calculation). Solve Stoichiometry Calculation: The product of the volume and concentration of each solution gives the number of moles of each reactant present before the neutralization: L soln2a mol C 3COO b = 5.00 * 10-3 mol C 1 L soln 3 COO L soln2a mol NaO b = 4.50 * 10-3 mol NaO 1 L soln The 4.50 * 10-3 of NaO consumes 4.50 * 10-3 of C 3 COO: The total volume of the solution is C 3 COO1aq2 O - 1aq2 C 3 COO - 1aq2 2 O1l2 Before reaction (mol) 5.00 * * Change (limiting * * * 10-3 reactant) (mol) After reaction (mol) 0.50 * * ml 50.0 ml = 95.0 ml = L

73 section 17.3 Acid Base Titrations 743 The resulting molarities of C 3 COO and C 3 COO - after the reaction are therefore 3C 3 COO4 = 0.50 * 10-3 mol L = M Equilibrium Calculation: The equilibrium between C 3 COO and C 3 COO - must obey the equilibrium-constant expression for C 3 COO: Solving for 3 4 gives 3C 3 COO - 4 = 4.50 * 10-3 mol L K a = 3 43C 3 COO - 4 3C 3 COO4 = 1.8 * 10-5 = M 3 4 = K a * 3C 3COO4 3C 3 COO - 4 = 11.8 * * a b = 2.0 * 10-6 M p = -log12.0 * = 5.70 Comment We could have solved for p equally well using the enderson asselbalch equation in the last step. Practice Exercise 1 If you think carefully about what happens during the course of a weak acid strong base titration, you can learn some very interesting things. For example, let s look back at Figure 17.9 and pretend you did not know that acetic acid was the acid being titrated. You can figure out the pk a of a weak acid just by thinking about the definition of K a and looking at the right place on the titration curve! Which of the following choices is the best way to do this? (a) At the equivalence point, p = pk a. (b) alfway to the equivalence point, p = pk a. (c) Before any base is added, p = pk a. (d) At the top of the graph with excess base added, p = pk a. Practice Exercise 2 (a) Calculate the p in the solution formed by adding 10.0 ml of M NaO to 40.0 ml of M benzoic acid 1C 6 5 COO, K a = 6.3 * (b) Calculate the p in the solution formed by adding 10.0 ml of M Cl to 20.0 ml of M N 3. In order to further monitor the evolution of p as a function of added base, we can calculate the p at the equivalence point. Sample Exercise 17.9 Calculating the p at the Equivalence Point Calculate the p at the equivalence point in the titration of 50.0 ml of M C 3 COO with M NaO. Solution Analyze We are asked to determine the p at the equivalence point of the titration of a weak acid with a strong base. Because the neutralization of a weak acid produces its anion, a conjugate base that can react with water, we expect the p at the equivalence point to be greater than 7. Plan The initial number of moles of acetic acid equals the number of moles of acetate ion at the equivalence point. We use the volume of the solution at the equivalence point to calculate the concentration of acetate ion. Because the acetate ion is a weak base, we can calculate the p using K b and 3C 3 COO - 4. Solve The number of moles of acetic acid in the initial solution is obtained from the volume and molarity of the solution: Moles = M * L = mol>l L2 = 5.00 * 10-3 mol C 3 COO ence, 5.00 * 10-3 mol of C 3 COO - is formed. It will take 50.0 ml of NaO to reach the equivalence point (Figure 17.9). The volume of this salt solution at the equivalence point is the sum of the volumes of the acid and base, 50.0 ml 50.0 ml = ml = L. Thus, the concentration of C 3 COO - is The C 3 COO - ion is a weak base: 3C 3 COO - 4 = 5.00 * 10-3 mol L = M C 3 COO - 1aq2 2 O1l2 C 3 COO1aq2 O - 1aq2

74 744 chapter 17 Additional Aspects of Aqueous Equilibria The K b for C 3 COO - can be calculated from the K a value of its conjugate acid, K b = K w >K a = 11.0 * >11.8 * = 5.6 * Using the K b expression, we have K b = 3C 3COO43O - 4 3C 3 COO - 4 = 1x21x x = 5.6 * Making the approximation that x , and then solving for x, we have x = 3O - 4 = 5.3 * 10-6 M, which gives po = 5.28 and p = Check The p is above 7, as expected for the salt of a weak acid and strong base. Practice Exercise 1 Why is p at the equivalence point larger than 7 when you titrate a weak acid with a strong base? (a) There is excess strong base at the equivalence point. (b) There is excess weak acid at the equivalence point. (c) The conjugate base that is formed at the equivalence point is a strong base. (d) The conjugate base that is formed at the equivalence point reacts with water. (e) This statement is false: the p is always 7 at an equivalence point in a p titration. Practice Exercise 2 Calculate the p at the equivalence point when (a) 40.0 ml of M benzoic acid 1C 6 5 COO, K a = 6.3 * is titrated with M NaO and (b) 40.0 ml of M N 3 is titrated with M Cl. The titration curve for a weak acid strong base titration (Figure 17.9) differs from the curve for a strong acid strong base titration (Figure 17.7) in three noteworthy ways: 1. The solution of the weak acid has a higher initial p than a solution of a strong acid of the same concentration. 2. The p change in the rapid-rise portion of the curve near the equivalence point is smaller for the weak acid than for the strong acid. 3. The p at the equivalence point is above 7.00 for the weak acid titration. Go Figure ow does the p at the equivalence point change as the acid being titrated becomes weaker? ow does the volume of NaO(aq) needed to reach the equivalence point change? p Equivalence point Initial p increases as K a decreases K a = 10 8 K a = 10 6 K a = 10 4 K a = 10 2 Strong acid ml NaO p at equivalence point (marked with circles) increases as K a decreases Figure A set of curves showing the effect of acid strength on the characteristics of the titration curve when a weak acid is titrated by a strong base. Each curve represents titration of 50.0 ml of 0.10 M acid with 0.10 M NaO. Give It Some Thought Describe the reasons why statement 3 above is true. The weaker the acid, the more pronounced these differences become. To illustrate this consider the family of titration curves shown in Figure Notice that as the acid becomes weaker (that is, as K a becomes smaller), the initial p increases and the p change near the equivalence point becomes less marked. Furthermore, the p at the equivalence point steadily increases as K a decreases, because the strength of the conjugate base of the weak acid increases. It is virtually impossible to determine the equivalence point when pk a is 10 or higher because the p change is too small and gradual. Titrating with an Acid Base Indicator Oftentimes in an acid base titration, an indicator is used rather than a p meter. An indicator is a compound that changes color in solution over a specific p range. Optimally, an indicator should change color at the equivalence point in a titration. In practice, however, an indicator need not precisely mark the equivalence point. The p changes very rapidly near the equivalence point, and in this region one drop of titrant can change the p by several units. Thus, an indicator beginning and ending its color change anywhere on the rapid-rise portion of the titration curve gives a sufficiently accurate measure of the titrant volume needed to reach the equivalence point. The point in a titration where the indicator changes color is called the end point to distinguish it from the equivalence point that it closely approximates.

75 section 17.3 Acid Base Titrations 745 Figure shows the curve for titration of a strong base (NaO) with a strong acid (Cl). We see from the vertical part of the curve that the p changes rapidly from roughly 11 to 3 near the equivalence point. Consequently, an indicator for this titration can change color anywhere in this range. Most strong acid strong base titrations are carried out using phenolphthalein as an indicator because it changes color in this range (see Figure 16.8, page 684). Several other indicators would also be satisfactory, including methyl red, which, as the lower color band in Figure shows, changes color in the p range from about 4.2 to 6.0 (see Figure 16.8, page 684). As noted in our discussion of Figure 17.11, because the p change near the equivalence point becomes smaller as K a decreases, the choice of indicator for a weak acid strong base titration is more critical than it is for titrations where both acid and base are strong. When M C 3 COO 1K a = 1.8 * is titrated with M NaO, for example, the p increases rapidly only over the p range from about 7 to 11 ( Figure 17.13). Phenolphthalein is therefore an ideal indicator because it changes color from p 8.3 to 10.0, close to the p at the equivalence point. Methyl red is a poor choice, however, because its color change, from 4.2 to 6.0, begins well before the equivalence point is reached. Titration of a weak base (such as M N 3 ) with a strong acid solution (such as M Cl) leads to the titration curve shown in Figure In this example, the equivalence point occurs at p Thus, methyl red is an ideal indicator but phenolphthalein would be a poor choice. Go Figure Is methyl red a suitable indicator when you are titrating a strong acid with a strong base? Explain your answer. p Equivalence point ml Cl Phenolphthalein color-change interval Methyl red color-change interval Figure Using color indicators for titration of a strong base with a strong acid. Both phenolphthalein and methyl red change color in the rapid-rise portion of the titration curve. 14 Phenolphthalein indicator Color-change interval 8.3 < p < Methyl red indicator Color-change interval 4.2 < p < Equivalence point 10 Equivalence point p 8 6 p ml NaO ml NaO Suitable indicator for titration of a weak acid with a strong base because equivalence point falls within the color-change interval Unsatisfactory indicator for titration of a weak acid with a strong base because color changes before reaching equivalence point Figure Good and poor indicators for titration of a weak acid with a strong base.

76 746 chapter 17 Additional Aspects of Aqueous Equilibria 14 Phenolphthalein indicator Color-change interval 8.3 < p < Methyl red indicator Color-change interval 4.2 < p < p 8 6 p Equivalence point 2 Equivalence point ml Cl ml Cl Unsatisfactory indicator for titration of a weak base with a strong acid because color changes before reaching equivalence point Suitable indicator for titration of a weak base with a strong acid because equivalence point falls within the color-change interval Figure Good and poor indicators for titration of a weak base with a strong acid. Give It Some Thought Why is the choice of indicator more crucial for a weak acid strong base titration than for a strong acid strong base titration? Titrations of Polyprotic Acids When weak acids contain more than one ionizable atom, the reaction with O - occurs in a series of steps. Neutralization of 3 PO 3, for example, proceeds in two steps (the third is bonded to the P and does not ionize): 3 PO 3 1aq2 O - 1aq2 2 PO - 3 1aq2 2 O1l2 [17.13] 2 PO - 3 1aq2 O - 2 1aq2 PO - 3 1aq2 2 O1l2 [17.14] When the neutralization steps of a polyprotic acid or polybasic base are sufficiently separated, the titration has multiple equivalence points. Figure shows the two equivalence points corresponding to Equations and Give It Some Thought Sketch an approximate titration curve for the titration of Na 2 CO 3 with Cl. You can use titration data such as that shown in Figure to figure out the pk a s of the weak polyprotic acid. For example, let s write the K a1 and K a2 reactions for phosphorous acid: 3 PO 3 1aq2 2 PO 3-1aq2 1aq2 K a1 = 3 2PO PO 3 2 PO 3-1aq2 PO 3 2-1aq2 1aq2 K a2 = 3PO PO 3 -

77 section 17.3 Acid Base Titrations PO 3 is the dominant species 14 2 PO 3 is the dominant species PO 3 2 is the dominant species p ml NaO Figure Titration curve for a diprotic acid. The curve shows the p change when 50.0 ml of 0.10 M 3 PO 3 is titrated with 0.10 M NaO. If we rearrange these equilibrium expressions, you see that we obtain enderson asselbalch equations: p = pk a1 log 3 2PO PO 3 4 p = pk a2 log 3PO PO 3-4 Therefore, if the concentrations of the each acid and base conjugate pairs were identical for each equilibrium, log 112 = 0 and so p = pk a. When does this happen during the titration? At the beginning of the titration, the acid is 3 PO 3 initially; at the first equivalence point, it is all converted to 2 PO 3 -. Therefore, halfway to the first equivalence point, half of the 3 PO 3 is converted to 2 PO 3 -. Thus, halfway to the equivalence point, the concentration of 3 PO 3 is equal to that of 2 PO 3 -, and at that point, p = pk a1. Similar logic is true for the second equilibrium reaction: halfway toward its equivalence point, p = pk a2. We can then just look at titration data and estimate the pk a s for the polyprotic acid directly from the titration curve. This procedure is especially useful if you are trying to identify an unknown polyprotic acid. In Figure 17.15, for instance, the first equivalence point occurs for 50 ml NaO added. alfway to the equivalence point corresponds to 25 ml NaO. Because the p at 25 ml NaO is about 1.5 we may estimate pk a1 = 1.5 for phosphorous acid. The second equivalence point occurs at 100 ml NaO added; halfway there (from the first equivalence point) is at 75 ml NaO added. The graph indicates the p at 75 ml NaO added is about 6.5, and we therefore estimate that pk a2 for phosphorous acid is 6.5. The actual values for the two pk a s are pk a1 = 1.3 and pk a2 = 6.7 (close to our estimates).

78 748 chapter 17 Additional Aspects of Aqueous Equilibria 17.4 Solubility Equilibria The equilibria we have considered thus far in this chapter have involved acids and bases. Furthermore, they have been homogeneous; that is, all the species have been in the same phase. Through the rest of the chapter, we will consider the equilibria involved in the dissolution or precipitation of ionic compounds. These reactions are heterogeneous. Dissolution and precipitation occur both within us and around us. Tooth enamel dissolves in acidic solutions, for example, causing tooth decay, and the precipitation of certain salts in our kidneys produces kidney stones. The waters of Earth contain salts dissolved as water passes over and through the ground. Coral reefs are principally made of CaCO 3, as we saw in the beginning of this chapter. Precipitation of CaCO 3 from groundwater is responsible for the formation of stalactites and stalagmites within limestone caves. In our earlier discussion of precipitation reactions, we considered general rules for predicting the solubility of common salts in water. (Section 4.2) These rules give us a qualitative sense of whether a compound has a low or high solubility in water. By considering solubility equilibria, however, we can make quantitative predictions about solubility. The Solubility-Product Constant, K sp Recall that a saturated solution is one in which the solution is in contact with undissolved solute. (Section 13.2) Consider, for example, a saturated aqueous solution of BaSO 4 in contact with solid BaSO 4. Because the solid is an ionic compound, it is a strong electrolyte and yields Ba 2 2 1aq2 and SO - 4 1aq2 ions when dissolved in water, readily establishing the equilibrium BaSO 4 1s2 Ba 2 2 1aq2 SO - 4 1aq2 [17.15] As with any other equilibrium, the extent to which this dissolution reaction occurs is expressed by the magnitude of the equilibrium constant. Because this equilibrium equation describes the dissolution of a solid, the equilibrium constant indicates how soluble the solid is in water and is referred to as the solubility-product constant (or simply the solubility product). It is denoted K sp, where sp stands for solubility product. The equilibrium-constant expression for the equilibrium between a solid and an aqueous solution of its component ions 1K sp 2 is written according to the rules that apply to any other equilibrium-constant expression. Remember, however, that solids do not appear in the equilibrium-constant expressions for heterogeneous equilibrium. (Section 15.4) Thus, the solubility-product expression for BaSO 4, which is based on Equation 17.15, is K sp = 3Ba 2 43SO [17.16] In general, the solubility product K sp of a compound equals the product of the concentration of the ions involved in the equilibrium, each raised to the power of its coefficient in the equilibrium equation. The coefficient for each ion in the equilibrium equation also equals its subscript in the compound s chemical formula. The values of K sp at 25 C for many ionic solids are tabulated in Appendix D. The value of K sp for BaSO 4 is 1.1 * 10-10, a very small number indicating that only a very small amount of the solid dissolves in 25 C water. Sample Exercise Writing Solubility-Product 1K sp 2 Expressions Write the expression for the solubility-product constant for CaF 2, and look up the corresponding K sp value in Appendix D. Solution Analyze We are asked to write an equilibrium-constant expression for the process by which CaF 2 dissolves in water. Plan We apply the general rules for writing an equilibrium-constant expression, excluding the solid reactant from the expression. We assume that the compound dissociates completely into its component ions: CaF 2 1s2 Ca 2 1aq2 2 F - 1aq2

79 section 17.4 Solubility Equilibria 749 Solve The expression for K sp is K sp = 3Ca 2 43F Appendix D gives 3.9 * for this K sp. Practice Exercise 1 Which of these expressions correctly expresses the solubility-product constant for Ag 3 PO 4 in water? (a) 3Ag43PO 4 4, (b) 3Ag 43PO 4 3-4, (c) 3Ag 4 3 3PO 4 3-4, (d) 3Ag 43PO , (e) 3Ag 4 3 3PO Practice Exercise 2 Give the solubility-product-constant expressions and K sp values (from Appendix D) for (a) barium carbonate and (b) silver sulfate. Solubility and K sp It is important to distinguish carefully between solubility and the solubility-product constant. The solubility of a substance is the quantity that dissolves to form a saturated solution. (Section 13.2) Solubility is often expressed as grams of solute per liter of solution 1g>L2. Molar solubility is the number of moles of solute that dissolve in forming 1 L of saturated solution of the solute 1mol>L2. The solubility-product constant 1K sp 2 is the equilibrium constant for the equilibrium between an ionic solid and its saturated solution and is a unitless number. Thus, the magnitude of K sp is a measure of how much of the solid dissolves to form a saturated solution. Give It Some Thought Mass solubility of compound (g/l) Formula weight Molar solubility of compound (mol/l) Without doing a calculation, predict which of these compounds has the greatest molar solubility in water: AgCl 1K sp = 1.8 * , AgBr 1K sp = 5.0 * , or AgI 1K sp = 8.3 * Empirical formula K sp Molar concentration of ions Solubility equilibrium Figure Procedure for converting between solubility and K sp. Starting from the mass solubility, follow the green arrows to determine K sp. Starting from K sp, follow the red arrows to determine either molar solubility or mass solubility. The solubility of a substance can change considerably in response to a number of factors. For example, the solubilities of hydroxide salts, like Mg1O2 2, are dependent upon the p of the solution. The solubility is also affected by concentrations of other ions in solution, especially common ions. In other words, the numeric value of the solubility of a given solute does change as the other species in solution change. In contrast, the solubility-product constant, K sp, has only one value for a given solute at any specific temperature.* Figure summarizes the relationships among various expressions of solubility and K sp. Sample Exercise Calculating K sp from Solubility Solid silver chromate is added to pure water at 25 C, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag 2 CrO 4 1s2 and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3 * 10-4 M. Assuming that the Ag 2 CrO 4 solution is saturated and that there are no other important equilibria involving Ag or CrO ions in the solution, calculate K sp for this compound. Solution Analyze We are given the equilibrium concentration of Ag in a saturated solution of Ag 2 CrO 4 and asked to determine the value of K sp for Ag 2 CrO 4. Plan The equilibrium equation and the expression for K sp are Ag 2 CrO 4 1s2 2 Ag 2 1aq2 CrO - 4 1aq2 K sp = 3Ag CrO To calculate K sp, we need the equilibrium concentrations of Ag and CrO We know that at equilibrium 3Ag 4 = 1.3 * 10-4 M. All the Ag and CrO ions in the solution come from the Ag 2 CrO 4 that dissolves. Thus, we can use 3Ag 4 to calculate 3CrO *This is strictly true only for very dilute solutions, for K sp values change somewhat when the concentration of ionic substances in water is increased. owever, we will ignore these effects, which are taken into consideration only for work that requires exceptional accuracy.

80 750 chapter 17 Additional Aspects of Aqueous Equilibria Solve From the chemical formula of silver chromate, we know that there must be two Ag ions in solution for each CrO ion in solution. Consequently, the concentration of CrO is half the concentration of Ag : 2 3CrO = a 1.3 * 10-4 mol Ag b a 1 mol CrO L 2 mol Ag b = 6.5 * 10-5 M and K sp is K sp = 3Ag CrO = 11.3 * * = 1.1 * Check We obtain a small value, as expected for a slightly soluble salt. Furthermore, the calculated value agrees well with the one given in Appendix D, 1.2 * Practice Exercise 1 You add 10.0 grams of solid copper(ii) phosphate, Cu 3 1PO 4 2 2, to a beaker and then add ml of water to the beaker at T = 298 K. The solid does not appear to dissolve. You wait a long time, with occasional stirring and eventually measure the equilibrium concentration of Cu 2 1aq2 in the water to be 5.01 * 10-8 M. What is the K sp of copper(ii) phosphate? (a) 5.01 * 10-8 (b) 2.50 * (c) 4.20 * (d) 3.16 * (e) 1.40 * Practice Exercise 2 A saturated solution of Mg1O2 2 in contact with undissolved Mg1O2 2 1s2 is prepared at 25 C. The p of the solution is found to be Assuming that there are no other simultaneous equilibria involving the Mg 2 or O - ions, calculate K sp for this compound. Sample Exercise Calculating Solubility from K sp In principle, it is possible to use the K sp value of a salt to calculate solubility under a variety of conditions. In practice, great care must be taken in doing so for the reasons indicated in A Closer Look: Limitations of Solubility Products at the end of this section. Agreement between the measured solubility and that calculated from K sp is usually best for salts whose ions have low charges (1 and 1-) and do not react with water. The K sp for CaF 2 is 3.9 * at 25 C. Assuming equilibrium is established between solid and dissolved CaF 2, and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF 2 in grams per liter. Solution Analyze We are given K sp for CaF 2 and asked to determine solubility. Recall that the solubility of a substance is the quantity that can dissolve in solvent, whereas the solubility-product constant, K sp, is an equilibrium constant. Plan To go from K sp to solubility, we follow the steps indicated by the red arrows in Figure We first write the chemical equation for the dissolution and set up a table of initial and equilibrium concentrations. We then use the equilibrium-constant expression. In this case we know K sp, and so we solve for the concentrations of the ions in solution. Once we know these concentrations, we use the formula weight to determine solubility in g>l. Solve Assume that initially no salt has dissolved, and then allow x mol/l of CaF 2 to dissociate completely when equilibrium is achieved: The stoichiometry of the equilibrium dictates that 2x mol>l of F - are produced for each x mol>l of CaF 2 that dissolve. We now use the expression for K sp and substitute the equilibrium concentrations to solve for the value of x: CaF 2 1s2 Ca 2 1aq2 2 F - 1aq2 Initial concentration (M) 0 0 Change (M) x 2x Equilibrium concentration (M) x 2x K sp = 3Ca 2 43F = 1x212x2 2 = 4x 3 = 3.9 * (Remember that 2y = y 1>3.) Thus, the molar solubility of CaF 2 is 2.1 * 10-4 mol>l. x = A * = 2.1 * The mass of CaF 2 that dissolves in water to form 1 L of solution is a 2.1 * 10-4 mol CaF 2 b a 78.1 g CaF 2 b = 1.6 * 10-2 g CaF 1 L soln 1 mol CaF 2 >L soln 2 Check We expect a small number for the solubility of a slightly soluble salt. If we reverse the calculation, we should be able to recalculate the solubility product: K sp = 12.1 * * = 3.7 * 10-11, close to the value given in the problem statement, 3.9 * Comment Because F - is the anion of a weak acid, you might expect hydrolysis of the ion to affect the solubility of CaF 2. The basicity of F - is so small 1K b = 1.5 * , however, that the hydrolysis occurs to only a slight extent and does not significantly influence the solubility. The reported solubility is g/l at 25 C, in good agreement with our calculation.

