Chapter 16 Acid Base Equilibria

Chapter 16 Acid Base Equilibria 2015 Pearson Education, Inc.

Acid Base Equilibria 16.1 : A Brief Review 16.2 Brønsted Lowry 16.3 The Autoionization of Water 16.4 The ph Scale 16.5 Strong Balsamic Vinegar 16.6 Weak 16.7 Weak 16.8 Relationship between K a K b 16.9 Acid Base Properties of Salt Solutions 16.10 Acid Base Behavior Chemical Structure 16.11 Lewis 2014 Pearson Education, Inc.

Arrhenius Arrhenius Acid Base An acid is a substance that, when dissolved in water, increases the concentration of hydrogen ions. A base is a substance that, when dissolved in water, increases the concentration of hydroxide ions. 2015 Pearson Education

Brønsted Lowry Acid Base Proton donor/proton acceptor. A Brønsted Lowry acid must have at least one removable (acidic) proton (H + ) to donate. A Brønsted Lowry base must have at least one nonbonding pair of electrons to accept a proton (H + ). 2015 Pearson Education Hydronium ion

Proton-Transfer Reaction 2015 Pearson Education

What Is Different about Water? Water can act as a Brønsted Lowry base accept a proton (H + ) from an acid, as on the previous slide. It can also donate a proton act as an acid, as is seen below. This makes water amphiprotic. 2015 Pearson Education

Conjugate The term conjugate means joined together as a pair. Reactions between acids bases always yield their conjugate bases acids. 2015 Pearson Education

Sample Exercise 16.1 Identifying Conjugate (a) What is the conjugate base of HClO 4, H 2 S, PH 4+, HCO 3? (b) What is the conjugate acid of CN, SO 4 2, H 2 O, HCO 3? Solution Analyze We are asked to give the conjugate base for several acids the conjugate acid for several bases. Plan The conjugate base of a substance is simply the parent substance minus one proton, the conjugate acid of a substance is the parent substance plus one proton. Solve (a) If we remove a proton from HClO 4, we obtain ClO 4, which is its conjugate base. The other conjugate bases are HS, PH 3, CO 2 3. (b) If we add a proton to CN, we get HCN, its conjugate acid. The other conjugate acids are HSO 4, H 3 O +, H 2 CO 3. Notice that the hydrogen carbonate ion (HCO 3 ) is amphiprotic. It can act as either an acid or a base. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Sample Exercise 16.2 Writing Equations for Proton-Transfer Reactions The hydrogen sulfite ion (HSO 3 ) is amphiprotic. Write an equation for the reaction of HSO 3 with water (a) in which the ion acts as an acid (b) in which the ion acts as a base. In both cases identify the conjugate acid base pairs. Solution Analyze Plan We are asked to write two equations representing reactions between HSO 3 water, one in which HSO 3 should donate a proton to water, thereby acting as a Brønsted Lowry acid, one in which HSO 3 should accept a proton from water, thereby acting as abase. We are also asked to identify the conjugate pairs in each equation. Solve (a) HSO 3 (aq)+ H 2 O(l) SO 2 3 (aq)+ H 3 O + (aq) The conjugate pairs in this equation are HSO 3 (acid) SO 2 3 (conjugate base), H 2 O (base) H 3 O + (conjugate acid). (b) HSO 3 (aq)+ H 2 O(l) H 2 SO 3 (aq)+ OH (aq) The conjugate pairs in this equation are H 2 O (acid) OH (conjugate base), HSO 3 (base) H 2 SO 3 (conjugate acid). Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Relative Strengths of The stronger an acid, the weaker its conjugate base, the stronger a base, the weaker its conjugate acid. 2015 Pearson Education

Acid Base Strength In every acid base reaction, equilibrium favors transfer of the proton from the stronger acid to the stronger base to form the weaker acid the weaker base. HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl (aq) H 2 O is a much stronger base than Cl, so the equilibrium lies far to the right (K >> 1). CH 3 COOH(aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO (aq) Acetate is a stronger base than H 2 O, so the equilibrium favors the left side (K < 1). 2015 Pearson Education

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 (K c < 1) or to the right (K c > 1): HSO 4 (aq)+ CO 2 3 (aq) SO 2 4 (aq)+ HCO 3 (aq) Solution Analyze We are asked to predict whether an equilibrium lies to the right, favoring products, or to the left, favoring reactants. Plan This is a proton-transfer reaction, the position of the equilibrium will favor the proton going to the stronger of two bases. The two bases in the equation are CO 3 2, the base in theforward reaction, SO 4 2, the conjugate base of HSO 4. We can find the relative positions of these two bases in Figure 16.4 to determine which is the stronger base. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Sample Exercise 16.3 Predicting the Position of a Proton-Transfer Equilibrium Continued Solve The CO 3 2 ion appears lower in the right-h column in Figure 16.4 is therefore a stronger base than SO 4 2. Therefore, CO 3 2 will get the proton preferentially to become HCO 3, while SO 4 2 will remain mostly unprotonated. The resulting equilibrium lies to the right, favoring products (that is, K c > 1): Comment Of the two acids HSO 4 HCO 3, the stronger one (HSO 4 ) gives up a proton more readily, the weaker one (HCO 3 ) tends to retain its proton. Thus, the equilibrium favors the direction in which the proton moves from the stronger acid becomes bonded to the stronger base. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Autoionization of Water Water is amphoteric. In pure water, a few molecules act as bases a few act as acids (2 per 10 9 ). This is referred to as autoionization. 2015 Pearson Education

Ion-Product Constant The equilibrium expression for this process is K c = [H 3 O + ][OH ] This special equilibrium constant is referred to as the ion-product constant for water, K w. At 25 C, K w = 1.0 10 14 2015 Pearson Education

Aqueous Solutions Can Be Acidic, Basic, or Neutral If a solution is neutral, [H + ] = [OH ]. If a solution is acidic, [H + ] > [OH ]. If a solution is basic, [H + ] < [OH ]. 2015 Pearson Education

Sample Exercise 16.5 Calculating [H + ]from [OH ] Calculate the concentration of H + (aq)in (a) a solution in which [OH ] is 0.010 M, (b) a solution in which [OH ] is 1.8 10 9 M. Note: In this problem all that follow, we assume, unless stated otherwise, that the temperature is 25. Solution Analyze We are asked to calculate the [H + ] concentration in an aqueous solution where the hydroxide concentration is known. Plan We can use the equilibrium-constant expression for the autoionization of water the value of K w to solve for each unknown concentration. Solve (a) Using Equation 16.16, we have This solution is basic because (b) In this instance This solution is acidic because Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

The ph Scale ph = log[h + ] poh = log[oh - ] The p in ph tells us to take the log of a quantity (in this case, hydrogen ions). 2015 Pearson Education

Sample Exercise 16.6 Calculating ph from [H + ] Calculate the ph values for the two solutions of Sample Exercise 16.5. Solution Analyze We are asked to determine the ph of aqueous solutions for which we have already calculated [H + ]. Plan We can calculate ph using its defining equation, Equation 16.17. Solve (a) In the first instance we found [H+] to be 1.0 10 12 M, so that ph = log(1.0 10 12 ) = ( 12.00) = 12.00 Because 1.0 10 12 has two significant figures, the ph has two decimal places, 12.00. (b) For the second solution, [H + ] = 5.6 10 6 M. Before performing the calculation, it is helpful to estimate the ph. To do so, we note that [H + ] lies between 1 10 6 1 10 5. Thus, we expect the ph to lie between 6.0 5.0. We use Equation 16.17 to calculate the ph: ph = log(5.6 10 6 ) = 5.25 Check After calculating a ph, it is useful to compare it to your estimate. In this case the ph, as we predicted, falls between 6 5. Had the calculated ph the estimate not agreed, we should have reconsidered our calculation or estimate or both. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Relating ph poh Because [H 3 O + ][OH ] = K w = 1.0 10 14 we can take the log of the equation log[h 3 O + ] + log[oh ] = log K w = 14.00 which results in ph + poh = pk w = 14.00 2015 Pearson Education

Sample Exercise 16.7 Calculating [H + ] from poh A sample of freshly pressed apple juice has a poh of 10.24. Calculate [H + ]. Solution Analyze We need to calculate [H + ] from poh. Plan We will first use Equation 16.20, ph + poh = 14.00, to calculate ph from poh. Then we will use Equation 16.17 to determine the concentration of H +. Solve From Equation 16.20, we have ph = 14.00 poh ph = 14.00 10.24 = 3.76 Next we use Equation 16.17: ph = log[h + ] = 3.76 Thus, log[h + ] = 3.76 To find [H + ], we need to determine the antilogarithm of 3.76. Your calculator will show this comm as 10 x or INV log (these functions are usually above the log key). We use this function to perform the calculation: [H + ] = antilog ( 3.76) = 10 3.76 = 1.7 10 4 M Comment The number of significant figures in [H + ] is two because the number of decimal places in the ph is two. Check Because the ph is between 3.0 4.0, we know that [H + ] will be between 1.0 10 3 M 1.0 10 4 M. Our calculated [H + ] falls within this estimated range. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