81 section 17.5 Factors That Affect Solubility 751 Practice Exercise 1 Of the five salts listed below, which has the highest concentration of its cation in water? Assume that all salt solutions are saturated and that the ions do not undergo any additional reactions in water. (a) lead (II) chromate, K sp = 2.8 * (b) cobalt(ii) hydroxide, K sp = 1.3 * (c) cobalt(ii) sulfide, K sp = 5 * (d) chromium(iii) hydroxide, K sp = 1.6 * (e) silver sulfide, K sp = 6 * Practice Exercise 2 The K sp for LaF 3 is 2 * What is the solubility of LaF 3 in water in moles per liter? A Closer Look Limitations of Solubility Products Ion concentrations calculated from K sp values sometimes deviate appreciably from those found experimentally. In part, these deviations are due to electrostatic interactions between ions in solution, which can lead to ion pairs. (Section 13.5, The van t off Factor ) These interactions increase in magnitude both as the concentrations of the ions increase and as their charges increase. The solubility calculated from K sp tends to be low unless corrected to account for these interactions. As an example of the effect of these interactions, consider CaCO 3 (calcite), whose solubility product, 4.5 * 10-9, gives a calculated solubility of 6.7 * 10-5 mol>l; correcting for ionic interactions in the solution yields 7.3 * 10-5 mol>l. The reported solubility, however, is 1.4 * 10-4 mol>l, indicating that there must be additional factors involved. Another common source of error in calculating ion concentrations from K sp is ignoring other equilibria that occur simultaneously in the solution. It is possible, for example, that acid base equilibria take place simultaneously with solubility equilibria. In particular, both basic anions and cations with high charge-to-size ratios undergo hydrolysis reactions that can measurably increase the solubilities of their salts. For example, CaCO 3 contains the basic carbonate ion 1K b = 1.8 * , which reacts with water: CO 3 2-1aq2 2 O1l2 CO 3-1aq2 O - 1aq2 If we consider the effect of ion ion interactions as well as simultaneous solubility and K b equilibria, we calculate a solubility of 1.4 * 10-4 mol>l, in agreement with the measured value for calcite. Finally, we generally assume that ionic compounds dissociate completely when they dissolve, but this assumption is not always valid. When MgF 2 dissolves, for example, it yields not only Mg 2 and F - ions but also MgF ions Factors That Affect Solubility Solubility is affected by temperature and by the presence of other solutes. The presence of an acid, for example, can have a major influence on the solubility of a substance. In Section 17.4, we considered the dissolving of ionic compounds in pure water. In this section, we examine three factors that affect the solubility of ionic compounds: (1) presence of common ions, (2) solution p, and (3) presence of complexing agents. We will also examine the phenomenon of amphoterism, which is related to the effects of both p and complexing agents. Common-Ion Effect The presence of either Ca 2 1aq2 or F - 1aq2 in a solution reduces the solubility of CaF 2, shifting the equilibrium concentrations to the left: CaF 2 1s2 Ca 2 1aq2 2 F - 1aq2 Addition of Ca 2 or F shifts equilibrium concentrations, reducing solubility This reduction in solubility is another manifestation of the common-ion effect we looked at in Section In general, the solubility of a slightly soluble salt is decreased by the presence of a second solute that furnishes a common ion, as Figure shows for CaF 2. Molar solubility of CaF 2 (mol/l) Solubility of CaF 2 decreases sharply as a common ion (F ) is added to the solution Pure water Concentration of NaF (mol/l) Figure Common-ion effect. Notice that the CaF 2 solubility is on a logarithmic scale.

82 752 chapter 17 Additional Aspects of Aqueous Equilibria Sample Exercise Calculating the Effect of a Common Ion on Solubility Calculate the molar solubility of CaF 2 at 25 C in a solution that is (a) M in Ca1NO and (b) M in NaF. Solution Analyze We are asked to determine the solubility of CaF 2 in the presence of two strong electrolytes, each containing an ion common to CaF 2. In (a) the common ion is Ca 2, and NO 3 - is a spectator ion. In (b) the common ion is F -, and Na is a spectator ion. Plan Because the slightly soluble compound is CaF 2, we need to use K sp for this compound, which Appendix D gives as 3.9 * The value of K sp is unchanged by the presence of additional solutes. Because of the common-ion effect, however, the solubility of the salt decreases in the presence of common ions. We use our standard equilibrium techniques of starting with the equation for CaF 2 dissolution, setting up a table of initial and equilibrium concentrations, and using the K sp expression to determine the concentration of the ion that comes only from CaF 2. Solve (a) The initial concentration of Ca 2 is M because of the dissolved Ca1NO : Substituting into the solubility-product expression gives CaF 2 1s2 Ca 2 1aq2 2 F - 1aq2 Initial Concentration (M) Change (M) x 2x Equilibrium Concentration (M) x2 2x K sp = 3.9 * = 3Ca 2 43F = x212x2 2 If we assume that x is small compared to 0.010, we have 3.9 * = x2 2 x 2 = 3.9 * = 9.8 * x = 29.8 * = 3.1 * 10-5 M This very small value for x validates the simplifying assumption we made. Our calculation indicates that 3.1 * 10-5 mol of solid CaF 2 dissolves per liter of M Ca1NO solution. (b) The common ion is F -, and at equilibrium we have 3Ca 2 4 = x and 3F - 4 = x Assuming that 2x is much smaller than M (that is, x 0.010), we have 3.9 * = 1x x2 2 x x = 3.9 * = 3.9 * 10-7 M Thus, 3.9 * 10-7 mol of solid CaF 2 should dissolve per liter of M NaF solution. Comment The molar solubility of CaF 2 in water is 2.1 * 10-4 M (Sample Exercise 17.12). By comparison, our calculations here give a CaF 2 solubility of 3.1 * 10-5 M in the presence of M Ca 2 and 3.9 * 10-7 M in the presence of M F - ion. Thus, the addition of either Ca 2 or F - to a solution of CaF 2 decreases the solubility. owever, the effect of F - on the solubility is more pronounced than that of Ca 2 because 3F - 4 appears to the second power in the K sp expression for CaF 2, whereas 3Ca 2 4 appears to the first power. Practice Exercise 1 Consider a saturated solution of the salt MA 3, in which M is a metal cation with a 3 charge and A is an anion with a 1- charge, in water at 298 K. Which of the following will affect the K sp of MA 3 in water? (a) The addition of more M 3 to the solution. (b) The addition of more A - to the solution. (c) Diluting the solution. (d) Raising the temperature of the solution. (e) More than one of the above factors. Practice Exercise 2 For manganese(ii) hydroxide, Mn1O2 2, K sp = 1.6 * Calculate the molar solubility of Mn1O2 2 in a solution that contains M NaO.

83 Solubility and p The p of a solution affects the solubility of any substance whose anion is basic. Consider Mg1O2 2, for which the solubility equilibrium is Mg1O2 2 1s2 Mg 2 1aq2 2 O - 1aq2 K sp = 1.8 * [17.17] A saturated solution of Mg1O2 2 has a calculated p of and its Mg 2 concentration is 1.7 * 10-4 M. Now suppose that solid Mg1O2 2 is equilibrated with a solution buffered at p 9.0. The po, therefore, is 5.0, so 3O - 4 = 1.0 * Inserting this value for 3O - 4 into the solubility-product expression, we have K sp = 3Mg 2 43O = 1.8 * Mg * = 1.8 * Mg 2 4 = 1.8 * * = 0.18 M Thus, the Mg1O2 2 dissolves until 3Mg 2 4 = 0.18 M. It is apparent that Mg1O2 2 is much more soluble in this solution. section 17.5 Factors That Affect Solubility 753 Chemistry and Life Ocean Acidification Seawater is a weakly basic solution, with p values typically between 8.0 and 8.3. This p range is maintained through a carbonic acid buffer system similar to the one in blood (see Equation 17.10). Because the p of seawater is higher than that of blood ( ), the second dissociation of carbonic acid cannot be neglected and CO becomes an important aqueous species. The availability of carbonate ions plays an important role in shell formation for a number of marine organisms, including stony corals ( Figure 17.18). These organisms, which are referred to as marine calcifiers and play an important role in the food chains of nearly all oceanic ecosystems, depend on dissolved Ca 2 and CO ions to form their shells and exoskeletons. The relatively low solubilityproduct constant of CaCO 3, CaCO 3 1s2 Ca 2 1aq2 CO 3 2-1aq2 K sp = 4.5 * 10-9 and the fact that the ocean contains saturated concentrations of Ca 2 and CO mean that CaCO 3 is usually quite stable once formed. In fact, calcium carbonate skeletons of creatures that died millions of years ago are not uncommon in the fossil record. Just as in our bodies, the carbonic acid buffer system can be perturbed by removing or adding CO 2 1g2. The concentration of dissolved CO 2 in the ocean is sensitive to changes in atmospheric CO 2 levels. As discussed in Chapter 18, the atmospheric CO 2 concentration has risen by approximately 30% over the past three centuries to the present level of 400 ppm. uman activity has played a prominent role in this increase. Scientists estimate that one-third to one-half of the CO 2 emissions resulting from human activity have been absorbed by Earth s oceans. While this absorption helps mitigate the greenhouse gas effects of CO 2, the extra CO 2 in the ocean produces carbonic acid, which lowers the p. Because CO is the conjugate base of the weak acid CO 3 -, the carbonate ion readily combines with the hydrogen ion: CO 3 2-1aq2 1aq2 CO 3-1aq2 This consumption of carbonate ion shifts the CaCO 3 dissolution equilibrium to the right, increasing the solubility of CaCO 3, which can lead to partial dissolution of calcium carbonate shells and exoskeletons. If the amount of atmospheric CO 2 continues to increase at the present rate, scientists estimate that seawater p will fall to 7.9 sometime over the next 50 years. While this change might sound small, it has dramatic ramifications for oceanic ecosystems. Related Exercises: Figure Marine calcifiers. Many sea-dwelling organisms use CaCO 3 for their shells and exoskeletons. Examples include stony coral, crustaceans, some phytoplankton, and echinoderms, such as sea urchins and starfish.

84 754 chapter 17 Additional Aspects of Aqueous Equilibria If 3O - 4 were reduced further by making the solution even more acidic, the Mg 2 concentration would have to increase to maintain the equilibrium condition. Thus, a sample of Mg1O2 2 1s2 dissolves completely if sufficient acid is added, as we saw in Figure 4.9 (page 137). The solubility of almost any ionic compound is affected if the solution is made sufficiently acidic or basic. The effects are noticeable, however, only when one (or both) ions in the compound are at least moderately acidic or basic. The metal hydroxides, such as Mg1O2 2, are examples of compounds containing a strongly basic ion, the hydroxide ion. In general, the solubility of a compound containing a basic anion (that is, the anion of a weak acid) increases as the solution becomes more acidic. As we have seen, the solubility of Mg1O2 2 greatly increases as the acidity of the solution increases. The solubility of PbF 2 increases as the solution becomes more acidic, too, because F - is a base (it is the conjugate base of the weak acid F). As a result, the solubility equilibrium of PbF 2 is shifted to the right as the concentration of F - is reduced by protonation to form F. Thus, the solution process can be understood in terms of two consecutive reactions: PbF 2 1s2 Pb 2 1aq2 2 F - 1aq2 [17.18] F - 1aq2 1aq2 F1aq2 [17.19] The equation for the overall process is PbF 2 1s2 2 1aq2 Pb 2 1aq2 2 F1aq2 [17.20] The processes responsible for the increase in solubility of PbF 2 in acidic solution are illustrated in Figure 17.19(a). Other salts that contain basic anions, such as CO 3 2 -, PO 4 3 -, CN -, or S 2 -, behave similarly. These examples illustrate a general rule: The solubility of slightly soluble salts containing basic anions increases as 3 4 increases (as p is lowered). The more basic the anion, the more the solubility is influenced by p. The solubility of salts with anions of negligible basicity (the anions of strong acids), such as Cl -, Br -, I -, and NO 3 -, is unaffected by p changes, as shown in Figure 17.19(b). Salt whose anion is conjugate base of weak acid: Solubility increases as p decreases Salt whose anion is conjugate base of strong acid: Solubility unaffected by changes in p F Add Add I F F I Pb 2 F F Pb 2 Pb 2 No reaction Pb 2 [F ] decreases [Pb 2 ] increases PbF 2 PbF 2 PbI 2 PbI 2 (a) (b) Figure Response of two ionic compounds to addition of a strong acid. (a) The solubility of PbF 2 increases upon addition of acid. (b) The solubility of Pbl 2 is not affected by addition of acid. The water molecules and the anion of the strong acid have been omitted for clarity.

85 section 17.5 Factors That Affect Solubility 755 Sample Exercise Predicting the Effect of Acid on Solubility Which of these substances are more soluble in acidic solution than in basic solution: (a) Ni1O2 2 1s2, (b) CaCO 3 1s2, (c) BaF 2 1s2, (d) AgCl(s)? Solution Analyze The problem lists four sparingly soluble salts, and we are asked to determine which are more soluble at low p than at high p. Plan We will identify ionic compounds that dissociate to produce a basic anion, as these are especially soluble in acid solution. Solve (a) Ni1O2 2 1s2 is more soluble in acidic solution because of the basicity of O - ; the reacts with the O - ion, forming water: Ni1O2 2 1s2 Ni 2 1aq2 2 O - 1aq2 2 O - 1aq2 2 1aq2 2 2 O1l2 Overall: Ni1O2 2 1s2 2 1aq2 Ni 2 1aq2 2 2 O1l2 (b) Similarly, CaCO 3 1s2 dissolves in acid solutions because CO is a basic anion: Overall: CaCO 3 1s2 Ca 2 1aq2 CO 3 2-1aq2 CO 3 2-1aq2 2 1aq2 2 CO 3 1aq2 2 CO 3 1aq2 CO 2 1g2 2 O1l2 CaCO 3 1s2 2 1aq2 Ca 2 1aq2 CO 2 1g2 2 O1l2 2 - The reaction between CO 3 and - occurs in steps, with CO 3 forming first and 2 CO 3 forming in appreciable amounts only when 3 4 is sufficiently high. (c) The solubility of BaF 2 is enhanced by lowering the p because F - is a basic anion: Overall: BaF 2 1s2 Ba 2 1aq2 2 F - 1aq2 2 F - 1aq2 2 1aq2 2 F1aq2 BaF 2 1s2 2 1aq2 Ba 2 1aq2 2 F1aq2 (d) The solubility of AgCl is unaffected by changes in p because Cl - is the anion of a strong acid and therefore has negligible basicity. Practice Exercise 1 Which of the following actions will increase the solubility of AgBr in water? (a) increasing the p, (b) decreasing the p, (c) adding NaBr, (d) adding NaNO 3, (e) none of the above. Practice Exercise 2 Write the net ionic equation for the reaction between a strong acid and (a) CuS, (b) Cu1N Chemistry and Life Tooth Decay and Fluoridation Tooth enamel consists mainly of the mineral hydroxyapatite, Ca 10 1PO O2 2, the hardest substance in the body. Tooth cavities form when acids dissolve tooth enamel: Ca 10 1PO O2 2 1s2 8 1aq2 10 Ca 2 2 1aq2 6 PO - 4 1aq2 2 2 O1l2 The Ca 2 and PO ions diffuse out of the enamel and are washed away by saliva. The acids that attack the hydroxyapatite are formed by the action of bacteria on sugars and other carbohydrates present in the plaque adhering to the teeth. Fluoride ion, which is added to municipal water systems and toothpastes, can react with hydroxyapatite to form fluoroapatite, Ca 10 1PO F 2. This mineral, in which F - has replaced O -, is much more resistant to attack by acids because the fluoride ion is a much weaker Brønsted Lowry base than the hydroxide ion. The usual concentration of F - in municipal water systems is 1 mg>l 11 ppm2. The compound added may be NaF or Na 2 SiF 6. The silicon-fluorine anion reacts with water to release fluoride ions: SiF 6 2-1aq2 2 2 O1l2 6 F - 1aq2 4 1aq2 SiO 2 1s2 About 80% of all toothpastes now sold in the United States contain fluoride compounds, usually at the level of 0.1% fluoride by mass. The most common compounds in toothpastes are sodium fluoride (NaF), sodium monofluorophosphate 1Na 2 PO 3 F2, and stannous fluoride 1SnF 2 2. Related Exercises: ,

86 756 chapter 17 Additional Aspects of Aqueous Equilibria Formation of Complex Ions A characteristic property of metal ions is their ability to act as Lewis acids toward water molecules, which act as Lewis bases. (Section 16.11) Lewis bases other than water can also interact with metal ions, particularly transition-metal ions. Such interactions can dramatically affect the solubility of a metal salt. For example, AgCl 1K sp = 1.8 * dissolves in the presence of aqueous ammonia because Ag interacts with the Lewis base N 3, as shown in Figure This process can be viewed as the sum of two reactions: AgCl1s2 Ag 1aq2 Cl - 1aq2 [17.21] Ag 1aq2 2 N 3 1aq2 Ag1N aq2 [17.22] Overall: AgCl1s2 2 N 3 1aq2 Ag1N aq2 Cl - 1aq2 [17.23] The presence of N 3 drives the reaction, the dissolution of AgCl, to the right as Ag 1aq2 is consumed to form Ag1N 3 2 2, which is a very soluble species. N 3 reacts with Ag, forming Ag(N 3 ) 2 Addition of sufficient N 3 leads to complete dissolution of AgCl Ag Cl Cl AgCl Reaction with N 3 reduces concentration of free Ag and increases solubility of AgCl Ag(N 3 ) 2 N 3 AgCl(s) 2 N 3 (aq) Ag(N 3 ) 2 (aq) Cl (aq) Figure Concentrated N 3 1aq2 dissolves AgCl(s), which otherwise has very low solubility in water. For a Lewis base such as N 3 to increase the solubility of a metal salt, the base must be able to interact more strongly with the metal ion than water does. In other words, the N 3 must displace solvating 2 O molecules (Sections 13.1 and 16.11) in order to form 3Ag1N : Ag 1aq2 2 N 3 1aq2 Ag1N aq2 [17.24]