How Do We Measure ph? Indicators, including litmus paper, are used for less accurate measurements; an indicator is one color in its acid form another color in its basic form. ph meters are used for accurate measurement of ph; electrodes indicate small changes in voltage to detect ph. 2015 Pearson Education

2015 Pearson Education

Strong You will recall that the seven strong acids are HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 3, HClO 4. These are, by definition, strong electrolytes exist totally as ions in aqueous solution; e.g., HA + H 2 O H 3 O + + A So, for the monoprotic strong acids, [H 3 O + ] = [acid] 2015 Pearson Education

Sample Exercise 16.8 Calculating the ph of a Strong Acid What is the ph of a 0.040 M solution of HClO 4? Solution Analyze Plan Because HClO 4 is a strong acid, it is completely ionized, giving [H + ] = [ClO 4 ] = 0.040 M. Solve ph = log(0.040) = 1.40 Check Because [H + ] lies between 1 10 2 1 10 1, the ph will be between 2.0 1.0. Our calculated ph falls within the estimated range. Furthermore, because the concentration has two significant figures, the ph has two decimal places. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Strong Strong bases are the soluble hydroxides, which are the alkali metal heavier alkaline earth metal hydroxides (Ca 2+, Sr 2+, Ba 2+ ). Again, these substances dissociate completely in aqueous solution; e.g., NaOH(aq) Na + (aq) + OH (aq) Sr(OH) 2 (aq) Sr 2+ (aq) + 2 OH (aq) 2015 Pearson Education

Sample Exercise 16.9 Calculating the ph of a Strong Base What is the ph of (a) a 0.028 M solution of NaOH, (b) a 0.0011 M solution of Ca(OH) 2? Solution Analyze We are asked to calculate the ph of two solutions of strong bases. Plan We can calculate each ph by either of two equivalent methods. First, we could use Equation 16.16 to calculate [H + ] then use Equation 16.17 to calculate the ph. Alternatively, we could use [OH ] to calculate poh then use Equation 16.20 to calculate the ph. Solve (a) NaOH dissociates in water to give one OH ion per formula unit. Therefore, the OH concentration for the solution in (a) equals the stated concentration of NaOH, namely 0.028 M. Method 1: Method 2: poh = log(0.028) = 1.55 ph = 14.00 poh = 12.45 Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Sample Exercise 16.9 Calculating the ph of a Strong Base Continued (b) Ca(OH) 2 is a strong base that dissociates in water to give two OH ions per formula unit. Thus, the concentration of OH (aq) for the solution in part (b) is 2 (0.0011 M) = 0.0022 M. Method 1: Method 2: poh = log(0.0022) = 2.66 ph = 14.00 poh = 11.34 Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Weak Strong acids completely dissociate to ions. Weak acids only partially dissociate to ions. HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) 2015 Pearson Education

Strong vs. Weak Strong Acid: [H + ] eq = [HA] init Weak Acid: [H + ] eq < [HA] init This creates a difference in conductivity in rates of chemical reactions. 2015 Pearson Education

Weak For a weak acid, the equation for its dissociation is HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) Since it is an equilibrium, there is an equilibrium constant related to it, called the acid-dissociation constant. 2015 Pearson Education The greater the value of K a, the stronger is the acid.