87 section 17.5 Factors That Affect Solubility 757 Table 17.1 Formation Constants for Some Metal Complex Ions in Water at 25 C Complex Ion K f Equilibrium Equation Ag1N * 10 7 Ag 1aq2 2 N 3 1aq2 Ag1N aq2 - Ag1CN2 2 1 * Ag 1aq2 2 CN - - 1aq2 Ag1CN2 2 1aq2 Ag1S 2 O * Ag 1aq2 2 S 2 O 3 2-1aq2 Ag1S 2 O aq2 2 - CdBr 4 5 * 10 3 Cd 2 1aq2 4 Br - 2-1aq2 CdBr 4 1aq2 - Cr1O2 4 8 * Cr 3 1aq2 4 O - - 1aq2 Cr1O2 4 1aq2 Co1SCN * 10 3 Co 2 1aq2 4 SCN - 1aq2 Co1SCN aq2 Cu1N * Cu 2 1aq2 4 N 3 1aq2 Cu1N aq2 Cu1CN * Cu 2 1aq2 4 CN - 1aq2 Cu1CN aq2 2 Ni1N * 10 9 Ni 2 2 1aq2 6 N 3 1aq2 Ni1N aq2 4 - Fe1CN2 6 1 * Fe 2 1aq2 6 CN - 4 1aq2 Fe1CN aq2 Fe1CN * Fe 3 1aq2 6 CN - 1aq2 Fe1CN aq2 An assembly of a metal ion and the Lewis bases bonded to it, such as Ag1N 3 2 2, is called a complex ion. Complex ions are very soluble in water. The stability of a complex ion in aqueous solution can be judged by the size of the equilibrium constant for its formation from the hydrated metal ion. For example, the equilibrium constant for Equation is K f = 3Ag1N Ag 43N = 1.7 * 107 [17.25] Note that the equilibrium constant for this kind of reaction is called a formation constant, K f. The formation constants for several complex ions are listed in Table Sample Exercise Evaluating an Equilibrium Involving a Complex Ion Calculate the concentration of Ag present in solution at equilibrium when concentrated ammonia is added to a M solution of AgNO 3 to give an equilibrium concentration of 3N 3 4 = 0.20 M. Neglect the small volume change that occurs when N 3 is added. Solution Analyze Addition of N 3 1aq2 to Ag 1aq2 forms Ag1N 3 2 2, as shown in Equation We are asked to determine what concentration of Ag 1aq2 remains uncombined when the N 3 concentration is brought to 0.20 M in a solution originally M in AgNO 3. Plan We assume that the AgNO 3 is completely dissociated, giving M Ag. Because K f for the formation of Ag1N is quite large, we assume that essentially all the Ag is converted to Ag1N and approach the problem as though we are concerned with the dissociation of Ag1N rather than its formation. To facilitate this approach, we need to reverse Equation and make the corresponding change to the equilibrium constant: Ag1N aq2 Ag 1aq2 2 N 3 1aq2 1 1 = 7 = 5.9 * 10-8 K f 1.7 * 10 Solve If 3Ag 4 is M initially, 3Ag1N will be M following addition of the N 3. We construct a table to solve this equilibrium problem. Note that the N 3 concentration given in the problem is an equilibrium concentration rather than an initial concentration. Ag1N aq2 Ag 1aq2 2 N 3 1aq2 Initial (M) Change (M) -x x Equilibrium (M) (0.010 x) x 0.20

88 758 chapter 17 Additional Aspects of Aqueous Equilibria Because 3Ag 4 is very small, we can assume x is small compared to Substituting these values into the equilibrium-constant expression for the dissociation of Ag1N 3 2 2, we obtain 3Ag 43N Ag1N = 1x = 5.9 * x = 1.5 * 10-8 M = 3Ag 4 Formation of the Ag1N complex drastically reduces the concentration of free Ag ion in solution. Practice Exercise 1 You have an aqueous solution of chromium(iii) nitrate that you titrate with an aqueous solution of sodium hydroxide. After a certain amount of titrant has been added, you observe a precipitate forming. You add more sodium hydroxide solution and the precipitate dissolves, leaving a solution again. What has happened? (a) The precipitate was sodium hydroxide, which redissolved in the larger volume. (b) The precipitate was chromium hydroxide, which dissolved once more solution was added, forming Cr 3 1aq2. (c) The precipitate was chromium hydroxide, which then reacted with more hydroxide to produce a soluble complex ion, Cr1O (d) The precipitate was sodium nitrate, which reacted with more nitrate to produce the soluble complex ion Na1NO Practice Exercise 2 Calculate 3Cr 3 4 in equilibrium with Cr1O2 4 - when mol of Cr1NO is dissolved in 1 L of solution buffered at p The general rule is that the solubility of metal salts increases in the presence of suitable Lewis bases, such as N 3, CN -, or O -, provided the metal forms a complex with the base. The ability of metal ions to form complexes is an extremely important aspect of their chemistry. Amphoterism Some metal oxides and hydroxides that are relatively insoluble in water dissolve in strongly acidic and strongly basic solutions. These substances, called amphoteric oxides and amphoteric hydroxides,* are soluble in strong acids and bases because they themselves are capable of behaving as either an acid or base. Examples of amphoteric substances include the oxides and hydroxides of Al 3, Cr 3, Zn 2, and Sn 2. Like other metal oxides and hydroxides, amphoteric species dissolve in acidic solutions because their anions, O 2 - or O -, react with acids. What makes amphoteric oxides and hydroxides special, though, is that they also dissolve in strongly basic solutions. This behavior results from the formation of complex anions containing several (typically four) hydroxides bound to the metal ion ( Figure 17.21): Al1O2 3 1s2 O - 1aq2 Al1O2-4 1aq2 [17.26] Al(O) 3 (s) O (aq) Al(O) 4 (aq) Al(O) 3 dissolves in strongly acidic solutions due to an acid base reaction Add O Add Al(O) 3 dissolves in strongly basic solutions due to complex ion formation Al 3 (aq) 3 2 O(l) 3 (aq) Al(O) 3 (s) Figure Amphoterism. Some metal oxides and hydroxides, such as Al1O2 3, are amphoteric, which means they dissolve in both strongly acidic and strongly basic solutions. *Notice that the term amphoteric is applied to the behavior of insoluble oxides and hydroxides that dissolve in acidic or basic solutions. The similar term amphiprotic (Section 16.2) relates more generally to any molecule or ion that can either gain or lose a proton.

89 section 17.6 Precipitation and Separation of Ions 759 The extent to which an insoluble metal hydroxide reacts with either acid or base varies with the particular metal ion involved. Many metal hydroxides such as Ca1O2 2, Fe1O2 2, and Fe1O2 3 :are capable of dissolving in acidic solution but do not react with excess base. These hydroxides are not amphoteric. The purification of aluminum ore in the manufacture of aluminum metal provides an interesting application of amphoterism. As we have seen, Al1O2 3 is amphoteric, whereas Fe1O2 3 is not. Aluminum occurs in large quantities as the ore bauxite, which is essentially hydrated Al 2 O 3 contaminated with Fe 2 O 3. When bauxite is added to a strongly basic solution, the Al 2 O 3 dissolves because the aluminum forms complex ions, such as Al1O The Fe 2 O 3 impurity, however, is not amphoteric and remains as a solid. The solution is filtered, getting rid of the iron impurity. Aluminum hydroxide is then precipitated by addition of acid. The purified hydroxide receives further treatment and eventually yields aluminum metal. Give It Some Thought What is the difference between an amphoteric substance and an amphiprotic substance? 17.6 precipitation and Separation of Ions Equilibrium can be achieved starting with the substances on either side of a chemical equation. For example, the equilibrium that exists between BaSO 4 1s2, Ba 2 1aq2, and 2 SO - 4 1aq2 (Equation 17.15), can be achieved either by starting with BaSO 4 1s2 or by starting with solutions containing Ba 2 2 and SO - 4. If we mix, say, a BaCl 2 aqueous solution with a Na 2 SO 4 aqueous solution, BaSO 4 might precipitate out. ow can we predict whether a precipitate will form under various conditions? Recall that we used the reaction quotient Q in Section 15.6 to determine the direction in which a reaction must proceed to reach equilibrium. The form of Q is the same as the equilibrium-constant expression for a reaction, but instead of only equilibrium concentrations, you can use whatever concentrations are being considered. The direction in which a reaction proceeds to reach equilibrium depends on the relationship between Q and K for the reaction. If Q 6 K, the product concentrations are too low and reactant concentrations are too high relative to the equilibrium concentrations, and so the reaction will proceed to the right (toward products) to achieve equilibrium. If Q 7 K, product concentrations are too high and reactant concentrations are too low, and so the reaction will proceed to the left to achieve equilibrium. If Q = K, the reaction is at equilibrium. For solubility-product equilibria, the relationship between Q and K sp is exactly like that for other equilibria. For K sp reactions, products are always the soluble ions, and the reactant is always the solid. Therefore, for solubility equilibria, If Q = K sp, the system is at equilibrium, which means the solution is saturated; this is the highest concentration the solution can have without precipitating. If Q 6 K sp, the reaction will proceed to the right, towards the soluble ions; no precipitate will form. If Q 7 K sp, the reaction will proceed to the left, towards the solid; precipitate will form. For the case of the barium sulfate solution, then we would calculate Q = 3Ba SO - 4 4, and compare this quantity to the K sp for barium sulfate.

90 760 chapter 17 Additional Aspects of Aqueous Equilibria Sample Exercise Predicting Whether a Precipitate Forms Does a precipitate form when 0.10 L of 8.0 * 10-3 M Pb1NO is added to 0.40 L of 5.0 * 10-3 M Na 2 SO 4? Solution Analyze The problem asks us to determine whether a precipitate forms when two salt solutions are combined. Plan We should determine the concentrations of all ions just after the solutions are mixed and compare the value of Q with K sp for any potentially insoluble product. The possible metathesis products are PbSO 4 and NaNO 3. Like all sodium salts NaNO 3 is soluble, but PbSO 4 has a K sp of 6.3 * 10-7 (Appendix D) and will precipitate if the Pb 2 and SO concentrations are high enough for Q to exceed K sp. Solve When the two solutions are mixed, the volume is 0.10 L 0.40 L = 0.50 L. The number of moles of Pb 2 in 0.10 L of 8.0 * 10-3 M Pb1NO is L2a 8.0 * 10-3 mol b = 8.0 * 10-4 mol L The concentration of Pb 2 in the 0.50-L mixture is therefore 3Pb 2 4 = 8.0 * 10-4 mol 0.50 L = 1.6 * 10-3 M 2 - The number of moles of SO 4 in 0.40 L of 5.0 * 10-3 M Na 2 SO 4 is L2a 5.0 * 10-3 mol b = 2.0 * 10-3 mol L Therefore 3SO = 2.0 * 10-3 mol 0.50 L = 4.0 * 10-3 M and Q = 3Pb 2 43SO = 11.6 * * = 6.4 * 10-6 Because Q 7 K sp, PbSO 4 precipitates. Practice Exercise 1 An insoluble salt MA has a K sp of 1.0 * Two solutions, MNO 3 and NaA are mixed, to yield a final solution that is 1.0 * 10-8 M in M 1aq2 and 1.00 * 10-7 M in A - (aq). Will a precipitate form? (a) Yes. (b) No. Practice Exercise 2 Does a precipitate form when L of 2.0 * 10-2 M NaF is mixed with L of 1.0 * 10-2 M Ca1NO 3 2 2? Sample Exercise Selective Precipitation Selective Precipitation of Ions Ions can be separated from each other based on the solubilities of their salts. Consider a solution containing both Ag and Cu 2. If Cl is added to the solution, AgCl 1K sp = 1.8 * precipitates, while Cu 2 remains in solution because CuCl 2 is soluble. Separation of ions in an aqueous solution by using a reagent that forms a precipitate with one or more (but not all) of the ions is called selective precipitation. A solution contains 1.0 * 10-2 M Ag and 2.0 * 10-2 M Pb 2. When Cl - is added, both AgCl 1K sp = 1.8 * and PbCl 2 1K sp = 1.7 * can precipitate. What concentration of Cl - is necessary to begin the precipitation of each salt? Which salt precipitates first? Solution Analyze We are asked to determine the concentration of Cl - necessary to begin the precipitation from a solution containing Ag and Pb 2 ions, and to predict which metal chloride will begin to precipitate first. Plan We are given K sp values for the two precipitates. Using these and the metal ion concentrations, we can calculate what Cl - concentration is necessary to precipitate each salt. The salt requiring the lower Cl - ion concentration precipitates first.

91 section 17.6 Precipitation and Separation of Ions 761 Solve For AgCl we have K sp = 3Ag 43Cl - 4 = 1.8 * Because 3Ag 4 = 1.0 * 10-2 M, the greatest concentration of Cl - that can be present without causing precipitation of AgCl can be calculated from the K sp expression: K sp = 11.0 * Cl - 4 = 1.8 * Cl * = 1.0 * 10-2 = 1.8 * 10-8 M Any Cl - in excess of this very small concentration will cause AgCl to precipitate from solution. Proceeding similarly for PbCl 2, we have K sp = 3Pb 2 43Cl = 1.7 * * Cl = 1.7 * Cl = 1.7 * 10-5 = 8.5 * * 10 3Cl - 4 = 28.5 * 10-4 = 2.9 * 10-2 M Thus, a concentration of Cl - in excess of 2.9 * 10-2 M causes PbCl 2 to precipitate. Comparing the Cl - concentration required to precipitate each salt, we see that as Cl - is added, AgCl precipitates first because it requires a much smaller concentration of Cl -. Thus, Ag can be separated from Pb 2 by slowly adding Cl - so that the chloride ion concentration remains between 1.8 * 10-8 M and 2.9 * 10-2 M. Comment Precipitation of AgCl will keep the Cl - concentration low until the number of moles of Cl - added exceeds the number of moles of Ag in the solution. Once past this point, 3Cl - 4 rises sharply and PbCl 2 will soon begin to precipitate. Practice Exercise 1 Under what conditions does an ionic compound precipitate from a solution of the constituent ions? (a) always, (b) when Q = K sp, (c) when Q exceeds K sp, (d) when Q is less than K sp, (e) never, if it is very soluble. Practice Exercise 2 A solution consists of M Mg 2 and Cu 2. Which ion precipitates first as O - is added? What concentration of O - is necessary to begin the precipitation of each cation? 3K sp = 1.8 * for Mg1O2 2, and K sp = 4.8 * for Cu1O2 2.4 Sulfide ion is often used to separate metal ions because the solubilities of sulfide salts span a wide range and depend greatly on solution p. For example, Cu 2 and Zn 2 can be separated by bubbling 2 S gas through an acidified solution containing these two cations. Because CuS 1K sp = 6 * is less soluble than ZnS 1K sp = 2 * , CuS precipitates from an acidified solution p 1 while ZnS does not ( Figure 17.22): Cu 2 1aq2 2 S1aq2 CuS1s2 2 1aq2 [17.27] Go Figure What would happen if the p were raised to 8 first and then 2 S were added? Add 2 S Remove CuS and increase p Cu 2 p 1 p 1 p 8 Zn 2 2 S S Solution containing Zn 2 (aq) and Cu 2 (aq) CuS(s) When 2 S is added to a solution whose p exceeds 0.6, CuS precipitates ZnS(s) After CuS is removed, the p is increased, allowing ZnS to precipitate Figure Selective precipitation. In this example, Cu 2 ions are separated from Zn 2 ions.

92 762 chapter 17 Additional Aspects of Aqueous Equilibria The CuS can be separated from the Zn 2 solution by filtration. The separated CuS can then be dissolved by raising the concentration of even further, shifting the equilibrium concentrations of the compounds in Equation to the left Qualitative Analysis for Metallic Elements In this final section, we look at how solubility equilibria and complex-ion formation can be used to detect the presence of particular metal ions in solution. Before the development of modern analytical instrumentation, it was necessary to analyze mixtures of metals in a sample by what were called wet chemical methods. For example, an ore sample that might contain several metallic elements was dissolved in a concentrated acid solution that was then tested in a systematic way for the presence of various metal ions. Qualitative analysis determines only the presence or absence of a particular metal ion relative to some threshold, whereas quantitative analysis determines how much of a given substance is present. Even though wet methods of qualitative analysis have become less important in the chemical industry, they are frequently used in general chemistry laboratory programs to illustrate equilibria, to teach the properties of common metal ions in solution, and to develop laboratory skills. Typically, such analyses proceed in three stages: (1) The ions are separated into broad groups on the basis of solubility properties. (2) The ions in each group are separated by selectively dissolving members in the group. (3) The ions are identified by means of specific tests. A scheme in general use divides the common cations into five groups ( Figure 17.23). The order in which reagents are added is important in this scheme. The most selective separations those that involve the smallest number of ions are carried out first. The reactions used must proceed so far toward completion that any concentration of cations remaining in the solution is too small to interfere with subsequent tests. Let s look at each of these five groups of cations, briefly examining the logic used in this qualitative analysis scheme. Group 1. Insoluble chlorides: Of the common metal ions, only Ag, g 2 2, and Pb 2 form insoluble chlorides. When Cl is added to a mixture of cations, therefore, only AgCl, g 2 Cl 2, and PbCl 2 precipitate, leaving the other cations in solution. The absence of a precipitate indicates that the starting solution contains no Ag 2, g 2, or Pb 2. Group 2. Acid-insoluble sulfides: After any insoluble chlorides have been removed, the remaining solution, now acidic from Cl treatment, is treated with 2 S. Since 2 S is a weak acid compared to Cl, its role here is to act as a source for small amounts of sulfide. Only the most insoluble metal sulfides CuS, Bi 2 S 3, CdS, PbS, gs, As 2 S 3, Sb 2 S 3, and SnS 2 precipitate. (Note the very small values of K sp for some of these sulfides in Appendix D.) Those metal ions whose sulfides are somewhat more soluble for example, ZnS or NiS remain in solution. Group 3. Base-insoluble sulfides and hydroxides: After the solution is filtered to remove any acid-insoluble sulfides, it is made slightly basic, and 1N S is added. In basic solutions the concentration of S 2 - is higher than in acidic solutions. Under these conditions, the ion products for many of the more soluble sulfides exceed their K sp values and thus precipitation occurs. The metal ions precipitated at this stage are Al 3, Cr 3, Fe 3, Zn 2, Ni 2, Co 2, and Mn 2. (The Al 3, Fe 3, and Cr 3 ions do not form insoluble sulfides; instead they precipitate as insoluble hydroxides, as Figure shows.)

93 section 17.7 Qualitative Analysis for Metallic Elements 763 Go Figure If a solution contained a mixture of Cu 2 and Zn 2 ions, would this separation scheme work? After which step would the first precipitate be observed? Solution containing unknown metal cations Add 6 M Cl Precipitate Group 1 Insoluble chlorides: AgCl, PbCl 2, g 2 Cl 2 Decantate Remaining cations Add 2 S and 0.2 M Cl Precipitate Group 2 Acid-insoluble sulfides: CuS, CdS, Bi 2 S 3, PbS, gs, As 2 S 3, Sb 2 S 3, SnS 2 Decantate Remaining cations Add (N 4 ) 2 S at p = 8 Precipitate Group 3 Base-insoluble sulfides and hydroxides: Al(O) 3, Fe(O) 3, Cr(O) 3, ZnS, NiS, MnS, CoS Decantate Remaining cations Add (N 4 ) 2 PO 4 and N 3 Precipitate Decantate Group 4 Insoluble phosphates: Ca 3 (PO 4 ) 2, Sr 3 (PO 4 ) 2, Ba 3 (PO 4 ) 2, MgN 4 PO 4 Group 5 Alkali metal ions and N 4 Figure Qualitative analysis. A flowchart showing a common scheme for identifying cations. Group 4. Insoluble phosphates: At this point, the solution contains only metal ions from groups 1A and 2A of the periodic table. Adding 1N PO 4 to a basic solution precipitates the group 2A elements Mg 2, Ca 2, Sr 2, and Ba 2 because these metals form insoluble phosphates. Group 5. The alkali metal ions and N 4 : The ions that remain after removing the insoluble phosphates are tested for individually. A flame test can be used to determine the presence of K, for example, because the flame turns a characteristic violet color if K is present (Figure 7.22). Give It Some Thought If a precipitate forms when Cl is added to an aqueous solution, what conclusions can you draw about the contents of the solution?

94 764 chapter 17 Additional Aspects of Aqueous Equilibria Sample Integrative Exercise Putting Concepts Together A sample of 1.25 L of Cl gas at 21 C and atm is bubbled through L of M N 3 solution. Calculate the p of the resulting solution assuming that all the Cl dissolves and that the volume of the solution remains L. Solution The number of moles of Cl gas is calculated from the ideal-gas law, n = PV RT = atm L L@atm>mol@K21294 K2 = mol Cl The number of moles of N 3 in the solution is given by the product of the volume of the solution and its concentration, Moles N 3 = L mol N 3 >L2 = mol N 3 The acid Cl and base N 3 react, transferring a proton from Cl to N 3, producing N 4 and Cl - ions, Cl1g2 N 3 1aq2 N 4 1aq2 Cl - 1aq2 To determine the p of the solution, we first calculate the amount of each reactant and each product present at the completion of the reaction. Because you can assume this neutralization reaction proceeds as far toward the product side as possible, this is a limiting reactant problem. Cl1g2 N 3 1aq2 N 4 1aq2 Cl - 1aq2 Before reaction (mol) Change (limiting reactant) (mol) After reaction (mol) Thus, the reaction produces a solution containing a mixture of N 3, N 4, and Cl -. The N 3 is a weak base 1K b = 1.8 * , N 4 is its conjugate acid, and Cl - is neither acidic nor basic. Consequently, the p depends on 3N 3 4 and 3N 4 4, 3N 3 4 = mol N L soln = M 3N 4 4 = mol N L soln = M We can calculate the p using either K b for N 3 or K a for N 4. Using the K b expression, we have N 3 1aq2 2 O1l2 N 4 1aq2 O - 1aq2 Initial (M) Change (M) -x x x Equilibrium (M) x x2 x K b = 3N 4 43O - 4 3N 3 4 = x21x x2 _ x = 1.8 * 10-5 x = 3O - 4 = * = 9.4 * 10-6 M ence, po = -log19.4 * = 5.03 and p = po = = 8.97.