Sample Exercise 16.10 Calculating K a from Measured ph A student prepared a 0.10 M solution of formic acid (HCOOH) found its ph at 25 to be 2.38. Calculate K a for formic acid at this temperature. Solution Analyze We are given the molar concentration of an aqueous solution of weak acid the ph of the solution, 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 a tabulation of initial 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 HCOOH(aq) H + (aq) + HCOO (aq) The equilibrium-constant expression is From the measured ph, we can calculate [H + ]: ph = log [H + ] = 2.38 log[h + ] = 2.38 [H + ] = 10 2.38 = 4.2 10 3 M Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Sample Exercise 16.10 Calculating K a from Measured ph Continued To determine the concentrations of the species involved in the equilibrium, we imagine that the solution is initially 0.10 M in HCOOH molecules. We then consider the ionization of the acid into H + HCOO. For each HCOOH molecule that ionizes, one H + ion one HCOO ion are produced in solution. Because the ph measurement indicates that [H + ] = 4.2 10 3 M at equilibrium, we can construct the following table: Notice that we have neglected the very small concentration of H + (aq) due to H 2 O autoionization. Notice also that the amount of HCOOH 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: (0.10 4.2 10 3 ) M 0.10 M We can now insert the equilibrium concentrations into the expression for K a : Check The magnitude of our answer is reasonable because K a for a weak acid is usually between 10 2 10 10. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Calculating Percent Ionization [H Percent ionization = 3 O + ] eq 100 In this example, [HA] initial [H 3 O + ] eq = 4.2 10 3 M [HCOOH] initial = 0.10 M 4.2 10 Percent ionization = 3 100 0.10 2015 Pearson Education = 4.2%

Method to Follow to Calculate ph Using K a 1) Write the chemical equation for the ionization equilibrium. 2) Write the equilibrium constant expression. 3) Set up a table for Initial/Change in/equilibrium Concentration to determine equilibrium concentrations as a function of change (x). 4) Substitute equilibrium concentrations into the equilibrium constant expression solve for x. (Make assumptions if you can!) 5) As a general rule, if x is more than about 5% of the initial concentration value, it is better to use the quadratic formula. 2015 Pearson Education

Sample Exercise 16.12 Using K a to Calculate ph Calculate the ph of a 0.20 M solution of HCN. (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 are asked for the ph. From Table 16.2, K a for HCN is 4.9 10 10. Plan We proceed as in the example just worked in the text, writing the chemical equation constructing a table of initial equilibrium concentrations in which the equilibrium concentration of H + is our unknown. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Sample Exercise 16.12 Using K a to Calculate ph Continued Solve Writing both the chemical equation for the ionization reaction that forms H + (aq) the equilibrium-constant (K a ) expression for the reaction: Next, we tabulate the concentrations of the species involved in the equilibrium reaction, letting x = [H + ] at equilibrium: Substituting the equilibrium concentrations into the equilibrium-constant 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, 0.20 x 0.20. Thus, Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Sample Exercise 16.12 Using K a to Calculate ph Continued Solving for x, we have A concentration of 9.9 10 6 M is much smaller than 5% of 0.20, the initial HCN concentration. Our simplifying approximation is therefore appropriate. We now calculate the ph of the solution: As a general rule, if x is more than about 5% of the initial concentration value, it is better to use the quadratic formula. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Sample Exercise 16.13 Using the Quadratic Equation to Calculate ph Percent Ionization Calculate the ph percentage of HF molecules ionized in a 0.10 M HF solution. Solution Analyze We are asked to calculate the percent ionization of a solution of HF. From Appendix D, we find K a = 6.8 10 4. Plan We approach this problem as for previous equilibrium problems: We write the chemical equation for the equilibrium tabulate the known unknown concentrations of all species. We then substitute the equilibrium concentrations into the equilibrium-constant expression solve for the unknown concentration of H +. Solve The equilibrium reaction equilibrium concentrations are as follows: The equilibrium-constant expression is Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Sample Exercise 16.13 Using the Quadratic Equation to Calculate ph Percent Ionization Continued When we try solving this equation using the approximation 0.10 x 0.10 (that is, by neglecting the concentration of acid that ionizes), we obtain x = 8.2 10 3 M Because this approximation is greater than 5% of 0.10 M, however, we should work the problem in stard quadratic form. Rearranging, we have x 2 = (0.10 x)(6.8 10 4 ) = 6.8 10 5 (6.8 10 4 )x x 2 + (6.8 10 4 )x 6.8 10 5 = 0 Substituting these values in the stard quadratic formula gives Of the two solutions, only the positive value for x is chemically reasonable. From that value, we can determine [H + ] hence the ph x = [H + ] = [F ] = 7.9 10 3 M, so ph = log[h + ] = 2.10 Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Sample Exercise 16.13 Using the Quadratic Equation to Calculate ph Percent Ionization Continued From our result, we can calculate the percent of molecules ionized: Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Polyprotic Polyprotic acids have more than one acidic proton. It is always easier to remove the first proton than any successive proton. If the factor in the K a values for the first second dissociation has a difference of 3 or greater, the ph generally depends only on the first dissociation. 2015 Pearson Education

Sample Exercise 16.14 Calculating the ph of a Solution of a Polyprotic Acid The solubility of CO 2 in water at 25 0.1 atm is 0.0037 M. The common practice is to assume that all the dissolved CO 2 is in the form of carbonic acid (H 2 CO 3 ), which is produced in the reaction CO 2 (aq) + H 2 O(l) H 2 CO 3 (aq) What is the ph of a 0.0037 M solution of H 2 CO 3? Solution Analyze We are asked to determine the ph of a 0.0037 M solution of a polyprotic acid. Plan H 2 CO 3 is a diprotic acid; the two acid-dissociation constants, K a1 K a2 (Table 16.3), differ by more than a factor of 10 3. Consequently, the ph can be determined by considering only K a1, thereby treating the acid as if it were a monoprotic acid. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Sample Exercise 16.14 Calculating the ph of a Solution of a Polyprotic Acid Continued Solve Proceeding as in Sample Exercises 16.12 16.13, we can write the equilibrium reaction equilibrium concentrations as The equilibrium-constant expression is 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 = 4.0 10 5 M 0.037 x 0.0037 Thus, Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Sample Exercise 16.14 Calculating the ph of a Solution of a Polyprotic Acid Continued Solving for x, we have Because we get the same value (to 2 significant figures) our simplifying assumption was justified. The ph is therefore Comment If we were asked for [CO 3 2 ] we would need to use K a2. Let s illustrate that calculation. Using our calculated values of [HCO 3 ] [H + ] setting [CO 3 2 ] = y, we have Assuming that y is small relative to 4.0 10 5, we have Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Sample Exercise 16.14 Calculating the ph of a Solution of a Polyprotic Acid Continued 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 HCO 3 is negligible relative to that of H 2 CO 3, as far as production of H + is concerned. However, 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 H 2 CO 3, only a small fraction ionizes to form H + HCO 3, an even smaller fraction ionizes to give CO 3 2. Notice also that [CO 3 2 ] is numerically equal to K a2. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Weak Ammonia, NH 3, is a weak base. Like weak acids, weak bases have an equilibrium constant called the base-dissociation constant. Equilibrium calculations work the same as for acids, using the base dissociation constant instead. 2015 Pearson Education

Base Dissociation Constants 2015 Pearson Education

Sample Exercise 16.15 Using K b to Calculate OH Calculate the concentration of OH in a 0.15 M solution of NH 3. Solution Analyze We are given the concentration of a weak base asked to determine the concentration of OH. 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 tabulate initial equilibrium concentrations. Solve The ionization reaction equilibrium-constant expression are Ignoring the concentration of H 2 O because it is not involved in the equilibrium-constant expression, the equilibrium concentrations are Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Sample Exercise 16.15 Using K b to Calculate OH Continued Inserting these quantities into the equilibriumconstant expression gives Because K b is small, the amount of NH 3 that reacts with water is much smaller than the NH 3 concentration, so we can neglect x relative to 0.15 M. Then we have Check The value obtained for x is only about 1% of the NH 3 concentration, 0.15 M. Therefore, neglecting x relative to 0.15 was justified. Comment You may be asked to find the ph of a solution of a weak base. Once you have found [OH ], you can proceed as in Sample Exercise 16.9, where we calculated the ph of a strong base. In the present sample exercise, we have seen that the 0.15 M solution of NH 3 contains [OH ] = 1.6 10 3 M. Thus, poh = log(1.6 10 3 ) = 2.80, ph = 14.00 2.80 = 11.20. The ph of the solution is above 7 because we are dealing with a solution of a base. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Types of Weak Neutral substances with an atom that has a nonbonding pair of electrons that can accept H + (like ammonia the amines) Anions (conjugated bases) of weak acids 2015 Pearson Education