95 Learning Outcomes 765 Chapter Summary and Key Terms The Common Ion Effect (Section 17.1) In this chapter, we have considered several types of important equilibria that occur in aqueous solution. Our primary emphasis has been on acid base equilibria in solutions containing two or more solutes and on solubility equilibria. The dissociation of a weak acid or weak base is repressed by the presence of a strong electrolyte that provides an ion common to the equilibrium (the common-ion effect). Buffers (Section 17.2) A particularly important type of acid base mixture is that of a weak conjugate acid base pair that functions as a buffered solution (buffer). Addition of small amounts of a strong acid or a strong base to a buffered solution causes only small changes in p because the buffer reacts with the added acid or base. (Strong acid strong base, strong acid weak base, and weak acid strong base reactions proceed essentially to completion.) Buffered solutions are usually prepared from a weak acid and a salt of that acid or from a weak base and a salt of that base. Two important characteristics of a buffered solution are its buffer capacity and its p range. The optimal p of a buffer is equal to pk a (or pk b ) of the acid (or base) used to prepare the buffer. The relationship between p, pk a, and the concentrations of an acid and its conjugate base can be expressed by the enderson asselbalch equation. It is important to realize that the enderson asselbalch equation is an approximation, and more detailed calculations may need to be performed to obtain equilibrium concentrations. Acid Base Titrations (Section 17.3) The plot of the p of an acid (or base) as a function of the volume of added base (or acid) is called a p titration curve. The titration curve of a strong acid strong base titration exhibits a large change in p in the immediate vicinity of the equivalence point; at the equivalence point for such a titration p = 7. For strong acid weak base or weak acid strong base titrations, the p change in the vicinity of the equivalence point is not as large as the strong acid strong base titration, nor will the p equal 7 at the equivalence point in these cases. Instead, what determines the p at the equivalence point is the conjugate base or acid salt solution that results from the neutralization reaction. For this reason, it is important to choose an indicator whose color change is near the p at the equivalence point for titrations involving either weak acids or weak bases. It is possible to calculate the p at any point of the titration curve by first considering the effects of the acid base reaction on solution concentrations and then examining equilibria involving the remaining solute species. Solubility Equilibria (Section 17.4) The equilibrium between a solid compound and its ions in solution provides an example of heterogeneous equilibrium. The solubility-product constant (or simply the solubility product), K sp, is an equilibrium constant that expresses quantitatively the extent to which the compound dissolves. The K sp can be used to calculate the solubility of an ionic compound, and the solubility can be used to calculate K sp. Factors That Affect Solubility (Section 17.5) Several experimental factors, including temperature, affect the solubilities of ionic compounds in water. The solubility of a slightly soluble ionic compound is decreased by the presence of a second solute that furnishes a common ion (the common-ion effect). The solubility of compounds containing basic anions increases as the solution is made more acidic (as p decreases). Salts with anions of negligible basicity (the anions of strong acids) are unaffected by p changes. The solubility of metal salts is also affected by the presence of certain Lewis bases that react with metal ions to form stable complex ions. Complex-ion formation in aqueous solution involves the displacement by Lewis bases (such as N 3 and CN - ) of water molecules attached to the metal ion. The extent to which such complex formation occurs is expressed quantitatively by the formation constant for the complex ion. Amphoteric oxides and hydroxides are those that are only slightly soluble in water but dissolve on addition of either acid or base. Precipitation and Separation of Ions (Section 17.6) Comparison of the reaction quotient, Q, with the value of K sp can be used to judge whether a precipitate will form when solutions are mixed or whether a slightly soluble salt will dissolve under various conditions. Precipitates form when Q 7 K sp. If two salts have sufficiently different solubilities, selective precipitation can be used to precipitate one ion while leaving the other in solution, effectively separating the two ions. Qualitative Analysis for Metallic Elements (Section 17.7) Metallic elements vary a great deal in the solubilities of their salts, in their acid base behavior, and in their tendencies to form complex ions. These differences can be used to separate and detect the presence of metal ions in mixtures. Qualitative analysis determines the presence or absence of species in a sample, whereas quantitative analysis determines how much of each species is present. The qualitative analysis of metal ions in solution can be carried out by separating the ions into groups on the basis of precipitation reactions and then analyzing each group for individual metal ions. Learning Outcomes After studying this chapter, you should be able to: Describe the common-ion effect. (Section 17.1) Explain how a buffer functions. (Section 17.2) Calculate the p of a buffered solution. (Section 17.2) Calculate the p of a buffer after the addition of small amounts of a strong acid or a strong base. (Section 17.2) Calculate the appropriate quantities of compounds to make a buffer at a given p. (Section 17.2) Calculate the p at any point in a strong acid strong base titration. (Section 17.3) Calculate the p at any point in a weak acid strong base or weak base strong acid titration. (Section 17.3) Describe the differences between the titration curves for a strong acid strong base titration and those when either the acid or base is weak. (Section 17.3) Estimate the pk a for monoprotic or polyprotic acids from titration curves. (Section 17.3) Given either K sp, molar solubility or mass solubility for a substance, calculate the other two quantities. (Section 17.4)

96 766 chapter 17 Additional Aspects of Aqueous Equilibria Calculate molar solubility in the presence of a common ion. (Section 17.5) Predict the effect of p on solubility. (Section 17.5) Predict whether a precipitate will form when solutions are mixed, by comparing Q and K sp. (Section 17.6) Calculate the ion concentrations required to begin precipitation. (Section 17.6) Explain the effect of complex-ion formation on solubility. (Section 17.6) Explain the logic of identification of metal ions in aqueous solution by a series of reactions. (Section 17.7) Key Equations p = pk a log 3base4 3acid4 [17.9] The enderson asselbalch equation, used to estimate the p of a buffer from the concentrations of a conjugate acid base pair Exercises Visualizing Concepts 17.1 The following boxes represent aqueous solutions containing a weak acid, A and its conjugate base, A -. Water molecules, hydronium ions, and cations are not shown. Which solution has the highest p? Explain. [Section 17.1] three represents the buffer after the addition of a strong base? (c) Which of the three represents a situation that cannot arise from the addition of either an acid or a base? [Section 17.2] = A = A A A A A A A A A (1) (2) (3) 17.2 The beaker on the right contains 0.1 M acetic acid solution with methyl orange as an indicator. The beaker on the left contains a mixture of 0.1 M acetic acid and 0.1 M sodium acetate with methyl orange. (a) Using Figure 16.7, which solution has a higher p? (b) Which solution is better able to maintain its p when small amounts of NaO are added? Explain. [Sections 17.1 and 17.2] 17.5 The following figure represents solutions at various stages of the titration of a weak acid, A, with NaO. (The Na ions and water molecules have been omitted for clarity.) To which of the following regions of the titration curve does each drawing correspond: (a) before addition of NaO, (b) after addition of NaO but before the equivalence point, (c) at the equivalence point, (d) after the equivalence point? [Section 17.3] = A = A = O (i) (ii) (iii) (iv) 17.3 A buffer contains a weak acid, A, and its conjugate base. The weak acid has a pk a of 4.5, and the buffer has a p of 4.3. Without doing a calculation, state which of these possibilities are correct. (a) 3A4 = 3A - 4, (b) 3A4 7 3A - 4, or (c) 3A4 6 3A - 4. [Section 17.2] 17.4 The following diagram represents a buffer composed of equal concentrations of a weak acid, A, and its conjugate base, A -. The heights of the columns are proportional to the concentrations of the components of the buffer. (a) Which of the three drawings, (1), (2), or (3), represents the buffer after the addition of a strong acid? (b) Which of the p 17.6 Match the following descriptions of titration curves with the diagrams: (a) strong acid added to strong base, (b) strong base added to weak acid, (c) strong base added to strong acid, (d) strong base added to polyprotic acid. [Section 17.3] ml titrant (i) p ml titrant (ii) p ml titrant (iii) p ml titrant (iv)

97 Exercises Equal volumes of two acids are titrated with 0.10 M NaO resulting in the two titration curves shown in the following figure. (a) Which curve corresponds to the more concentrated acid solution? (b) Which corresponds to the acid with the larger K a? [Section 17.3] p ml NaO 17.8 A saturated solution of Cd1O2 2 is shown in the middle beaker. If hydrochloric acid solution is added, the solubility of Cd 2 (aq) Add Cl(aq) Beaker A Saturated solution O (aq) Add Cl(aq) Beaker B Cd(O) 2 (s) (i) 50 ml 1.0 M Cl(aq) (iii) 50 ml 1.0 M CaCl 2 (aq) (ii) 50 ml 1.0 M NaCl(aq) (iv) 50 ml 0.10 M CaCl 2 (aq) The graph below shows the solubility of a salt as a function of p. Which of the following choices explain the shape of this graph? (a) None; this behavior is not possible. (b) A soluble salt reacts with acid to form a precipitate, and additional acid reacts with this product to dissolve it. (c) A soluble salt forms an insoluble hydroxide, then additional base reacts with this product to dissolve it. (d) The solubility of the salt increases with p then decreases because of the heat generated from the neutralization reactions. [Section 17.5] Cd(O) 2 (s) Cd(O) 2 (s) Solubility Cd1O2 2 will increase, causing additional solid to dissolve. Which of the two choices, Beaker A or Beaker B, accurately represents the solution after equilibrium is reestablished? (The water molecules and CI - ions are omitted for clarity.) [Sections 17.4 and 17.5] 17.9 The following graphs represent the behavior of BaCO 3 under different circumstances. In each case, the vertical axis indicates the solubility of the BaCO 3 and the horizontal axis represents the concentration of some other reagent. (a) Which graph represents what happens to the solubility of BaCO 3 as NO 3 is added? (b) Which graph represents what happens to the BaCO 3 solubility as Na 2 CO 3 is added? (c) Which represents what happens to the BaCO 3 solubility as NaNO 3 is added? [Section 17.5] p Three cations, Ni 2, Cu 2, and Ag, are separated using two different precipitating agents. Based on Figure 17.23, what two precipitating agents could be used? Using these agents, indicate which of the cations is A, which is B, and which is C. [Section 17.7] Mixture of cations A,B,C Add 1st precipitating agent Cation A removed Decant liquid Cations B,C Add 2nd precipitating agent Cation B removed Decant liquid Cation C Solubility Solubility Solubility Conc Conc Conc Cation A Cation B Cation C Ca1O2 2 has a K sp of 6.5 * (a) If g of Ca1O2 2 is added to 500 ml of water and the mixture is allowed to come to equilibrium, will the solution be saturated? (b) If 50 ml of the solution from part (a) is added to each of the beakers shown here, in which beakers, if any, will a precipitate form? In those cases where a precipitate forms, what is its identity? [Section 17.6] The Common-Ion Effect (Section 17.1) Which of these statements about the common-ion effect is most correct? (a) The solubility of a salt MA is decreased in a solution that already contains either M or A -. (b) Common ions alter the equilibrium constant for the reaction of an ionic

98 768 chapter 17 Additional Aspects of Aqueous Equilibria solid with water. (c) The common-ion effect does not apply to 2 unusual ions like SO - 3. (d) The solubility of a salt MA is affected equally by addition of either A - or a non-common ion Consider the equilibrium B1aq2 2 O1l2 B 1aq2 O - 1aq2. Suppose that a salt of B is added to a solution of B at equilibrium. (a) Will the equilibrium constant for the reaction increase, decrease, or stay the same? (b) Will the concentration of B(aq) increase, decrease, or stay the same? (c) Will the p of the solution increase, decrease, or stay the same? Use information from Appendix D to calculate the p of (a) a solution that is M in potassium propionate 1C 2 5 COOK or KC 3 5 O 2 2 and M in propionic acid 1C 2 5 COO K or C 3 5 O 2 2; (b) a solution that is M in trimethylamine, 1C N, and 0.10 M in trimethylammonium chloride, 1C NCl; (c) a solution that is made by mixing 50.0 ml of 0.15 M acetic acid and 50.0 ml of 0.20 M sodium acetate Use information from Appendix D to calculate the p of (a) a solution that is M in sodium formate (COONa) and M in formic acid (COO), (b) a solution that is M in pyridine 1C 5 5 N2 and M in pyridinium chloride 1C 5 5 NCl2, (c) a solution that is made by combining 55 ml of M hydrofluoric acid with 125 ml of 0.10 M sodium fluoride (a) Calculate the percent ionization of M butanoic acid 1K a = 1.5 * (b) Calculate the percent ionization of M butanoic acid in a solution containing M sodium butanoate (a) Calculate the percent ionization of M lactic acid 1K a = 1.4 * (b) Calculate the percent ionization of M lactic acid in a solution containing M sodium lactate. Buffers (Section 17.2) Which of the following solutions is a buffer? (a) 0.10 M C 3 COO and 0.10 M C 3 COONa, (b) 0.10 M C 3 COO, (c) 0.10 M Cl and 0.10 M NaCl, (d) both a and c, (e) all of a, b, and c Which of the following solutions is a buffer? (a) A solution made by mixing 100 ml of M C 3 COO and 50 ml of M NaO, (b) a solution made by mixing 100 ml of M C 3 COO and 500 ml of M NaO, (c) A solution made by mixing 100 ml of M C 3 COO and 50 ml of M Cl, (d) A solution made by mixing 100 ml of M C 3 COOK and 50 ml of M KCl (a) Calculate the p of a buffer that is 0.12 M in lactic acid and 0.11 M in sodium lactate. (b) Calculate the p of a buffer formed by mixing 85 ml of 0.13 M lactic acid with 95 ml of 0.15 M sodium lactate (a) Calculate the p of a buffer that is M in NaCO 3 and M in Na 2 CO 3. (b) Calculate the p of a solution formed by mixing 65 ml of 0.20 M NaCO 3 with 75 ml of 0.15 M Na 2 CO A buffer is prepared by adding 20.0 g of sodium acetate 1C 3 COONa2 to 500 ml of a M acetic acid 1C 3 COO2 solution. (a) Determine the p of the buffer. (b) Write the complete ionic equation for the reaction that occurs when a few drops of hydrochloric acid are added to the buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of sodium hydroxide solution are added to the buffer A buffer is prepared by adding 10.0 g of ammonium chloride 1N 4 Cl2 to 250 ml of 1.00 M N 3 solution. (a) What is the p of this buffer? (b) Write the complete ionic equation for the reaction that occurs when a few drops of nitric acid are added to the buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of potassium hydroxide solution are added to the buffer You are asked to prepare a p = 3.00 buffer solution starting from 1.25 L of a 1.00 M solution of hydrofluoric acid (F) and any amount you need of sodium fluoride (NaF). (a) What is the p of the hydrofluoric acid solution prior to adding sodium fluoride? (b) ow many grams of sodium fluoride should be added to prepare the buffer solution? Neglect the small volume change that occurs when the sodium fluoride is added You are asked to prepare a p = 4.00 buffer starting from 1.50 L of M solution of benzoic acid 1C 6 5 COO2 and any amount you need of sodium benzoate 1C 6 5 COONa2. (a) What is the p of the benzoic acid solution prior to adding sodium benzoate? (b) ow many grams of sodium benzoate should be added to prepare the buffer? Neglect the small volume change that occurs when the sodium benzoate is added A buffer contains 0.10 mol of acetic acid and 0.13 mol of sodium acetate in 1.00 L. (a) What is the p of this buffer? (b) What is the p of the buffer after the addition of 0.02 mol of KO? (c) What is the p of the buffer after the addition of 0.02 mol of NO 3? A buffer contains 0.15 mol of propionic acid 1C 2 5 COO2 and 0.10 mol of sodium propionate 1C 2 5 COONa2 in 1.20 L. (a) What is the p of this buffer? (b) What is the p of the buffer after the addition of 0.01 mol of NaO? (c) What is the p of the buffer after the addition of 0.01 mol of I? (a) What is the ratio of CO - 3 to 2 CO 3 in blood of p 7.4? (b) What is the ratio of CO - 3 to 2 CO 3 in an exhausted marathon runner whose blood p is 7.1? A buffer, consisting of 2 PO - 4 and PO 2-4, helps control the p of physiological fluids. Many carbonated soft drinks also use this buffer system. What is the p of a soft drink in which the major buffer ingredients are 6.5 g of Na 2 PO 4 and 8.0 g of Na 2 PO 4 per 355 ml of solution? You have to prepare a p 3.50 buffer, and you have the following 0.10 M solutions available: COO, C 3 COO, 3 PO 4, COONa, C 3 COONa, and Na 2 PO 4. Which solutions would you use? ow many milliliters of each solution would you use to make approximately 1 L of the buffer? You have to prepare a p 5.00 buffer, and you have the following 0.10 M solutions available: COO, COONa, C 3 COO, C 3 COONa, CN, and NaCN. Which solutions would you use? ow many milliliters of each solution would you use to make approximately 1 L of the buffer? Acid Base Titrations (Section 17.3) The accompanying graph shows the titration curves for two monoprotic acids. (a) Which curve is that of a strong acid? (b) What is the approximate p at the equivalence point of

99 Exercises 769 each titration? (c) 40.0 ml of each acid was titrated with M base. Which acid is more concentrated? (d) Estimate the pk a of the weak acid p A B ml NaO Compare the titration of a strong, monoprotic acid with a strong base to the titration of a weak, monoprotic acid with a strong base. Assume the strong and weak acid solutions initially have the same concentrations. Indicate whether the following statements are true or false. (a) More base is required to reach the equivalence point for the strong acid than the weak acid. (b) The p at the beginning of the titration is lower for the weak acid than the strong acid. (c) The p at the equivalence point is 7 no matter which acid is titrated The samples of nitric and acetic acids shown here are both titrated with a M solution of NaO(aq) ml of 1.0 M NO 3 (aq) 25.0 ml of 1.0 M C 3 COO(aq) Determine whether each of the following statements concerning these titrations is true or false. (a) A larger volume of NaO1aq2 is needed to reach the equivalence point in the titration of NO 3. (b) The p at the equivalence point in the NO 3 titration will be lower than the p at the equivalence point in the C 3 COO titration. (c) Phenolphthalein would be a suitable indicator for both titrations Determine whether each of the following statements concerning the titrations in Problem is true or false. (a) The p at the beginning of the two titrations will be the same. (b) The titration curves will both be essentially the same after passing the equivalence point. (c) Methyl red would be a suitable indicator for both titrations Predict whether the equivalence point of each of the following titrations is below, above, or at p 7: (a) NaCO 3 titrated with NaO, (b) N 3 titrated with Cl, (c) KO titrated with Br Predict whether the equivalence point of each of the following titrations is below, above, or at p 7: (a) formic acid titrated with NaO, (b) calcium hydroxide titrated with perchloric acid, (c) pyridine titrated with nitric acid As shown in Figure 16.8, the indicator thymol blue has two color changes. Which color change will generally be more suitable for the titration of a weak acid with a strong base? Assume that 30.0 ml of a 0.10 M solution of a weak base B that accepts one proton is titrated with a 0.10 M solution of the monoprotic strong acid A. (a) ow many moles of A have been added at the equivalence point? (b) What is the predominant form of B at the equivalence point? (c) Is the p 7, less than 7, or more than 7 at the equivalence point? (d) Which indicator, phenolphthalein or methyl red, is likely to be the better choice for this titration? ow many milliliters of M NaO are required to titrate each of the following solutions to the equivalence point: (a) 40.0 ml of M NO 3, (b) 35.0 ml of M C 3 COO, (c) 50.0 ml of a solution that contains 1.85 g of Cl per liter? ow many milliliters of M Cl are needed to titrate each of the following solutions to the equivalence point: (a) 45.0 ml of M NaO, (b) 22.5 ml of M N 3, (c) ml of a solution that contains 1.35 g of NaO per liter? A 20.0-mL sample of M Br solution is titrated with M NaO solution. Calculate the p of the solution after the following volumes of base have been added: (a) 15.0 ml, (b) 19.9 ml, (c) 20.0 ml, (d) 20.1 ml, (e) 35.0 ml A 20.0-mL sample of M KO is titrated with M ClO 4 solution. Calculate the p after the following volumes of acid have been added: (a) 20.0 ml, (b) 23.0 ml, (c) 24.0 ml, (d) 25.0 ml, (e) 30.0 ml A 35.0-mL sample of M acetic acid 1C 3 COO2 is titrated with M NaO solution. Calculate the p after the following volumes of base have been added: (a) 0 ml, (b) 17.5 ml, (c) 34.5 ml, (d) 35.0 ml, (e) 35.5 ml, (f) 50.0 ml Consider the titration of 30.0 ml of M N 3 with M Cl. Calculate the p after the following volumes of titrant have been added: (a) 0 ml, (b) 20.0 ml, (c) 59.0 ml, (d) 60.0 ml, (e) 61.0 ml, (f) 65.0 ml Calculate the p at the equivalence point for titrating M solutions of each of the following bases with M Br: (a) sodium hydroxide (NaO), (b) hydroxylamine 1N 2 O2, (c) aniline 1C 6 5 N Calculate the p at the equivalence point in titrating M solutions of each of the following with M NaO: (a) hydrobromic acid (Br), (b) chlorous acid 1ClO 2 2, (c) benzoic acid 1C 6 5 COO2. Solubility Equilibria and Factors Affecting Solubility (Sections 17.4 and 17.5) For each statement, indicate whether it is true or false. (a) The solubility of a slightly soluble salt can be expressed in units of moles per liter. (b) The solubility product of a slightly soluble salt is simply the square of the solubility.

100 770 chapter 17 Additional Aspects of Aqueous Equilibria (c) The solubility of a slightly soluble salt is independent of the presence of a common ion. (d) The solubility product of a slightly soluble salt is independent of the presence of a common ion The solubility of two slightly soluble salts of M 2, MA and MZ 2, are the same, 4 * 10-4 mol/l. (a) Which has the larger numerical value for the solubility product constant? (b) In a saturated solution of each salt in water, which has the higher concentration of M 2? (c) If you added an equal volume of a solution saturated in MA to one saturated in MZ 2, what would be the equilibrium concentration of the cation, M 2? (a) Why is the concentration of undissolved solid not explicitly included in the expression for the solubility-product constant? (b) Write the expression for the solubility-product constant for each of the following strong electrolytes: AgI, SrSO 4, Fe1O2 2, and g 2 Br (a) True or false: solubility and solubility-product constant are the same number for a given compound. (b) Write the expression for the solubility-product constant for each of the following ionic compounds: MnCO 3, g1o2 2, and Cu 3 1PO (a) If the molar solubility of CaF 2 at 35 C i s 1.24 * 10-3 mol>l, what is K sp at this temperature? (b) It is found that 1.1 * 10-2 g SrF 2 dissolves per 100 ml of aqueous solution at 25 C. Calculate the solubility product for SrF 2. (c) The K sp of Ba1IO at 25 C is 6.0 * What is the molar solubility of Ba1IO 3 2 2? (a) The molar solubility of PbBr 2 at 25 C is 1.0 * 10-2 mol>l. Calculate K sp. (b) If g of AgIO 3 dissolves per liter of solution, calculate the solubility-product constant. (c) Using the appropriate K sp value from Appendix D, calculate the p of a saturated solution of Ca1O A 1.00-L solution saturated at 25 C with calcium oxalate 1CaC 2 O 4 2 contains g of CaC 2 O 4. Calculate the solubility-product constant for this salt at 25 C A 1.00-L solution saturated at 25 C with lead(ii) iodide contains 0.54 g of PbI 2. Calculate the solubility-product constant for this salt at 25 C Using Appendix D, calculate the molar solubility of AgBr in (a) pure water, (b) 3.0 * 10-2 M AgNO 3 solution, (c) 0.10 M NaBr solution Calculate the solubility of LaF 3 in grams per liter in (a) pure water, (b) M KF solution, (c) M LaCl 3 solution Consider a beaker containing a saturated solution of CaF 2 in equilibrium with undissolved CaF 2 1s2. Solid CaCl 2 is then added to the solution. (a) Will the amount of solid CaF 2 at the bottom of the beaker increase, decrease, or remain the same? (b) Will the concentration of Ca 2 ions in solution increase or decrease? (c) Will the concentration of F - ions in solution increase or decrease? Consider a beaker containing a saturated solution of Pbl 2 in equilibrium with undissolved Pbl 2 1s2. Now solid KI is added to this solution. (a) Will the amount of solid Pbl 2 at the bottom of the beaker increase, decrease, or remain the same? (b) Will the concentration of Pb 2 ions in solution increase or decrease? (c) Will the concentration of I - ions in solution increase or decrease? Calculate the solubility of Mn1O2 2 in grams per liter when buffered at p (a) 7.0, (b) 9.5, (c) Calculate the molar solubility of Ni1O2 2 when buffered at p (a) 8.0, (b) 10.0, (c) Which of the following salts will be substantially more soluble in acidic solution than in pure water: (a) ZnCO 3, (b) ZnS, (c) BiI 3, (d) AgCN, (e) Ba 3 1PO 4 2 2? For each of the following slightly soluble salts, write the net ionic equation, if any, for reaction with a strong acid: (a) MnS, (b) PbF 2, (c) AuCl 3, (d) g 2 C 2 O 4, (e) CuBr From the value of K f listed in Table 17.1, calculate the concentration of Ni 2 in 1.0 L of a solution that contains a total of 1 * 10-3 mol of nickel(ii) ion and that is 0.20 M in N To what final concentration of N 3 must a solution be adjusted to just dissolve mol of NiC 2 O 4 1K sp = 4 * in 1.0 L of solution? (int: You can neglect the hydrolysis of C 2 O 2-4 because the solution will be quite basic.) Use values of K sp for AgI and K f for Ag1CN2-2 to (a) calculate the molar solubility of AgI in pure water, (b) calculate the equilibrium constant for the reaction AgI1s2 2 CN - 1aq2 Ag1CN2-2 1aq2 I - 1aq2, (c) determine the molar solubility of AgI in a M NaCN solution Using the value of K sp for Ag 2 S, K a1 and K a2 for 2 S, and K f = 1.1 * 10 5 for AgCl - 2, calculate the equilibrium constant for the following reaction: Ag 2 S1s2 4 Cl - 1aq2 2 1aq2 2 AgCl 2-1aq2 2 S1aq2 Precipitation and Separation of Ions (Section 17.6) (a) Will Ca1O2 2 precipitate from solution if the p of a M solution of CaCl 2 is adjusted to 8.0? (b) Will Ag 2 SO 4 precipitate when 100 ml of M AgNO 3 is mixed with 10 ml of 5.0 * 10-2 M Na 2 SO 4 solution? (a) Will Co1O2 2 precipitate from solution if the p of a M solution of Co1NO is adjusted to 8.5? (b) Will AgIO 3 precipitate when 20 ml of M AgIO 3 is mixed with 10 ml of M NaIO 3? 1K sp of AgIO 3 is 3.1 * Calculate the minimum p needed to precipitate Mn1O2 2 so completely that the concentration of Mn 2 is less than 1 mg per liter [1 part per billion (ppb)] Suppose that a 10-mL sample of a solution is to be tested for I - ion by addition of 1 drop (0.2 ml) of 0.10 M Pb1NO What is the minimum number of grams of I - that must be present for Pbl 2 1s2 to form? A solution contains 2.0 * 10-4 M Ag and 1.5 * 10-3 M Pb 2. If NaI is added, will AgI 1K sp = 8.3 * or PbI 2 1K sp = 7.9 * precipitate first? Specify the concentration of I - needed to begin precipitation A solution of Na 2 SO 4 is added dropwise to a solution that is M in Ba 2 and M in Sr 2. (a) What concentration of SO 4 is necessary to begin precipitation? (Ne- 2 - glect volume changes. BaSO 4 : K sp = 1.1 * ; SrSO 4 : K sp = 3.2 * 10-7.) (b) Which cation precipitates first? 2 - (c) What is the concentration of SO 4 when the second cation begins to precipitate? A solution contains three anions with the following concentrations: M CrO - 2 4, 0.10 M CO - 3, and M Cl -. If a dilute

101 additional Exercises 771 AgNO 3 solution is slowly added to the solution, what is the first compound to precipitate: Ag 2 CrO 4 1K sp = 1.2 * , Ag 2 CO 3 1K sp = 8.1 * , or AgCl 1K sp = 1.8 * ? A 1.0 M Na 2 SO 4 solution is slowly added to 10.0 ml of a solution that is 0.20 M in Ca 2 and 0.30 M in Ag. (a) Which compound will precipitate first: CaSO 4 1K sp = 2.4 * or Ag 2 SO 4 1K sp = 1.5 * ? (b) ow much Na 2 SO 4 solution must be added to initiate the precipitation? Qualitative Analysis for Metallic Elements (Section 17.7) A solution containing an unknown number of metal ions is treated with dilute Cl; no precipitate forms. The p is adjusted to about 1, and 2 S is bubbled through. Again, no precipitate forms. The p of the solution is then adjusted to about 8. Again, 2 S is bubbled through. This time a precipitate forms. The filtrate from this solution is treated with 1N PO 4. No precipitate forms. Which metal ions discussed in Section 17.7 are possibly present? Which are definitely absent within the limits of these tests? An unknown solid is entirely soluble in water. On addition of dilute Cl, a precipitate forms. After the precipitate is filtered off, the p is adjusted to about 1 and 2 S is bubbled in; a precipitate again forms. After filtering off this precipitate, the p is adjusted to 8 and 2 S is again added; no precipitate forms. No precipitate forms upon addition of 1N PO 4. (See Figure 7.23.) The remaining solution shows a yellow color in a flame test (see Figure 7.22). Based on these observations, which of the following compounds might be present, which are definitely present, and which are definitely absent: CdS, Pb1NO 3 2 2, go, ZnSO 4, Cd1NO 3 2 2, and Na 2 SO 4? In the course of various qualitative analysis procedures, the following mixtures are encountered: (a) Zn 2 and Cd 2, (b) Cr1O2 3 and Fe1O2 3, (c) Mg 2 and K, (d) Ag and Mn 2. Suggest how each mixture might be separated Suggest how the cations in each of the following solution mixtures can be separated: (a) Na and Cd 2, (b) Cu 2 and Mg 2, (c) Pb 2 and Al 3, (d) Ag and g (a) Precipitation of the group 4 cations of Figure requires a basic medium. Why is this so? (b) What is the most significant difference between the sulfides precipitated in group 2 and those precipitated in group 3? (c) Suggest a procedure that would serve to redissolve the group 3 cations following their precipitation A student who is in a great hurry to finish his laboratory work decides that his qualitative analysis unknown contains a metal ion from group 4 of Figure e therefore tests his sample directly with 1N PO 4, skipping earlier tests for the metal ions in groups 1, 2, and 3. e observes a precipitate and concludes that a metal ion from group 4 is indeed present. Why is this possibly an erroneous conclusion? Additional Exercises Derive an equation similar to the enderson asselbalch equation relating the po of a buffer to the pk b of its base component. [17.84] Rainwater is acidic because CO 2 1g2 dissolves in the water, creating carbonic acid, 2 CO 3. If the rainwater is too acidic, it will react with limestone and seashells (which are principally made of calcium carbonate, CaCO 3 ). Calculate the concentrations of carbonic acid, bicarbonate ion 1CO and carbonate ion 1CO that are in a raindrop that has a p of 5.60, 2 assuming that the sum of all three species in the raindrop is 1.0 * 10-5 M Furoic acid 1C 5 3 O 3 2 has a K a value of 6.76 * 10-4 at 25 C. Calculate the p at 25 C of (a) a solution formed by adding 25.0 g of furoic acid and 30.0 g of sodium furoate 1NaC 5 3 O 3 2 to enough water to form L of solution, (b) a solution formed by mixing 30.0 ml of M C 5 3 O 3 and 20.0 ml of 0.22 M NaC 5 3 O 3 and diluting the total volume to 125 ml, (c) a solution prepared by adding 50.0 ml of 1.65 M NaO solution to L of M C 5 3 O The acid base indicator bromcresol green is a weak acid. The yellow acid and blue base forms of the indicator are present in equal concentrations in a solution when the p is What is the pk a for bromcresol green? Equal quantities of M solutions of an acid A and a base B are mixed. The p of the resulting solution is 9.2. (a) Write the chemical equation and equilibrium-constant expression for the reaction between A and B. (b) If K a for A is 8.0 * 10-5, what is the value of the equilibrium constant for the reaction between A and B? (c) What is the value of K b for B? Two buffers are prepared by adding an equal number of moles of formic acid (COO) and sodium formate (COONa) to enough water to make 1.00 L of solution. Buffer A is prepared using 1.00 mol each of formic acid and sodium formate. Buffer B is prepared by using mol of each. (a) Calculate the p of each buffer. (b) Which buffer will have the greater buffer capacity? (c) Calculate the change in p for each buffer upon the addition of 1.0 ml of 1.00 M Cl. (d) Calculate the change in p for each buffer upon the addition of 10 ml of 1.00 M Cl A biochemist needs 750 ml of an acetic acid sodium acetate buffer with p Solid sodium acetate 1C 3 COONa2 and glacial acetic acid 1C 3 COO2 are available. Glacial acetic acid is 99% C 3 COO by mass and has a density of 1.05 g>ml. If the buffer is to be 0.15 M in C 3 COO, how many grams of C 3 COONa and how many milliliters of glacial acetic acid must be used? A sample of g of an unknown monoprotic acid was dissolved in 25.0 ml of water and titrated with M NaO. The acid required 27.4 ml of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After 15.0 ml of base had been added in the titration, the p was found to be What is the K a for the unknown acid? A sample of g of an unknown monoprotic acid was dissolved in 25.0 ml of water and titrated with M NaO. The acid required 15.5 ml of base to reach the equivalence point. (a) What is the molecular weight of the acid? (b) After 7.25 ml of base had been added in the titration, the p was found to be What is the K a for the unknown acid?

102 772 chapter 17 Additional Aspects of Aqueous Equilibria Mathematically prove that the p at the halfway point of a titration of a weak acid with a strong base (where the volume of added base is half of that needed to reach the equivalence point) is equal to pk a for the acid A weak monoprotic acid is titrated with M NaO. It requires 50.0 ml of the NaO solution to reach the equivalence point. After 25.0 ml of base is added, the p of the solution is Estimate the pka of the weak acid What is the p of a solution made by mixing 0.30 mol NaO, 0.25 mol Na 2 PO 4, and 0.20 mol 3 PO 4 with water and diluting to 1.00 L? Suppose you want to do a physiological experiment that calls for a p 6.50 buffer. You find that the organism with which you are working is not sensitive to the weak acid 2 A 1K a1 = 2 * 10-2 ; K a2 = 5.0 * or its sodium salts. You have available a 1.0 M solution of this acid and a 1.0 M solution of NaO. ow much of the NaO solution should be added to 1.0 L of the acid to give a buffer at p 6.50? (Ignore any volume change.) ow many microliters of M NaO solution must be added to ml of a M solution of lactic acid 3C 3 C1O2COO or C 3 5 O 3 4 to produce a buffer with p = 3.75? A person suffering from anxiety begins breathing rapidly and as a result suffers alkalosis, an increase in blood p. (a) Using Equation 17.10, explain how rapid breathing can cause the p of blood to increase. (b) One cure for this problem is breathing in a paper bag. Why does this procedure lower blood p? For each pair of compounds, use K sp values to determine which has the greater molar solubility: (a) CdS or CuS, (b) PbCO 3 or BaCrO 4, (c) Ni1O2 2 or NiCO 3, (d) AgI or Ag 2 SO The solubility of CaCO 3 is p dependent. (a) Calculate the molar solubility of CaCO 3 1K sp = 4.5 * neglecting the acid base character of the carbonate ion. (b) Use the K b 2 - expression for the CO 3 ion to determine the equilibrium constant for the reaction CaCO 3 1s2 2 O1l2 Ca 2 1aq2 CO - 3 1aq2 O - 1aq2 (c) If we assume that the only sources of Ca 2, CO 3 -, and O - ions are from the dissolution of CaCO 3, what is the molar solubility of CaCO 3 using the equilibrium expression from part (b)? (d) What is the molar solubility of CaCO 3 at the p of the ocean (8.3)? (e) If the p is buffered at 7.5, what is the molar solubility of CaCO 3? Tooth enamel is composed of hydroxyapatite, whose simplest formula is Ca 5 1PO O, and whose corresponding K sp = 6.8 * As discussed in the Chemistry and Life box on page 755, fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, Ca 5 1PO F, whose K sp = 1.0 * (a) Write the expression for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds Use the solubility-product constant for Cr1O2 3 1K sp = 6.7 * and the formation constant for - Cr1O2 4 from Table 17.1 to determine the concentration of - Cr1O2 4 in a solution that is buffered at p = 10.0 and is in equilibrium with solid Cr1O Calculate the solubility of Mg1O2 2 in 0.50 M N 4 Cl The solubility-product constant for barium permanganate, Ba1MnO 4 2 2, is 2.5 * Assume that solid Ba1MnO is in equilibrium with a solution of KMnO 4. What concentration of KMnO 4 is required to establish a concentration of 2.0 * 10-8 M for the Ba 2 ion in solution? Calculate the ratio of 3Ca 2 4 to 3Fe 2 4 in a lake in which the water is in equilibrium with deposits of both CaCO 3 and FeCO 3. Assume that the water is slightly basic and that the hydrolysis of the carbonate ion can therefore be ignored The solubility product constants of PbSO 4 and SrSO 4 are 6.3 * 10-7 and 3.2 * 10-7, respectively. What are the values 2 of 3SO - 4 4, 3Pb 2 4, and 3Sr 2 4 in a solution at equilibrium with both substances? A buffer of what p is needed to give a Mg 2 concentration of 3.0 * 10-2 M in equilibrium with solid magnesium oxalate? The value of K sp for Mg 3 1AsO is 2.1 * The 3 - AsO 4 ion is derived from the weak acid 3 AsO 4 1pK a1 = 2.22; pk a2 = 6.98; pk a3 = When asked to calculate the molar solubility of Mg 3 1AsO in water, a student used the K sp expression and assumed that 3Mg = 1.53AsO Why was this a mistake? The solubility product for Zn1O2 2 is 3.0 * The formation constant for the hydroxo complex, Zn1O2-4, is * What concentration of O - is required to dissolve mol of Zn1O2 2 in a liter of solution? The value of K sp for Cd1O2 2 is 2.5 * (a) What is the molar solubility of Cd1O2 2? (b) The solubility of Cd1O2 2 can be increased through formation of the complex ion CdBr - 4 1K f = 5 * If solid Cd1O2 2 is 2 added to a NaBr solution, what is the initial concentration of NaBr needed to increase the molar solubility of Cd1O2 2 to 1.0 * 10-3 mol/l? Integrative Exercises (a) Write the net ionic equation for the reaction that occurs when a solution of hydrochloric acid (Cl) is mixed with a solution of sodium formate 1NaCO 2 2. (b) Calculate the equilibrium constant for this reaction. (c) Calculate the equilibrium concentrations of Na, Cl -,, CO 2 -, and CO 2 when 50.0 ml of 0.15 M Cl is mixed with 50.0 ml of 0.15 M NaCO (a) A g sample of an unknown monoprotic acid requires ml of M NaO to reach the end point. What is the molecular weight of the unknown? (b) As the acid is titrated, the p of the solution after the addition of ml of the base is What is the K a for the acid? (c) Using Appendix D, suggest the identity of the acid A sample of 7.5 L of N 3 gas at 22 C and 735 torr is bubbled into a 0.50-L solution of 0.40 M Cl. Assuming that all the N 3 dissolves and that the volume of the solution remains 0.50 L, calculate the p of the resulting solution.

103 Design an Experiment Aspirin has the structural formula O O C O C C 3 O At body temperature 137 C2, K a for aspirin equals 3 * If two aspirin tablets, each having a mass of 325 mg, are dissolved in a full stomach whose volume is 1 L and whose p is 2, what percent of the aspirin is in the form of neutral molecules? What is the p at 25 C of water saturated with CO 2 at a partial pressure of 1.10 atm? The enry s law constant for CO 2 at 25 C is 3.1 * 10-2 mol>l@atm Excess Ca1O2 2 is shaken with water to produce a saturated solution. The solution is filtered, and a mL sample titrated with Cl requires ml of M Cl to reach the end point. Calculate K sp for Ca1O2 2. Compare your result with that in Appendix D. 25 C. Suggest a reason for any differences you find between your value and the one in Appendix D The osmotic pressure of a saturated solution of strontium sulfate at 25 C is 21 torr. What is the solubility product of this salt at 25 C? A concentration of parts per billion (by mass) of Ag is an effective disinfectant in swimming pools. owever, if the concentration exceeds this range, the Ag can cause adverse health effects. One way to maintain an appropriate concentration of Ag is to add a slightly soluble salt to the pool. Using K sp values from Appendix D, calculate the equilibrium concentration of Ag in parts per billion that would exist in equilibrium with (a) AgCl, (b) AgBr, (c) AgI Fluoridation of drinking water is employed in many places to aid in the prevention of tooth decay. Typically the F - ion concentration is adjusted to about 1 ppb. Some water supplies are also hard ; that is, they contain certain cations such as Ca 2 that interfere with the action of soap. Consider a case where the concentration of Ca 2 is 8 ppb. Could a precipitate of CaF 2 form under these conditions? (Make any necessary approximations.) Baking soda (sodium bicarbonate, NaCO 3 ) reacts with acids in foods to form carbonic acid 1 2 CO 3 2, which in turn decomposes to water and carbon dioxide gas. In a cake batter, the CO 2 1g2 forms bubbles and causes the cake to rise. (a) A rule of thumb in baking is that 1/2 teaspoon of baking soda is neutralized by one cup of sour milk. The acid component in sour milk is lactic acid, C 3 C1O2COO. Write the chemical equation for this neutralization reaction. (b) The density of baking soda is 2.16 g>cm 3. Calculate the concentration of lactic acid in one cup of sour milk (assuming the rule of thumb applies), in units of mol/l. 1One cup = ml = 48 teaspoons2. (c) If 1/2 teaspoon of baking soda is indeed completely neutralized by the lactic acid in sour milk, calculate the volume of carbon dioxide gas that would be produced at 1 atm pressure, in an oven set to 350 F In nonaqueous solvents, it is possible to react F to create 2 F. Which of these statements follows from this observation? (a) F can act like a strong acid in nonaqueous solvents, (b) F can act like a base in nonaqueous solvents, (c) F is thermodynamically unstable, (d) There is an acid in the nonaqueous medium that is a stronger acid than F. Design an Experiment The pk a of acetic acid is The pk a of chloroacetic acid, C 2 ClCOO, is The pk a of trichloroacetic acid, CCl 3 COO, is (a) Why might this be the case? For example, one hypothesis is that the O bond of trichloroacetic acid is significantly more polar than the O bond in acetic acid due to chlorine being more electron-withdrawing than hydrogen, making the O bond in trichloroacetic acid weak. Another hypothesis is that the chlorines thermodynamically stabilize the conjugate base forms of these acids, and the more chlorines there are, the more stable the conjugate bases. Design a set of experiments or calculations to test these hypotheses. (b) Would you predict that the differences in pk a s of these acids would lead to differences in aqueous solubilities of their sodium salts? Design an experiment to test this hypothesis.