Sample Exercise 16.16 Using ph 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 ph of 10.50. 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 + ClO ion. As such, it is a strong electrolyte that completely dissociates in solution into Na +, a spectator ion, 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 ph of 2.00-L of water to 10.50. Plan From the ph, we can determine the equilibrium concentration of OH. We can then construct a table of initial equilibrium concentrations in which the initial concentration of ClO is our unknown. We can calculate [ClO ] using the expression for K b. Solve We can calculate [OH ] by using either Equation 16.16 or Equation 16.20; we will use the latter method here: poh = 14.00 ph = 14.00 10.50 = 3.50 [OH ] = 10 3.50 = 3.2 10 4 M Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Sample Exercise 16.16 Using ph to Determine the Concentration of a Salt Continued This concentration is high enough that we can assume that Equation 16.37 is the only source of OH ; that is, we can neglect any OH produced by the autoionization of H 2 O. We now assume a value of x for the initial concentration of ClO solve the equilibrium problem in the usual way. We now use the expression for the basedissociation constant to solve for x: 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 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. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Relationship between K a K b For a conjugate acid base pair, K a K b are related in this way: K a K b = K w 2015 Pearson Education

Relationship between K a K b For a conjugate acid base pair, K a K b are related in this way: K a K b = K w Therefore, if you know one of them, you can calculate the other. 2015 Pearson Education

Sample Exercise 16.17 Calculating K a or K b for a Conjugate Acid Base Pair Calculate (a) K b for the fluoride ion, (b) K a for the ammonium ion. Solve (a) For the weak acid HF, Table 16.2 Appendix D give K a = 6.8 10 4. We can use Equation 16.40 to calculate K b for the conjugate base, F : (b) For NH 3, Table 16.4 in Appendix D give K b = 1.8 10 5, this value in Equation 16.40 gives us K a for the conjugate acid, NH 4+ : Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Acid Base Properties of Salts Many ions react with water to create H + or OH. The reaction with water is often called hydrolysis. To determine whether a salt is an acid or a base, you need to look at the cation anion separately. The cation can be acidic or neutral. The anion can be acidic, basic, or neutral. 2015 Pearson Education

Cation Group I or Group II (Ca2+, Sr2+, or Ba2+) metal cations are neutral. Polyatomic cations are typically the conjugate acids of a weak base; e.g., NH4+. Transition post-transition metal cations are acidic. 2015 Pearson Education

Anion Anions of strong acids are neutral. For example, Cl will not react with water, so OH can t be formed. Anions of weak acids are conjugate bases, so they create OH in water. Protonated anions from polyprotic acids can be acids or bases: If K a > K b, the anion will be acidic; if K b > K a, the anion will be basic. 2015 Pearson Education

Hydrated Cations Transition post-transition metals form hydrated cations. The water attached to the metal is more acidic than free water molecules, making the hydrated ions acidic. 2015 Pearson Education

Salt Solutions Neutral If the salt cation is the counterion of a strong base the anion is the conjugate base of a strong acid, it will form a neutral solution. If the salt contains an anion that does not react with water a cation that does not react with water, we expect the ph to be neutral. Such is the case when the anion is a conjugate base of a strong acid the cation is either from group 1A or one of the heavier members of group 2A (Ca 2+, Sr 2+, Ba 2+ ). Examples: NaCl, Ba(NO 3 ) 2, RbClO 4. 2015 Pearson Education

Salt Solutions Basic If the salt cation is the counterion of a strong base the anion is the conjugate base of a weak acid, it will form a basic solution. If the salt contains an anion that reacts with water to produce hydroxide ions a cation that does not react with water, we expect the ph to be basic. Such is the case when the anion is the conjugate base of a weak acid the cation is either from group 1A or one of the heavier members of group 2A (Ca 2+, Sr 2+, Ba 2+ ). Examples: NaClO, RbF, BaSO 3. 2015 Pearson Education

Salt Solutions Acidic If the salt cation is the conjugate acid of a weak base the anion is the conjugate base of a strong acid, it will form an acidic solution. If the salt contains a cation that reacts with water to produce hydronium ions an anion that does not react with water, we expect the ph 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: NH 4 NO 3, AlCl 3, Fe(NO 3 ) 3. 2015 Pearson Education

Salt Solutions Acidic, Basic, or Neutral? If the salt cation is the conjugate acid of a weak base the anion is the conjugate base of a weak acid, the ph of the solution depends on the relative strengths of the acid base. If the salt contains an anion a cation both capable of reacting with water, both hydroxide ions 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: NH 4 ClO, Al(CH3COO) 3, CrF 3. 2015 Pearson Education (NH 4 Cl) (Al(NO 3 ) 3 ) (NaCl) (NaF) (NH 4 F) (AlF 3 )