104 18 Chemistry of the Environment The richness of life on Earth, represented in the chapter-opening photograph, is made possible by our planet s supportive atmosphere, the energy received from the Sun, and an abundance of water. These are the signature environmental features believed to be necessary for life. As technology has advanced and the world human population has increased, humans have put new and greater stresses on the environment. Paradoxically, the very technology that can cause pollution also provides the tools to help understand and manage the environment in a beneficial way. Chemistry is often at the heart of environmental issues. The economic growth of both developed and developing nations depends critically on chemical processes that range from treatment of water supplies to industrial processes. Some of these processes produce products or by-products that are harmful to the environment. We are now in a position to apply the principles we have learned in preceding chapters to an understanding of how our environment operates and how human activities affect it. To understand and protect the environment in which we live, we must understand how human-made and natural chemical compounds interact on land and in the sea and sky. Our daily actions as consumers turn on the same choices made by leading experts and governmental leaders: Each decision should reflect the costs versus the benefits of our choices. Unfortunately, the environmental impacts of our decisions are often subtle and not immediately evident. Tahquomenon falls in the Upper Peninsula of Michigan, one of the largest waterfalls east of the Mississippi in the United States. What s Ahead 18.1 Earth s Atmosphere We begin with a look at the temperature profile, pressure profile, and chemical composition of Earth s atmosphere. We then examine photoionization and photodissociation, reactions that result from atmospheric absorption of solar radiation uman Activities and Earth s Atmosphere We next examine the effect human activities have on the atmosphere. We discuss how atmospheric ozone is depleted by reactions involving human-made gases and how acid rain and smog are the result of atmospheric reactions involving compounds produced by human activity Earth s Water We examine the global water cycle, which describes how water moves from the ground to surface to the atmosphere and back into the ground. We compare the chemical compositions of seawater, freshwater, and groundwater.

105 18.4 uman Activities and WATER QUALITY We consider how Earth s water is connected to the global climate and examine one measure of water quality: dissolved oxygen concentration. Water for drinking and for irrigation must be free of salts and pollutants Green Chemistry We conclude by examining green chemistry, an international initiative to make industrial products, processes, and chemical reactions compatible with a sustainable society and environment.

106 776 chapter 18 Chemistry of the Environment Go Figure At what altitude is the atmospheric temperature lowest? Altitude (km) Mesopause Stratopause Thermosphere Mesosphere Stratosphere 10 Tropopause Troposphere Temperature (K) 18.1 Earth s Atmosphere Because most of us have never been very far from Earth s surface, we often take for granted the many ways in which the atmosphere determines the environment in which we live. In this section we examine some of the important characteristics of our planet s atmosphere. The temperature of the atmosphere varies with altitude ( Figure 18.1), and the atmosphere is divided into four regions based on this temperature profile. Just above the surface, in the troposphere, the temperature normally decreases with increasing altitude, reaching a minimum of about 215 K at about 10 km. Nearly all of us live our entire lives in the troposphere. owling winds and soft breezes, rain, and sunny skies all that we normally think of as weather occur in this region. Commercial jet aircraft typically fly Pressure (torr) Figure 18.1 Temperature and pressure in the atmosphere vary as a function of altitude above sea level. Altitude (km) about 10 km (33,000 ft) above Earth, an altitude that defines the upper limit of the troposphere, which we call the tropopause. Above the tropopause, air temperature increases with altitude, reaching a maximum of about 275 K at about 50 km. The region from 10 km to 50 km is the stratosphere, and above it are the mesosphere and thermosphere. Notice in Figure 18.1 that the temperature extremes that form the boundaries between adjacent regions are denoted by the suffix -pause. The boundaries are important because gases mix across them relatively slowly. For example, pollutant gases generated in the troposphere pass through the tropopause and find their way into the stratosphere only very slowly. Atmospheric pressure decreases with increasing elevation (Figure 18.1), declining much more rapidly at lower elevations than at higher ones because of the atmosphere s compressibility. Thus, the pressure decreases from an average value of 760 torr (101 kpa) at sea level to 2.3 * 10-3 torr 13.1 * 10-4 kpa2 at 100 km, to only 1.0 * 10-6 torr 11.3 * 10-7 kpa2 at 200 km. The troposphere and stratosphere together account for 99.9% of the mass of the atmosphere, 75% of which is the mass in the troposphere. Nevertheless the thin upper atmosphere plays many important roles in determining the conditions of life at the surface. Composition of the Atmosphere Earth s atmosphere is constantly bombarded by radiation and energetic particles from the Sun. This barrage of energy has profound chemical and physical effects, especially in the upper regions of the atmosphere, above about 80 km ( Figure 18.2). In addition, because of Earth s gravitational field, heavier atoms and molecules tend to sink in the atmosphere, leaving lighter atoms and molecules at the top of the atmosphere. (This is why, as just noted, 75% of the atmosphere s mass is in the troposphere.) Because of all these factors, the composition of the atmosphere is not uniform. Table 18.1 shows the composition of dry air near sea level. Note that although traces of many substances are present, N 2 and O 2 make up about 99% of sea-level air. The noble gases and CO 2 make up most of the remainder.

107 section 18.1 Earth s Atmosphere 777 Table 18.1 The Major Components of Dry Air near Sea Level Component* Content (mole fraction) Molar Mass (g/mol) Nitrogen Oxygen Argon Carbon dioxide Neon elium Methane Krypton ydrogen Nitrous oxide Xenon *Ozone, sulfur dioxide, nitrogen dioxide, ammonia, and carbon monoxide are present as trace gases in variable amounts. igh-energy solar particles create excited N and O atoms; visible light results as electrons in these atoms fall from upper-level states to lower-level states Give It Some Thought ow would you expect the ratio of atmospheric helium to argon to differ at 100 km elevation as compared with sea level? Figure 18.2 The aurora borealis (northern lights). When applied to substances in aqueous solution, the concentration unit parts per million (ppm) refers to grams of substance per million grams of solution. (Section 13.4) When dealing with gases, however, 1 ppm means one part by volume in 1 million volumes of the whole. Because volume is proportional to number of moles of gas via the ideal-gas equation 1PV = nrt2, volume fraction and mole fraction are the same. Thus, 1 ppm of a trace constituent of the atmosphere amounts to 1 mol of that constituent in 1 million moles of air; that is, the concentration in parts per million is equal to the mole fraction times For example, Table 18.1 lists the mole fraction of CO 2 in the atmosphere as , which means its concentration in parts per million is * 10 6 = 400 ppm. Other minor constituents of the troposphere, in addition to CO 2, are listed in Table Before we consider the chemical processes that occur in the atmosphere, let s review some of the properties of the two major components, N 2 and O 2. Recall that the N 2 molecule possesses a triple bond between the nitrogen atoms. (Section 8.3) This very strong bond (bond energy 941 kj>mol) is largely responsible for the very low reactivity of N 2. The bond energy in O 2 is only 495 kj>mol, making O 2 much more reactive than N 2. For example, oxygen reacts with many substances to form oxides. The oxides of nonmetals, Table 18.2 Sources and Typical Concentrations of Some Minor Atmospheric Constituents Constituent Sources Typical Concentration Carbon dioxide, CO 2 Decomposition of organic matter, release from 400 ppm throughout troposphere oceans, fossil-fuel combustion Carbon monoxide, CO Decomposition of organic matter, industrial processes, fossil-fuel combustion 0.05 ppm in unpolluted air; 1 50 ppm in urban areas Methane, C 4 Decomposition of organic matter, natural-gas 1.82 ppm throughout troposphere seepage, livestock emissions Nitric oxide, NO Atmospheric electrical discharges, internal combustion engines, combustion of organic matter 0.01 ppm in unpolluted air; 0.2 ppm in smog Ozone, O 3 Sulfur dioxide, SO 2 Atmospheric electrical discharges, diffusion from the stratosphere, photochemical smog Volcanic gases, forest fires, bacterial action, fossil-fuel combustion, industrial processes ppm in unpolluted air; 0.5 ppm in photochemical smog ppm in unpolluted air; ppm in polluted urban areas

108 778 chapter 18 Chemistry of the Environment Sample Exercise 18.1 Calculating Concentration from Partial Pressure What is the concentration, in parts per million, of water vapor in a sample of air if the partial pressure of the water is 0.80 torr and the total pressure of the air is 735 torr? Solution Analyze We are given the partial pressure of water vapor and the total pressure of an air sample and asked to determine the water vapor concentration. Plan Recall that the partial pressure of a component in a mixture of gases is given by the product of its mole fraction and the total pressure of the mixture (Section 10.6): P 2 O = X 2 OP t Solve Solving for the mole fraction of water vapor in the mixture, X 2 O, gives X 2 O = P 2 O P t = 0.80 torr 735 torr = The concentration in ppm is the mole fraction times 10 6 : * 10 6 = 1100 ppm Practice Exercise 1 From the data in Table 18.1, the partial pressure of argon in dry air at an atmospheric pressure of 668 mm g is (a) 3.12 mm g, (b) 7.09 mm g, (c) 6.24 mm g, (d) 9.34 mm g, (e) 39.9 mm g. Practice Exercise 2 The concentration of CO in a sample of air is 4.3 ppm. What is the partial pressure of the CO if the total air pressure is 695 torr? such as SO 2, usually form acidic solutions when dissolved in water. The oxides of active metals, such as CaO, form basic solutions when dissolved in water. (Section 7.7) Photochemical Reactions in the Atmosphere Although the atmosphere beyond the stratosphere contains only a small fraction of the atmospheric mass, it forms the outer defense against the hail of radiation and high-energy particles that continuously bombard Earth. As the bombarding radiation passes through the upper atmosphere, it causes two kinds of chemical changes: photodissociation and photoionization. These processes protect us from high-energy radiation by absorbing most of the radiation before it reaches the troposphere. If it were not for these photochemical processes, plant and animal life as we know it could not exist on Earth. The Sun emits radiant energy over a wide range of wavelengths ( Figure 18.3). To understand the connection between the wavelength of radiation and its effect on Go Figure Why does not the solar spectrum at sea level perfectly match the solar spectrum outside the atmosphere? Solar spectrum outside atmosphere Solar spectrum at sea level Increasing flux Ultraviolet Visible Infrared Wavelength (nm) Figure 18.3 The solar spectrum above Earth s atmosphere compared to that at sea level. The more structured curve at sea level is due to gases in the atmosphere absorbing specific wavelengths of light. Flux, the unit on the vertical axis, is light energy per area per unit of time.

109 section 18.1 Earth s Atmosphere 779 atoms and molecules, recall that electromagnetic radiation can be pictured as a stream of photons. (Section 6.2) The energy of each photon is given by E = hn, where h is Planck constant and n is the radiation frequency. For a chemical change to occur when radiation strikes atoms or molecules, two conditions must be met. First, the incoming photons must have sufficient energy to break a chemical bond or remove an electron from the atom or molecule. Second, the atoms or molecules being bombarded must absorb these photons. When these requirements are met, the energy of the photons is used to do the work associated with some chemical change. The rupture of a chemical bond resulting from absorption of a photon by a molecule is called photodissociation. No ions are formed when the bond between two atoms is cleaved by photodissociation. Instead, half the bonding electrons stay with one atom and half stay with the other atom. The result is two electrically neutral particles. One of the most important processes occurring above an altitude of about 120 km is photodissociation of the oxygen molecule: O O hν O O [18.1] The minimum energy required to cause this change is determined by the bond energy (or dissociation energy) of O 2, 495 kj>mol. Sample Exercise 18.2 Calculating the Wavelength Required to Break a Bond What is the maximum wavelength of light, in nanometers, that has enough energy per photon to dissociate the O 2 molecule? Solution Analyze We are asked to determine the wavelength of a photon that has just enough energy to break the O O double bond in O 2. Plan We first need to calculate the energy required to break the O O double bond in one molecule and then find the wavelength of a photon of this energy. Solve The dissociation energy of O 2 is 495 kj>mol. Using this value and Avogadro s number, we can calculate the amount of energy needed to break the bond in a single O 2 molecule: a495 * 10 3 J mol b a 1 mol * molecules b = 8.22 * J molecule We next use the Planck relationship, E = hn, (Equation 6.2) to calculate the frequency n of a photon that has this amount of energy: n = E h = 8.22 * J * J@s = 1.24 * 1015 s -1 Finally, we use the relationship between frequency and wavelength (Section 6.1) to calculate the wavelength of the light: l = c n = a 3.00 * 108 m>s 1.24 * >s b a 109 nm b = 242 nm 1 m Thus, light of wavelength 242 nm, which is in the ultraviolet region of the electromagnetic spectrum, has sufficient energy per photon to photodissociate an O 2 molecule. Because photon energy increases as wavelength decreases, any photon of wavelength shorter than 242 nm will have sufficient energy to dissociate O 2. Practice Exercise 1 The bond dissociation energy of the Br Br bond is 193 kj>mol. What wavelength of light has just sufficient energy to cause Br Br bond dissociation? (a) 620 nm (b) 310 nm (c) 148 nm (d) 6200 nm (e) 563 nm Practice Exercise 2 The bond energy in N 2 is 941 kj>mol. What is the longest wavelength a photon can have and still have sufficient energy to dissociate N 2? Fortunately for us, O 2 absorbs much of the high-energy, short-wavelength radiation from the solar spectrum before that radiation reaches the lower atmosphere. As it does, atomic oxygen, O, is formed. The dissociation of O 2 is very extensive at higher elevations. At 400 km, for example, only 1% of the oxygen is in the form of O 2 ; 99% is atomic oxygen. At 130 km, O 2 and atomic oxygen are just about equally abundant. Below 130 km, O 2 is more abundant than atomic oxygen because most of the solar energy has been absorbed in the upper atmosphere. The dissociation energy of N 2 is very high, 941 kj>mol. As you should have seen in working out Practice Exercise 2 of Sample Exercise 18.2, only photons having a

110 780 chapter 18 Chemistry of the Environment wavelength shorter than 127 nm possess sufficient energy to dissociate N 2. Furthermore, N 2 does not readily absorb photons, even when they possess sufficient energy. As a result, very little atomic nitrogen is formed in the upper atmosphere by photodissociation of N 2. Other photochemical processes besides photodissociation occur in the upper atmosphere, although their discovery has taken many twists and turns. In 1901 Guglielmo Marconi received a radio signal in St. John s, Newfoundland, that had been transmitted from Land s End, England, 2900 km away. Because people at the time thought radio waves traveled in straight lines, they assumed that the curvature of Earth s surface would make radio communication over large distances impossible. Marconi s successful experiment suggested that Earth s atmosphere in some way substantially affects radio-wave propagation. is discovery led to intensive study of the upper atmosphere. In about 1924, the existence of electrons in the upper atmosphere was established by experimental studies. The electrons in the upper atmosphere result mainly from photoionization, which occurs when a molecule in the upper atmosphere absorbs solar radiation and the absorbed energy causes an electron to be ejected from the molecule. The molecule then becomes a positively charged ion. For photoionization to occur, therefore, a molecule must absorb a photon, and the photon must have enough energy to remove an electron. (Section 7.4) Notice that this is a very different process from photodissociation. Four important photoionization processes occurring in the atmosphere above about 90 km are shown in Table Photons of any wavelength shorter than the maximum lengths given in the table have enough energy to cause photoionization. A look back at Figure 18.3 shows you that virtually all of these high-energy photons are filtered out of the radiation reaching Earth because they are absorbed by the upper atmosphere. Give It Some Thought Explain the difference between photoionization and photodissociation. Ozone in the Stratosphere Although N 2, O 2, and atomic oxygen absorb photons having wavelengths shorter than 240 nm, ozone, O 3, is the key absorber of photons having wavelengths ranging from 240 to 310 nm, in the ultraviolet region of the electromagnetic spectrum. Ozone in the upper atmosphere protects us from these harmful high-energy photons, which would otherwise penetrate to Earth s surface. Let s consider how ozone forms in the upper atmosphere and how it absorbs photons. By the time radiation from the Sun reaches an altitude of 90 km above Earth s surface, most of the short-wavelength radiation capable of photoionization has been absorbed. At this altitude, however, radiation capable of dissociating the O 2 molecule is sufficiently intense for photodissociation of O 2 (Equation 18.1) to remain important down to an altitude of 30 km. In the region between 30 and 90 km, however, the concentration of O 2 is much greater than the concentration of atomic oxygen. From this Table 18.3 Photoionization Reactions for Four Components of the Atmosphere Process Ionization Energy 1kJ,mol2 L max 1nm2 N 2 hn N 2 e O 2 hn O 2 e O hn O e NO hn NO e

111 section 18.1 Earth s Atmosphere 781 finding, we conclude that the oxygen atoms formed by photodissociation of O 2 in this region frequently collide with O 2 molecules and form ozone: O O 2 O 3 * [18.2] The asterisk on O 3 denotes that the product contains an excess of energy, because the reaction is exothermic. The 105 kj/mol that is released must be transferred away from the O 3 * molecule quickly or else the molecule will fly apart into O 2 and atomic O a decomposition that is the reverse of the reaction by which O 3 * is formed. An energy-rich O 3 * molecule can release its excess energy by colliding with another atom or molecule and transferring some of the excess energy to it. Let s use M to represent the atom or molecule with which O 3 * collides. (Usually M is N 2 or O 2 because these are the most abundant molecules in the atmosphere.) The formation of O 3 * and the transfer of excess energy to M are summarized by the equations O1g2 O 2 1g2 O 3 *1g2 [18.3] O 3 *1g2 M1g2 O 3 1g2 M*1g2 [18.4] O1g2 O 2 1g2 M1g2 O 3 1g2 M*1g2 [18.5] The rate at which the reactions of Equations 18.3 and 18.4 proceed depends on two factors that vary in opposite directions with increasing altitude. First, the Equation 18.3 reaction depends on the presence of O atoms. At low altitudes, most of the radiation energetic enough to dissociate O 2 into O atoms has been absorbed; thus, O atoms are more plentiful at higher altitudes. Second, Equations 18.3 and 18.4 both depend on molecular collisions. (Section 14.5) The concentration of molecules is greater at low altitudes, and so the rates of both reactions are greater at lower altitudes. Because these two reactions vary with altitude in opposite directions, the highest rate of O 3 formation occurs in a band at an altitude of about 50 km, near the stratopause (Figure 18.1). Overall, roughly 90% of Earth s ozone is found in the stratosphere. Give It Some Thought Why do O 2 and N 2 molecules fail to filter out ultraviolet light with wavelengths between 240 and 310 nm? The photodissociation of ozone reverses the reaction that forms it. We thus have a cycle of ozone formation and decomposition, summarized as follows: O 2 1g2 hn O1g2 O1g2 O1g2 O 2 1g2 M1g2 O 3 1g2 M*1g2 1heat released2 O 3 1g2 hn O 2 1g2 O1g2 O1g2 O1g2 M1g2 O 2 1g2 M*1g2 1heat released2 The first and third processes are photochemical; they use a solar photon to initiate a chemical reaction. The second and fourth are exothermic chemical reactions. The net result of the four reactions is a cycle in which solar radiant energy is converted into thermal energy. The ozone cycle in the stratosphere is responsible for the rise in temperature that reaches its maximum at the stratopause (Figure 18.1). The reactions of the ozone cycle account for some, but not all, of the facts about the ozone layer. Many chemical reactions occur that involve substances other than oxygen. We must also consider the effects of turbulence and winds that mix up the stratosphere. A complicated picture results. The overall result of ozone formation and removal reactions, coupled with atmospheric turbulence and other factors, is to produce the upper-atmosphere ozone profile shown in Figure 18.4, with a maximum ozone concentration occurring at an altitude of about 25 km. This band of relatively high ozone concentration is referred to as the ozone layer or the ozone shield. Altitude (km) Go Figure Estimate the ozone concentration in moles per liter for the peak value in this graph Stratosphere Ozone concentration (molecules/cm 3 ) Figure 18.4 Variation in ozone concentration in the atmosphere as a function of altitude.

112 782 chapter 18 Chemistry of the Environment Photons with wavelengths shorter than about 300 nm are energetic enough to break many kinds of single chemical bonds. Thus, the ozone shield is essential for our continued well-being. The ozone molecules that form this essential shield against high-energy radiation represent only a tiny fraction of the oxygen atoms present in the stratosphere, however, because these molecules are continually destroyed even as they are formed human Activities and Earth s Atmosphere Both natural and anthropogenic (human-caused) events can modify Earth s atmosphere. One impressive natural event was the eruption of Mount Pinatubo in June 1991 ( Figure 18.5). The volcano ejected approximately 10 km 3 of material into the stratosphere, causing a 10% drop in the amount of sunlight reaching Earth s surface during the next 2 years. That drop in sunlight led to a temporary 0.5 C drop in Earth s surface temperature. The volcanic particles that made it to the stratosphere remained there for approximately 3 years, raising the temperature of the stratosphere by several degrees due to light absorption. Measurements of the stratospheric ozone concentration showed significantly increased ozone decomposition in this 3-year period. Figure 18.5 Mount Pinatubo erupts, June Total ozone (Dobson units) Figure 18.6 Ozone present in the Southern emisphere, Sept. 24, The data were taken from an orbiting satellite. This day had the lowest stratospheric ozone concentration yet recorded. One Dobson unit corresponds to 2.69 * ozone molecules in a 1 cm 2 column of atmosphere. The Ozone Layer and Its Depletion The ozone layer protects Earth s surface from the damaging ultraviolet (UV) radiation. Therefore, if the concentration of ozone in the stratosphere decreases substantially, more UV radiation will reach Earth s surface, causing unwanted photochemical reactions, including reactions correlated with skin cancer. Satellite monitoring of ozone, which began in 1978, has revealed a depletion of ozone in the stratosphere that is particularly severe over Antarctica, a phenomenon known as the ozone hole ( Figure 18.6). The first scientific paper on this phenomenon appeared in 1985, and the National Aeronautics and Space Administration (NASA) maintains an Ozone ole Watch website with daily updates and data from 1999 to the present. In 1995 the Nobel Prize in Chemistry was awarded to F. Sherwood Rowland, Mario Molina, and Paul Crutzen for their studies of ozone depletion. In 1970 Crutzen showed that naturally occurring nitrogen oxides catalytically destroy ozone. Rowland and Molina recognized in 1974 that chlorine from chlorofluorocarbons (CFCs) may deplete the ozone layer. These substances, principally CFCl 3 and CF 2 Cl 2, do not occur in nature and have been widely used as propellants in spray cans, as refrigerant and air-conditioner gases, and as foaming agents for plastics. They are virtually unreactive in the lower atmosphere. Furthermore, they are relatively insoluble in water and are therefore not removed from the atmosphere by rainfall or by dissolution in the oceans. Unfortunately, the lack of reactivity that makes them commercially useful also allows them to survive in the atmosphere and to diffuse into the stratosphere. It is estimated that several million tons of chlorofluorocarbons are now present in the atmosphere. As CFCs diffuse into the stratosphere, they are exposed to high-energy radiation, which can cause photodissociation. Because C Cl bonds are considerably weaker than C F bonds, free chlorine atoms are formed readily in the presence of light with wavelengths in the range from 190 to 225 nm, as shown in this typical reaction: CF 2 Cl 2 1g2 hn CF 2 Cl1g2 Cl1g2 [18.6] Calculations suggest that chlorine atom formation occurs at the greatest rate at an altitude of about 30 km, the altitude at which ozone is at its highest concentration. Atomic chlorine reacts rapidly with ozone to form chlorine monoxide and molecular oxygen: Cl1g2 O 3 1g2 ClO1g2 O 2 1g2 [18.7]

113 section 18.2 uman Activities and Earth s Atmosphere 783 This reaction follows a second-order rate law with a very large rate constant: Rate = k3cl43o 3 4 k = 7.2 * 10 9 M -1 s -1 at 298 K [18.8] Under certain conditions, the ClO generated in Equation 18.7 can react to regenerate free Cl atoms. One way that this can happen is by photodissociation of ClO: ClO1g2 hn Cl1g2 O1g2 [18.9] The Cl atoms generated in Equations 18.6 and 18.9 can react with more O 3, according to Equation The result is a sequence of reactions that accomplishes the Cl-catalyzed decomposition of O 3 to O 2 : 2 Cl1g2 2 O 3 1g2 2 ClO1g2 2 O 2 1g2 2 ClO1g2 hn 2 Cl1g2 2 O1g2 O1g2 O1g2 O 2 1g2 2 Cl1g2 2 O 3 1g2 2 ClO1g2 2 O1g2 2 Cl1g2 2 ClO1g2 3 O 2 1g2 2 O1g2 The equation can be simplified by eliminating like species from each side to give 2 O 3 1g2 Cl 3 O 2 1g2 [18.10] Because the rate of Equation 18.7 increases linearly with [Cl], the rate at which ozone is destroyed increases as the quantity of Cl atoms increases. Thus, the greater the amount of CFCs that diffuse into the stratosphere, the faster the destruction of the ozone layer. Even though troposphere-to-stratosphere diffusion rates are slow, a substantial thinning of the ozone layer over the South Pole has been observed, particularly during September and October (Figure 18.6). Give It Some Thought Since the rate of ozone destruction depends on [Cl], can Cl be considered a catalyst for the reaction of Equation 18.10? Because of the environmental problems associated with CFCs, steps have been taken to limit their manufacture and use. A major step was the signing in 1987 of the Montreal Protocol on Substances That Deplete the Ozone Layer, in which participating nations agreed to reduce CFC production. More stringent limits were set in 1992, when representatives of approximately 100 nations agreed to ban the production and use of CFCs by 1996, with some exceptions for essential uses. Since then, the production of CFCs has indeed dropped precipitously. Images such as that shown in Figure 18.6 taken annually reveal that the depth and size of the ozone hole has begun to decline. Nevertheless, because CFCs are unreactive and because they diffuse so slowly into the stratosphere, scientists estimate that ozone depletion will continue for many years to come. What substances have replaced CFCs? At this time, the main alternatives are hydrofluorocarbons (FCs), compounds in which C bonds replace the C Cl bonds of CFCs. One such compound in current use is C 2 FCF 3, known as FC-134a. While the FCs are a big improvement over the CFCs because they contain no C Cl bonds, it turns out that they are potent greenhouse warming gases, with which we will deal shortly. There are no naturally occurring CFCs, but some natural sources contribute chlorine and bromine to the atmosphere, and, just like halogens from CFC, these naturally occurring Cl and Br atoms can participate in ozone-depleting reactions. The principal natural sources are methyl bromide and methyl chloride, which are emitted from the oceans. It is estimated that these molecules contribute less than a third of the total Cl and Br in the atmosphere; the remaining two-thirds is a result of human activities. Volcanoes are a source of Cl, but generally the Cl they release reacts with water in the troposphere and does not make it to the upper atmosphere.

114 784 chapter 18 Chemistry of the Environment Table 18.4 Median Concentrations of Atmospheric Pollutants in a Typical Urban Atmosphere Concentration Pollutant (ppm) Carbon monoxide 10 ydrocarbons 3 Sulfur dioxide 0.08 Nitrogen oxides 0.05 Total oxidants 0.02 (ozone and others) Sulfur Compounds and Acid Rain Sulfur-containing compounds are present to some extent in the natural, unpolluted atmosphere. They originate in the bacterial decay of organic matter, in volcanic gases, and from other sources listed in Table The amount of these compounds released into the atmosphere worldwide from natural sources is about 24 * g per year, less than the amount from human activities, about 80 * g per year (principally related to combustion of fuels). Sulfur compounds, chiefly sulfur dioxide, SO 2, are among the most unpleasant and harmful of the common pollutant gases. Table 18.4 shows the concentrations of several pollutant gases in a typical urban environment (where by typical we mean one that is not particularly affected by smog). According to these data, the level of sulfur dioxide is 0.08 ppm or higher about half the time. This concentration is considerably lower than that of other pollutants, notably carbon monoxide. Nevertheless, SO 2 is regarded as the most serious health hazard among the pollutants shown, especially for people with respiratory difficulties. Combustion of coal accounts for about 65% of the SO 2 released annually in the United States, and combustion of oil accounts for another 20%. The majority of this amount is from coal-burning electrical power plants, which generate about 50% of our electricity. The extent to which SO 2 emissions are a problem when coal is burned depends on the amount of sulfur in the coal. Because of concern about SO 2 pollution, low-sulfur coal is in greater demand and is thus more expensive. Much of the coal from east of the Mississippi is relatively high in sulfur content, up to 6% by mass. Much of the coal from the western states has a lower sulfur content, but also a lower heat content per unit mass, so the difference in sulfur content per unit of heat produced is not as large as is often assumed. In 2010, the U.S. Environmental Protection Agency set new standards to reduce SO 2 emissions, the first change in nearly 40 years. The old standard of 140 parts per billion, measured over 24 h, has been replaced by a standard of 75 parts per billion, measured over 1 h. The impact of SO 2 emissions is not restricted to the United States, however. China, which generates about 70% of its energy from coal, is the world s largest generator of SO 2, producing about 35 * g annually from coal and other sources. As a result, that nation has a major problem with SO 2 pollution and has set targets to reduce emissions with some success. India, which is projected to surpass China as the largest importer of coal by 2014, is also concerned about increased SO 2 emissions. Nations will need to work together to address what has truly become a global issue. Sulfur dioxide is harmful to both human health and property; furthermore, atmospheric SO 2 can be oxidized to SO 3 by several pathways (such as reaction with O 2 or O 3 ). When SO 3 dissolves in water, it produces sulfuric acid: SO 3 1g2 2 O1l2 2 SO 4 1aq2 Many of the environmental effects ascribed to SO 2 are actually due to 2 SO 4. The presence of SO 2 in the atmosphere and the sulfuric acid it produces result in the phenomenon of acid rain. (Nitrogen oxides, which form nitric acid, are also major contributors to acid rain.) Uncontaminated rainwater generally has a p value of about 5.6. The primary source of this natural acidity is CO 2, which reacts with water to form carbonic acid, 2 CO 3. Acid rain typically has a p value of about 4. This shift toward greater acidity has affected many lakes in northern Europe, the northern United States, and Canada, reducing fish populations and affecting other parts of the ecological network in the lakes and surrounding forests. The p of most natural waters containing living organisms is between 6.5 and 8.5, but as Figure 18.7 shows, freshwater p values are far below 6.5 in many parts of the continental United States. At p levels below 4.0, all vertebrates, most invertebrates, and many microorganisms are destroyed. The lakes most susceptible to damage are those with low concentrations of basic ions, such as CO 3 -, that would act as a buffer to minimize changes in p. Some of these lakes are recovering as sulfur emissions from fossil

115 section 18.2 uman Activities and Earth s Atmosphere 785 Go Figure Why is the p found in freshwater sources in the Eastern half of the United States dramatically lower than found in the western half? Lab p < 4.3 Figure 18.7 Water p values from freshwater sites across the United States, The numbered dots indicate the locations of monitoring stations. (a) (b) Figure 18.8 Damage from acid rain. The right photograph, recently taken, shows how the statue has lost detail in its carvings. fuel combustion decrease, in part because of the Clean Air Act. In the period the average ambient air concentration of SO 2 nationwide has declined by 75%. Because acids react with metals and with carbonates, acid rain is corrosive both to metals and to stone building materials. Marble and limestone, for example, whose major constituent is CaCO 3, are readily attacked by acid rain ( Figure 18.8). Billions of dollars each year are lost because of corrosion due to SO 2 pollution.

116 786 chapter 18 Chemistry of the Environment Go Figure What is the major solid product resulting from removal of SO 2 from furnace gas? 1 Powdered limestone 3 CaO reacts with SO 2 from 4 CaSO 3 and unreacted SO 2 (CaCO 3 ) and air reaction S O 2 SO 2 removed by passing through injected into furnace to form CaSO 3 aqueous suspension of CaO 6 Cleaner air expelled through stack igh-sulfur coal CaCO 3 (s) CaO(s) CO 2 (g) CaO(s) SO 2 (g) CaSO 3 (s) Furnace 2 CaCO 3 decomposes to CaO (lime) and CO 2 Water slurry removed 5 CaSO 3 precipitated into watery slurry Figure 18.9 One method for removing SO 2 from combusted fuel. One way to reduce the quantity of SO 2 released into the environment is to remove sulfur from coal and oil before these fuels are burned. Although difficult and expensive, several methods have been developed. Powdered limestone 1CaCO 3 2, for example, can be injected into the furnace of a power plant, where it decomposes into lime (CaO) and carbon dioxide: CaCO 3 1s2 CaO1s2 CO 2 1g2 The CaO then reacts with SO 2 to form calcium sulfite: CaO1s2 SO 2 1g2 CaSO 3 1s2 The solid particles of CaSO 3, as well as much of the unreacted SO 2, can be removed from the furnace gas by passing it through an aqueous suspension of CaO ( Figure 18.9). Not all the SO 2 is removed, however, and given the enormous quantities of coal and oil burned worldwide, pollution by SO 2 will probably remain a problem for some time. Give It Some Thought What chemical behavior associated with sulfur oxides gives rise to acid rain? Figure Photochemical smog is produced largely by the action of sunlight on vehicle exhaust gases. Nitrogen Oxides and Photochemical Smog Nitrogen oxides are primary components of smog, a phenomenon with which city dwellers are all too familiar. The term smog refers to the pollution condition that occurs in certain urban environments when weather conditions produce a relatively stagnant air mass. The smog made famous by Los Angeles, but now common in many other urban areas as well, is more accurately described as photochemical smog because photochemical processes play a major role in its formation ( Figure 18.10). The majority of nitrogen oxide emissions (about 50%) comes from cars, buses, and other forms of transportation. Nitric oxide, NO, forms in small quantities in the cylinders of internal combustion engines in the reaction N 2 1g2 O 2 1g2 2 NO1g2 = kj [18.11] As noted in the Chemistry Put to Work box in Section 15.7, the equilibrium constant for this reaction increases from about at 300 K to about 0.05 at 2400 K

117 section 18.2 uman Activities and Earth s Atmosphere 787 (approximate temperature in the cylinder of an engine during combustion). Thus, the reaction is more favorable at higher temperatures. In fact, some NO is formed in any high-temperature combustion. As a result, electrical power plants are also major contributors to nitrogen oxide pollution. Before the installation of pollution-control devices on automobiles, typical emission levels of NO x were 4 g>mi. (The x is either 1 or 2 because both NO and NO 2 are formed, although NO predominates.) Starting in 2004, the auto emission standards for NO x called for a phased-in reduction to 0.07 g>mi by 2009, which was achieved. In air, nitric oxide is rapidly oxidized to nitrogen dioxide: 2 NO1g2 O 2 1g2 2 NO 2 1g2 = kj [18.12] The equilibrium constant for this reaction decreases from about at 300 K to about 10-5 at 2400 K. The photodissociation of NO 2 initiates the reactions associated with photochemical smog. Dissociation of NO 2 requires 304 kj>mol, which corresponds to a photon wavelength of 393 nm. In sunlight, therefore, NO 2 dissociates to NO and O: NO 2 1g2 hn NO1g2 O1g2 [18.13] The atomic oxygen formed undergoes several reactions, one of which gives ozone, as described earlier: O1g2 O 2 M1g2 O 3 1g2 M*1g2 [18.14] Although it is an essential UV screen in the upper atmosphere, ozone is an undesirable pollutant in the troposphere. It is extremely reactive and toxic, and breathing air that contains appreciable amounts of ozone can be especially dangerous for asthma sufferers, exercisers, and the elderly. We therefore have two ozone problems: excessive amounts in many urban environments, where it is harmful, and depletion in the stratosphere, where its presence is vital. In addition to nitrogen oxides and carbon monoxide, an automobile engine also emits unburned hydrocarbons as pollutants. These organic compounds are the principal components of gasoline and of many compounds we use as fuel (propane, C 3 8, and butane, C 4 10 ; for example), but are major ingredients of smog. A typical engine without effective emission controls emits about 10 to 15 g of hydrocarbons per mile. Current standards require that hydrocarbon emissions be less than g>mi. ydrocarbons are also emitted naturally from living organisms (see A Closer Look box later in this section). Reduction or elimination of smog requires that the ingredients essential to its formation be removed from automobile exhaust. Catalytic converters reduce the levels of NO x and hydrocarbons, two of the major ingredients of smog. (See the Chemistry Put to Work: Catalytic Converters in Section 14.7.) Give It Some Thought What photochemical reaction involving nitrogen oxides initiates the formation of photochemical smog? Greenhouse Gases: Water Vapor, Carbon Dioxide, and Climate In addition to screening out harmful short-wavelength radiation, the atmosphere is essential in maintaining a reasonably uniform and moderate temperature on Earth s surface. Earth is in overall thermal balance with its surroundings. This means that the planet radiates energy into space at a rate equal to the rate at which it absorbs energy from the Sun. Figure shows the distribution of radiation to and from Earth s surface, and Figure shows which portion of the infrared radiation leaving the surface is absorbed by atmospheric water vapor and carbon dioxide. In absorbing this radiation, these two atmospheric gases help maintain a livable uniform temperature at the surface by holding in, as it were, the infrared radiation, which we feel as heat.

118 788 chapter 18 Chemistry of the Environment Go Figure What fraction of the incoming solar radiation is absorbed by Earth s surface? = 100 watts for each square meter IR and longer wavelengths of radiation out through atmosphere 235 W/m 2 Back radiation Solar radiation reflected by surface 30 W/m 2 Solar radiation reflected by clouds, asteroids and atmosphere 77 W/m 2 Greenhouse gases Latent heat of condensation 78 W/m 2 IR absorbed by surface 324 W/m 2 Incoming solar radiation 342 W/m 2 Solar radiation absorbed by atmosphere and clouds 67 W/m 2 IR emitted from surface 390 W/m 2 Solar radiation absorbed by surface 168 W/m 2 Figure Earth s thermal balance. The amount of radiation reaching the surface of the planet is approximately equal to the amount radiated back into space. Wavelengths absorbed by atmospheric CO 2 and 2 O CO 2 Radiation intensity 2 O 10,000 20,000 30,000 Wavelength (nm) Infrared radiation emitted from Earth s surface Figure Portions of the infrared radiation emitted by Earth s surface that are absorbed by atmospheric CO 2 and 2 O.

119 section 18.2 uman Activities and Earth s Atmosphere 789 The influence of 2 O, CO 2, and certain other atmospheric gases on Earth s temperature is called the greenhouse effect because in trapping infrared radiation these gases act much like the glass of a greenhouse. The gases themselves are called greenhouse gases. Water vapor makes the largest contribution to the greenhouse effect. The partial pressure of water vapor in the atmosphere varies greatly from place to place and time to time but is generally highest near Earth s surface and drops off with increasing elevation. Because water vapor absorbs infrared radiation so strongly, it plays the major role in maintaining the atmospheric temperature at night, when the surface is emitting radiation into space and not receiving energy from the Sun. In very dry desert climates, where the watervapor concentration is low, it may be extremely hot during the day but very cold at night. In the absence of a layer of water vapor to absorb and then radiate part of the infrared radiation back to Earth, the surface loses this radiation into space and cools off very rapidly. Carbon dioxide plays a secondary but very important role in maintaining the surface temperature. The worldwide combustion of fossil fuels, principally coal and oil, on a prodigious scale in the modern era has sharply increased carbon dioxide levels in the atmosphere. To get a sense of the amount of CO 2 produced for example, by the combustion of hydrocarbons and other carbon-containing substances, which are the components of fossil fuels consider the combustion of butane, C Combustion of 1.00 g of C 4 10 produces 3.03 g of CO 2. (Section 3.6) Similarly, a gallon (3.78 L) of gasoline (density 0.7 g>ml, approximate composition C 8 18 ) produces about 8 kg of CO 2. Combustion of fossil fuels releases about 2.2 * g (24 billion tons) of CO 2 into the atmosphere annually, with the largest quantity coming from transportation vehicles. Much CO 2 is absorbed into oceans or used by plants. Nevertheless, we are now generating CO 2 much faster than it is being absorbed or used. Analysis of air trapped in ice cores taken from Antarctica and Greenland makes it possible to determine the atmospheric levels of CO 2 during the past 160,000 years. These measurements reveal that the level of CO 2 remained fairly constant from the last Ice Age, some 10,000 years ago, until roughly the beginning of the Industrial Revolution, about 300 years ago. Since that time, the concentration of CO 2 has increased by about 30% to a current high of about 400 ppm ( Figure 18.13). Climate scientists believe that the CO 2 level has not been this high since 3 to 5 million years ago. Go Figure What is the source of the slight but steady increase in slope of this curve over time? 400 Mauna Loa Observatory, awaii CO 2 concentration (ppm) Year Figure Rising CO 2 levels. The sawtooth shape of the graph is due to regular seasonal variations in CO 2 concentration for each year.