Determining the Overall Acidity or Basicity of Salt Solutions Determine if the solution formed by each salt is acidic, basic, or neutral. a. SrCl 2 b. AlBr 3 c. CH 3 NH 3 NO 3 d. NaCHO 2 e. NH 4 F Solution a. The Sr 2+ cation is the counterion of a strong base (Sr(OH) 2 ) is ph-neutral. The Cl anion is the conjugate base of a strong acid (HCl) is ph-neutral as well. The SrCl 2 solution is therefore ph-neutral (neither acidic nor basic). b. The Al 3+ cation is a small, highly charged metal ion (that is not an alkali metal or an alkaline earth metal) is a weak acid. The Br anion is the conjugate base of a strong acid (HBr) is ph-neutral. The AlBr 3 solution is therefore acidic. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.

Determining the Overall Acidity or Basicity of Salt Solutions Continued c. The CH 3 NH 3+ ion is the conjugate acid of a weak base (CH 3 NH 2 ) is acidic. The NO 3 anion is the conjugate base of a strong acid (HNO 3 ) is ph-neutral. The CH 3 NH 3 NO 3 solution is therefore acidic. d. The Na + cation is the counterion of a strong base is ph-neutral. The CHO 2 anion is the conjugate base of a weak acid is basic. The NaCHO 2 solution is therefore basic. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.

Determining the Overall Acidity or Basicity of Salt Solutions Continued e. The NH 4+ ion is the conjugate acid of a weak base (NH 3 ) is acidic. The F ion is the conjugate base of a weak acid is basic. To determine the overall acidity or basicity of the solution, compare the values of K a for the acidic cation K b for the basic anion. Obtain each value of K from the conjugate by using K a K b = K w. Since K a is greater than K b, the solution is acidic. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.

Sample Exercise 16.18 Determining Whether Salt Solutions Are Acidic, Basic, or Neutral Determine whether aqueous solutions of each of these salts are acidic, basic, or neutral: (a) Ba(CH 3 COO) 2, (b) NH 4 Cl, (c) CH 3 NH 3 Br, (d) KNO 3, (e) Al(ClO 4 ) 3. Solution Analyze We are given the chemical formulas of five ionic compounds (salts) 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 by assessing how each ion will affect the ph. Solve (a) This solution contains barium ions acetate ions. The cation is an ion of a heavy alkaline earth metal will therefore not affect the ph. The anion, CH 3 COO, is the conjugate base of the weak acid CH 3 COOH will hydrolyze to produce OH ions, thereby making the solution basic (combination 2). (b) In this solution, NH 4+ is the conjugate acid of a weak base (NH 3 ) is therefore acidic. Cl is the conjugate base of a strong acid (HCl) therefore has no influence on the ph of the solution. Because the solution contains an ion that is acidic (NH 4+ ) one that has no influence on ph (Cl ), the solution of NH 4 Cl will be acidic (combination 3). (c) Here CH 3 NH 3+ is the conjugate acid of a weak base (CH 3 NH 2, an amine) is therefore acidic, Br is the conjugate base of a strong acid (HBr) therefore ph neutral. Because the solution contains one ion that is acidic one that has no influence on ph, the solution of CH 3 NH 3 Br will be acidic (combination 3). Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Sample Exercise 16.18 Determining Whether Salt Solutions Are Acidic, Basic, or Neutral Continued (d) This solution contains the K + ion, which is a cation of group 1A, the NO 3 ion, which is the conjugate base of the strong acid HNO 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+ 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 1(HClO 4 ) therefore does not affect ph. Thus, the solution of Al(ClO 4 ) 3 will be acidic (combination 3). Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Sample Exercise 16.19 Predicting Whether the Solution of an Amphiprotic Anion Is Acidic or Basic Predict whether the salt Na 2 HPO 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 HPO 4 is acidic or basic. This substance is an ionic compound composed of Na + HPO 4 2 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 ph. Thus, our analysis of whether the solution is acidic or basic must focus on the behavior of the HPO 2 4 ion. We need to consider that HPO 2 4 can act as either an acid or a base: As acid HPO 2 4 (aq) H + (aq) + PO 3 4 (aq) [16.46] As base HPO 2 4 (aq) + H 2 O H 2 PO 4 (aq) + OH (aq) [16.47] Of these two reactions, the one with the larger equilibrium constant determines whether the solution is acidic or basic. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Sample Exercise 16.19 Predicting Whether the Solution of an Amphiprotic Anion Is Acidic or Basic Continued Solve The value of K a for Equation 16.46 is K a3 for H 3 PO 4 : 4.2 10 13 (Table 16.3). For Equation 16.47, we must calculate K b for the base HPO 4 2 from the value of K a for its conjugate acid, H 2 PO 4, the relationship K a K b = K w (Equation 16.40). The relevant value of K a for H 2 PO 4 is Ka 2 for H 3 PO 4 : 6.2 10 8 (from Table 16.3). We therefore have This K b value is more than 10 5 times larger than K a for HPO 4 2 ; thus, the reaction in Equation 16.47 predominates over that in Equation 16.46, the solution is basic. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Factors that Affect Acid Strength H A bond must be polarized with δ+ on the H atom δ on the A atom Bond strength: Weaker bonds can be broken more easily, making the acid stronger. Stability of A : More stable anion means stronger acid. 2015 Pearson Education

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Binary Binary acids consist of H one other element. Within a group, H A bond strength is generally the most important factor. Within a period, bond polarity is the most important factor to determine acid strength. 2015 Pearson Education

Oxyacids Oxyacids consist of H, O, one other element, which is a nonmetal. Generally, as the electronegativity of the nonmetal increases, the acidity increases for acids with the same structure. 2015 Pearson Education

Oxyacids with Same Other Element If an element can form more than one oxyacid, the oxyacid with more O atoms is more acidic; e.g., sulfuric acid versus sulfurous acid. If the oxidation number increases, the acidity increases. 2015 Pearson Education

Sample Exercise 16.20 Predicting Whether the Solution of an Amphiprotic Anion Is Acidic or Basic Arrange the compounds in each series in order of increasing acid strength: (a) AsH 3, HBr, KH, H 2 Se; (b) H 2 SO 4, H 2 SeO 3, H 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 H, in (b) the substances are oxyacids. Plan For the binary compounds, we will consider the electronegativities of As, Br, K, Se relative to the electronegativity of H. The higher the electronegativity of these atoms, the higher the partial positive charge on H so the more acidic the compound. For the oxyacids, we will consider both the electronegativities of the central atom 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. 348). As a result, the hydrogen in KH carries a negative charge. Thus, KH should be the least acidic (most basic) compound in the series. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Sample Exercise 16.20 Predicting Whether the Solution of an Amphiprotic Anion Is Acidic or Basic Continued Arsenic hydrogen have similar electronegativities, 2.0 2.1, respectively. This means that the As H bond is nonpolar, so AsH 3 has little tendency to donate a proton in aqueous solution. The electronegativity of Se is 2.4, that of Br is 2.8. Consequently, the H Br bond is more polar than the H Se bond, giving HBr the greater tendency to donate a proton. (This expectation is confirmed by Figure 16.18, where we see that H 2 Se is a weak acid HBr a strong acid.) Thus, the order of increasing acidity is KH < AsH 3 < H 2 Se < HBr. (b) The acids H 2 SO 4 H 2 SeO 4 have the same number of O atoms the same number of OH 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 H 2 SO 4 is more acidic than H 2 SeO 4. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Sample Exercise 16.20 Predicting Whether the Solution of an Amphiprotic Anion Is Acidic or Basic Continued For acids with the same central atom, the acidity increases as the number of oxygen atoms bonded to the central atom increases. Thus, H 2 SeO 4 should be a stronger acid than H 2 SeO 3. We predict the order of increasing acidity to be H 2 SeO 3 < H 2 SeO 4 <H 2 SO 4. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus 2015 Pearson Education, Inc.

Carboxylic Carboxylic acids are organic acids containing the COOH group. Factors contributing to their acidic behavior: Other O attached to C draws electron density from O H bond, increasing polarity. Its conjugate base (carboxylate anion) has resonance forms to stabilize the anion. 2015 Pearson Education

Lewis Acid/Base Chemistry Electron acceptor/electron donor. Lewis acids are electron pair acceptors. Lewis bases are electron pair donors. All Brønsted Lowry acids bases are also called Lewis acids bases. There are compounds which do not meet the Brønsted Lowry definition which meet the Lewis definition. 2015 Pearson Education

Ammonia s Reaction with H + BF 3 2015 Pearson Education

Examples of Lewis Acid Base Reactions Ag + (aq) + 2 :NH 3(aq) Ag(NH 3 ) 2 + (aq) Lewis Acid Lewis Base Adduct 2014 Pearson Education, Inc.