120 790 chapter 18 Chemistry of the Environment The consensus among climate scientists is that the increase in atmospheric CO 2 is perturbing Earth s climate and is very likely playing a role in the observed increase in the average global air temperature of C over the past century. Scientists often use the term climate change instead of global warming to refer to this effect because as the Earth s temperature increases, winds and ocean currents are affected in ways that cool some areas and warm others. On the basis of present and expected future rates of fossil-fuel use, the atmospheric CO 2 level is expected to double from its present level sometime between 2050 and Computer models predict that this increase will result in an average global temperature increase of 1 C to 3 C. Because so many factors go into determining climate, we cannot predict with certainty what changes will occur because of this warming. Clearly, however, humanity has acquired the potential, by changing the concentrations of CO 2 and other heat-trapping gases in the atmosphere, to substantially alter the climate of the planet. The climate change threat posed by atmospheric CO 2 has sparked considerable research into ways of capturing the gas at its largest combustion sources and storing it under ground or under the seafloor. There is also much interest in developing new ways to use CO 2 as a chemical feedstock. owever, the approximately 115 million tons of CO 2 used annually by the global chemical industry is but a small fraction of the approximately 24 billion tons of annual CO 2 emissions. The use of CO 2 as a raw material will probably never be great enough to significantly reduce its atmospheric concentration. Give It Some Thought Explain why nighttime temperatures remain higher in locations where there is higher humidity. A Closer Look Other Greenhouse Gases Although CO 2 receives most of the attention, other gases contribute to the greenhouse effect, including methane, C 4, hydrofluorocarbons (FCs), and chlorofluorocarbons (CFCs). FCs have replaced CFCs in a host of applications, including refrigerants and air-conditioner gases. Although they do not contribute to the depletion of the ozone layer, FCs are nevertheless potent greenhouse gases. For example, one of the byproduct molecules from production of FCs that are used in commerce is CF 3, which is estimated to have a global warming potential, gram for gram, more than 14,000 times that of CO 2. The total concentration of FCs in the atmosphere has been increasing about 10% per year. Thus, these substances are becoming increasingly important contributors to the greenhouse effect. Methane already makes a significant contribution to the greenhouse effect. Studies of atmospheric gas trapped long ago in the Greenland and Antarctic ice sheets show that the atmospheric methane concentration has increased from preindustrial values of 0.3 to 0.7 ppm to the present value of about 1.8 ppm. The major sources of methane are associated with agriculture and fossil-fuel use. Methane is formed in biological processes that occur in low-oxygen environments. Anaerobic bacteria, which flourish in swamps and landfills, near the roots of rice plants, and in the digestive systems of cows and other ruminant animals, produce methane ( Figure 18.14). It also leaks into the atmosphere during natural-gas extraction and transport. It is estimated that about two-thirds of present-day methane emissions, which are increasing by about 1% per year, are related to human activities. Methane has a half-life in the atmosphere of about 10 years, whereas CO 2 is much longer-lived. This might seem a good thing, Figure Methane production. Ruminant animals, such as cows and sheep, produce methane in their digestive systems. but there are indirect effects to consider. Methane is oxidized in the stratosphere, producing water vapor, a powerful greenhouse gas that is otherwise virtually absent from the stratosphere. In the troposphere, methane is attacked by reactive species such as O radicals or nitrogen oxides, eventually producing other greenhouse gases, such as O 3. It has been estimated that on a per-molecule level, the global warming potential of C 4 is about 21 times that of CO 2. Given this large contribution, important reductions of the greenhouse effect could be achieved by reducing methane emissions or capturing the emissions for use as a fuel. Related Exercises: 18.67, 18.69

121 section 18.3 Earth s Water Earth s Water Water covers 72% of Earth s surface and is essential to life. Our bodies are about 65% water by mass. Because of extensive hydrogen bonding, water has unusually high melting and boiling points and a high heat capacity. (Section 11.2) Water s highly polar character is responsible for its exceptional ability to dissolve a wide range of ionic and polarcovalent substances. Many reactions occur in water, including reactions in which 2 O itself is a reactant. Recall, for example, that 2 O can participate in acid base reactions as either a proton donor or a proton acceptor. (Section 16.3) All these properties play a role in our environment. The Global Water Cycle All the water on Earth is connected in a global water cycle ( Figure 18.15). Most of the processes depicted here rely on the phase changes of water. For instance, warmed by the Sun, liquid water in the oceans evaporates into the atmosphere as water vapor and condenses into liquid water droplets that we see as clouds. Water droplets in the clouds can crystallize to ice, which can precipitate as hail or snow. Once on the ground, the hail or snow melts to liquid water, which soaks into the ground. If conditions are right, it is also possible for ice on the ground to sublime to water vapor in the atmosphere. Give It Some Thought Consider the phase diagram for water shown in Figure (page 465). In what pressure range and in what temperature range must 2 O exist in order for 2 O1s2 to sublime to 2 O1g2? Go Figure Which processes shown in this figure involve the phase transition 2 O1l 2 2 O1g2? Water storage in ice and snow Precipitation Snowmelt runoff to streams Condensation Deposition Sublimation Evaporation Fog drip Surface runoff Evaporation, transpiration Water storage in the atmosphere Evaporation Groundwater discharge Freshwater storage Plant uptake Groundwater storage Water storage in oceans Figure The global water cycle.

122 792 chapter 18 Chemistry of the Environment Salt Water: Earth s Oceans and Seas The vast layer of salty water that covers so much of the planet is in actuality one large connected body and is generally constant in composition. For this reason, oceanographers speak of a world ocean rather than of the separate oceans we learn about in geography books. The world ocean is huge, having a volume of 1.35 * 10 9 km 3 and containing 97.2% of all the water on Earth. Of the remaining 2.8%, 2.1% is in the form of ice caps and glaciers. All the freshwater in lakes, in rivers, and in the ground amounts to only 0.6%. Most of the remaining 0.1% is in brackish (salty) water, such as that in the Great Salt Lake in Utah. Seawater is often referred to as saline water. The salinity of seawater is the mass in grams of dry salts present in 1 kg of seawater. In the world ocean, salinity averages about 35. To put it another way, seawater contains about 3.5% dissolved salts by mass. The list of elements present in seawater is very long. Most, however, are present only in very low concentrations. Table 18.5 lists the 11 ionic species most abundant in seawater. Seawater temperature varies as a function of depth ( Figure 18.16), as does salinity and density. Sunlight penetrates well only 200 m into the water; the region between 200 m and 1000 m deep is the twilight zone, where visible light is faint. Below 1000 m, the ocean is pitch-black and cold, about 4 C. The transport of heat, salt, and other chemicals throughout the ocean is influenced by these changes in the physical properties of seawater, and in turn the changes in the way heat and substances are transported affects ocean currents and the global climate. The sea is so vast that if the concentration of a substance in seawater is 1 part per billion (1 * 10-6 g>kg of water), there is 1 * kg of the substance in the world ocean. Nevertheless, because of high extracting costs, only three substances are obtained from seawater in commercially important amounts: sodium chloride, bromine (from bromide salts), and magnesium (from its salts). Absorption of CO 2 by the ocean plays a large role in global climate. Because carbon dioxide and water form carbonic acid, the 2 CO 3 concentration in the ocean increases as the water absorbs atmospheric CO 2. Most of the carbon in the ocean, however, is in the form of CO 3 - and CO 3 2- ions, which form a buffer system that maintains the ocean s p between 8.0 and 8.3. The p of the ocean is predicted to decrease as the concentration of CO 2 in the atmosphere increases, as discussed in the Chemistry and Life box on ocean acidification in Section Freshwater and Groundwater Freshwater is the term used to denote natural waters that have low concentrations (less than 500 ppm) of dissolved salts and solids. Freshwater includes the waters of Table 18.5 Ionic Constituents of Seawater Present in Concentrations Greater than g,kg 11 ppm2 Ionic Constituent Salinity Concentration (M) Chloride, Cl Sodium, Na Sulfate, SO Magnesium, Mg Calcium, Ca Potassium, K Carbon dioxide* * 10-3 Bromide, Br * 10-4 Boric acid, 3 BO * 10-4 Strontium, Sr * 10-5 Fluoride, F * 10-5 *CO 2 is present in seawater as CO 3 - and CO 3 2-.

123 section 18.3 Earth s Water 793 Go Figure ow would you expect the temperature variation to affect the density of seawater in the range 0 to 100 m depth? 20 Temperature ( C) Depth (m) Figure Typical average temperature of mid-latitude seawater as a function of depth. lakes, rivers, ponds, and streams. The United States is fortunate in its abundance of freshwater:1.7 * L (660 trillion gallons) is the estimated reserve, which is renewed by rainfall. An estimated 9 * L of freshwater is used every day in the United States. Most of this is used for agriculture (41%) and hydroelectric power (39%), with small amounts for industry (6%), household needs (6%), and drinking water (1%). An adult drinks about 2 L of water per day. In the United States, our daily use of water per person far exceeds this subsistence level, amounting to an average of about 300 L/ day for personal consumption and hygiene. We use about 8 L/person for cooking and drinking, about 120 L/person for cleaning (bathing, laundering, and housecleaning), 80 L>person for flushing toilets, and 80 L/person for watering lawns. The total amount of freshwater on Earth is not a very large fraction of the total water present. Indeed, freshwater is one of our most precious resources. It forms by evaporation from the oceans and the land. The water vapor that accumulates in the atmosphere is transported by global atmospheric circulation, eventually returning to Earth as rain, snow, and other forms of precipitation (Figure 18.15). As water runs off the land on its way to the oceans, it dissolves a variety of cations (mainly Na, K, Mg 2, Ca 2, and Fe 2 ), anions (mainly Cl -, SO 4 2-, and CO 3 - ), and gases (principally O 2, N 2, and CO 2 ). As we use water, it becomes laden with additional dissolved material, including the wastes of human society. As our population and output of environmental pollutants increase, ever-increasing amounts of money and resources must be spent to guarantee a supply of freshwater. The availability and cost of fresh water that is clean enough to sustain daily life varies greatly among nations. To illustrate, daily fresh water use in the United States approaches 600 L/person, whereas in the relatively underdeveloped nations of sub- Sahara Africa it is only about 30 L. To make matters worse, for many people, water is not only scarce, it is so contaminated that it is a continuing source of diseases. Approximately 20% of the world s freshwater is under the soil, in the form of groundwater. Groundwater resides in aquifers, which are layers of porous rock that hold water. The water in aquifers can be very pure, and accessible for human consumption if near the surface. Dense underground formations that do not allow water to readily penetrate can hold groundwater for years or even millennia. When their water is removed by drilling and pumping, such aquifers are slow to recharge via the diffusion of surface water.

124 794 chapter 18 Chemistry of the Environment A Closer Look The Ogallala Aquifer A shrinking resource The Ogallala Aquifer, also referred to as the igh Plains Aquifer, is an enormous underground water body of water lying beneath the Great Plains of the United States. One of the world s largest aquifers, it covers an area of approximately 450,000 km2 1170,000 mi22 encompassing portions of the eight states of South Dakota, Nebraska, Wyoming, Colorado, Kansas, Oklahoma, New Mexico, and Texas. ( Figure 18.17). The depth of the Ogallala formation that gives rise to the aquifer ranges from about 120 m, to more than 300 m, particularly in the northern portion. Before large-scale pumping in the modern era, the depth of water in the aquifer ranged up to more than 120 m in the northern portion. Anyone who has flown over the Great Plains is familiar with the view of huge circles made by the center pivot irrigators nearly covering the land. The center post irrigation system, developed in the post World War II era, permitted application of water onto large areas. As a result, the Great Plains became one of the most productive agricultural areas in the world. Unfortunately, the premise that the aquifer is an inexhaustible source of fresh water proved false. Recharge of the aquifer from surface water is slow, taking hundreds, perhaps thousands of years. Recently, water levels in many regions of the Ogallala have declined to the point where the costs of bringing water to the surface have become prohibitive. As the aquifer levels continue to drop, less water will be available for the needs of cities, residences and businesses. Related Exercises: 18.41, SOUT DAKOTA WYOMING Northern igh Plains Scottsbluff Cheyenne North Platte COLORADO NEBRASKA KANSAS Garden City Central igh Plains Wichita Guymon NEW MEXICO OKLAOMA Amarillo Elevation, in feet 8,000 Southern igh Plains 1,000 Lubbock TEXAS Odessa Midland miles kilometers Figure A map showing the extent of the Ogallala (igh Plains) aquifer. Note that the elevation of the land varies greatly. The aquifer follows the topography of the formations that underlie the area. The nature of the rock that contains the groundwater has a large influence on the water s chemical composition. If minerals in the rock are water soluble to some extent, ions can leach out of the rock and remain dissolved in the groundwater. Arsenic in the form of AsO42-, 2AsO4-, and 3AsO3 is found in many groundwater sources across the world, most infamously in Bangladesh, at concentrations poisonous to humans uman Activities and Water Quality All life on Earth depends on the availability of suitable water. Many human activities entail waste disposal into natural waters without any treatment. These practices result in contaminated water that is detrimental to both plant and animal aquatic life. Unfortunately people in many parts of the world do not have access to water that has been treated to remove harmful contaminants, including disease-bearing bacteria. Dissolved Oxygen and Water Quality The amount of O2 dissolved in water is an important indicator of water quality. Water fully saturated with air at 1 atm and 20 C contains about 9 ppm of O2. Oxygen

125 section 18.4 uman Activities and Water Quality 795 is necessary for fish and most other aquatic life. Cold-water fish require water containing at least 5 ppm of dissolved oxygen for survival. Aerobic bacteria consume dissolved oxygen to oxidize organic materials for energy. The organic material the bacteria are able to oxidize is said to be biodegradable. Excessive quantities of biodegradable organic materials in water are detrimental because they remove the oxygen necessary to sustain normal animal life. Typical sources of these biodegradable materials, which are called oxygen-demanding wastes, include sewage, industrial wastes from food-processing plants and paper mills, and liquid waste from meatpacking plants. In the presence of oxygen, the carbon, hydrogen, nitrogen, sulfur, and phosphorus in biodegradable material end up mainly as CO 2, CO 3 -, 2 O, NO 3 -, SO 4 2-, and phosphates. The formation of these oxidation products sometimes reduces the amount of dissolved oxygen to the point where aerobic bacteria can no longer survive. Anaerobic bacteria then take over the decomposition process, forming C 4, N 3, 2 S, P 3, and other products, several of which contribute to the offensive odors of some polluted waters. Plant nutrients, particularly nitrogen and phosphorus, contribute to water pollution by stimulating excessive growth of aquatic plants. The most visible results of excessive plant growth are floating algae and murky water. What is more significant, however, is that as plant growth becomes excessive, the amount of dead and decaying plant matter increases rapidly, a process called eutrophication ( Figure 18.18). The processes by which plants decay consumes O 2, and without sufficient oxygen, the water cannot sustain animal life. The most significant sources of nitrogen and phosphorus compounds in water are domestic sewage (phosphate-containing detergents and nitrogen-containing body wastes), runoff from agricultural land (fertilizers contain both nitrogen and phosphorus), and runoff from livestock areas (animal wastes contain nitrogen). Figure Eutrophication. This rapid accumulation of dead and decaying plant matter in a body of water uses up the water s oxygen supply, making the water unsuitable for aquatic animals. Go Figure What feature of this process is responsible for its being called reverse osmosis? Give It Some Thought If a test on a sample of polluted water shows a considerable decrease in dissolved oxygen over a five-day period, what can we conclude about the nature of the pollutants present? Water Purification: Desalination Because of its high salt content, seawater is unfit for human consumption and for most of the uses to which we put water. In the United States, the salt content of municipal water supplies is restricted by health codes to no more than about 0.05% by mass. This amount is much lower than the 3.5% dissolved salts present in seawater and the 0.5% or so present in brackish water found underground in some regions. The removal of salts from seawater or brackish water to make the water usable is called desalination. Water can be separated from dissolved salts by distillation because water is a volatile substance and the salts are nonvolatile. (Section 1.3, Separation of Mixtures ) The principle of distillation is simple enough, but carrying out the process on a large scale presents many problems. As water is distilled from seawater, for example, the salts become more and more concentrated and eventually precipitate out. Distillation is also an energy-intensive process. Seawater can also be desalinated using reverse osmosis. Recall that osmosis is the net movement of solvent molecules, but not solute molecules, through a semipermeable membrane. (Section 13.5) In osmosis, the solvent passes from the more dilute solution into the more concentrated one. owever, if sufficient external pressure is applied, osmosis can be stopped and, at still higher pressures, reversed. When reverse osmosis occurs, solvent passes from the more concentrated into the more dilute solution. In a modern reverse-osmosis facility, hollow fibers are used as the semipermeable membrane ( Figure 18.19). Saline water (water containing significant salts) is introduced under pressure into the fibers, and desalinated water is recovered. The world s largest desalination plant, in Jubail, Saudi Arabia, provides 50% of that country s drinking water by using reverse osmosis to desalinate seawater from the Persian Gulf. An even larger plant, which will produce 600 million L/day (160 million gallons) of drinking water, is scheduled for completion in Saudi Arabia in Such plants are becoming increasingly common in the United States. The largest, near ollow fibers of semipermeable membrane Water molecules pushed into hollow fibers Permeator Seawater pumped through at high pressure Desalinated water to collector Fiber Few solute particles enter hollow fibers Figure Reverse osmosis.

126 796 chapter 18 Chemistry of the Environment Go Figure What is the primary function of the aeration step in water treatment? Storage tank CaO, Al 2 (SO 4 ) 3 added Aeration Chlorine sterilizers To users Water intake Coarse filtration screen Sedimentation tanks Sand filter Figure Common steps in treating water for a public water system. Tampa Bay, Florida, has been operating since 2007 and produces 35 million gallons of drinking water a day by reverse osmosis. Small-scale, manually operated reverseosmosis desalinators are used in camping, traveling, and at sea. Carbon removes iodine smells and parasites Iodine-impregnated beads kill bacteria, viruses, and parasites 15-μm textile filter removes debris 100-μm textile filter removes debris Figure A LifeStraw purifies water as it is drunk. Water Purification: Municipal Treatment The water needed for domestic, agricultural, and industrial use is taken either from lakes, rivers, and underground sources or from reservoirs. Much of the water that finds its way into municipal water systems is used water, meaning it has already passed through one or more sewage systems or industrial plants. Consequently, this water must be treated before it is distributed to our faucets. Municipal water treatment usually involves five steps ( Figure 18.20). After coarse filtration through a screen, the water is allowed to stand in large sedimentation tanks where sand and other minute particles settle out. To aid in removing very small particles, the water may first be made slightly basic with CaO. Then Al 2 1SO is added and reacts with O - ions to form a spongy, gelatinous precipitate of Al1O2 3 1K sp = 1.3 * This precipitate settles slowly, carrying suspended particles down with it, thereby removing nearly all finely divided matter and most bacteria. The water is then filtered through a sand bed. Following filtration, the water may be sprayed into the air (aeration) to hasten oxidation of dissolved inorganic ions of iron and manganese, reduce concentrations of any 2 S or N 3 that may be present and reduce bacterial concentrations. The final step normally involves treating the water with a chemical agent to ensure the destruction of bacteria. Ozone is more effective, but chlorine is less expensive. Liquefied Cl 2 is dispensed from tanks through a metering device directly into the water supply. The amount used depends on the presence of other substances with which the chlorine might react and on the concentrations of bacteria and viruses to be removed. The sterilizing action of chlorine is probably due not to Cl 2 itself but to hypochlorous acid, which forms when chlorine reacts with water: Cl 2 1aq2 2 O1l2 ClO1aq2 1aq2 Cl - 1aq2 [18.15] It is estimated that about 800 million people worldwide lack access to clean water. According to the United Nations, 95% of the world s cities still dump raw sewage into their water supplies. Thus, it should come as no surprise that 80% of all the health maladies in developing countries can be traced to waterborne diseases associated with unsanitary water. One promising development is a device called the LifeStraw ( Figure 18.21). When a person sucks water through the straw, the water first encounters a textile

127 section 18.4 uman Activities and Water Quality 797 A Closer Look Fracking and Water Quality In recent years fracking, short for hydraulic fracturing, has become widely used to Waste water ponds greatly increase the availability of petroleum reserves. In fracking, a large volume of Fracking fluid water, typically two million gallons or more, Contaminants escape to surroundings mixed with various additives, is injected at high pressure into wellbores extended horizontally into rock formations ( Figure ). The water is laden with sand, ceramic materishallow aquifer als and other additives, including gels, foams, Impermeable layer and compressed gases, that serve to increase the yield in the process. The high pressure Deep aquifer fluid finds its way into tiny faults in geological formations, releasing petroleum and natural Migration of released gases Impermeable layer gas. Fracking has greatly increased petroleum reserves, particularly of natural gas, in many Gas-bearing formation parts of the world. The technique has been so Methane productive that more than 20,000 new wells are being drilled annually in the United States ydraulic fractures alone, in all areas of the country. Unfortunately, the potential for environmental damage from fracking is significant. The large volume of fracking fluid Figure A schematic of a well site employing fracking. The yellow arrows required to create a well must be returned indicate the avenues through which contaminants enter the environment. to the surface. Without purification the fluid is rendered unfit for other uses, and a variety of gases, including methane and other hydrocarbons, from becomes a large-scale environmental problem. Often the waste w ater the w ellheads contributes to air pollution. In a study published in is allowed to sit in open waste water pits. The 2005 Energy Policy act 2013, methane emissions to the atmosphere during hydraulic frackand other federal legislation exempts hydraulic fracturing o perations ing operations in Utah were estimated to be in the range of 6 12% from certain provisions of the Safe Drinking Water act and other of the amount of methane produced. As related in the Closer Look regulations. Some areas of the country that are already facing water box on page 790, methane is a potent greenhouse gas. shortages thus have one more large demand for a limited supply. The many environmental issues surrounding the practice of frack Because fracturing of rock formations increases the pathways for ing have generated widespread concern and adverse public reaction. flows of petroleum and various gases, bodies of underground water Fracking represents yet one more instance of the conflict between those that have been serving as municipal water supplies or wells for inwho advocate the availability of low cost energy and those who are dividual homes in some locales have become contaminated with more focused on sustaining long term the quality of the environment. petroleum, hydrogen sulfide and other toxic substances. The escape of filter with a mesh opening of 100 μm followed by a second textile filter with a mesh opening of 15 μm. These filters remove debris and even clusters of bacteria. The water next encounters a chamber of iodine-impregnated beads, where bacteria, viruses, and parasites are killed. Finally, the water passes through granulated active carbon, which removes the smell of iodine as well as the parasites that have not been taken by the filters or killed by the iodine. At present the Lifestraw is too costly to permit widespread use in underdeveloped countries, but there is hope that its cost can be greatly reduced. Water disinfection is one of the greatest public health innovations in human history. It has dramatically decreased the incidences of waterborne bacterial diseases such as cholera and typhus. owever, this great benefit comes at a price. In 1974 scientists in Europe and the United States discovered that chlorination of water produces a group of by-products previously undetected. These by-products are called trihalomethanes (TMs) because all have a single carbon atom and three halogen atoms: CCl3, CCl2Br, CClBr2, and CBr3. These and many other chlorineand bromine-containing organic substances are produced by the reaction of dissolved chlorine with the organic materials present in nearly all natural waters, as well as with