1. Strong Acids and Bases Equilibrium Acids and Bases 6 Examples of strong acids are H 2 SO 4, HNO 3, HCl, HBr, and HI. (There are a few others, e.g. HClO 4, which we will not discuss.) Since strong acids dissociate completely in water, the H + concentration (or the H 3 O + concentration) is equal to the concentration of acid. The exception is sulfuric acid which, since it is diprotic, gives an H + concentration which is almost twice the acid concentration (or would be twice the concentration if the second proton came off completely). Similarly strong bases, the hydroxides of Group IA and Group IIA metals, dissociate completely. So in a solution of potassium hydroxide, the hydroxide concentration is equal to the concentration of potassium hydroxide. In a solution of calcium hydroxide, Ca(OH) 2, the hydroxide concentration is twice the concentration of calcium hydroxide. Because of their complete dissociation, calculating the ph of a strong acid or a strong base solution is simple. Let us calculate, for example, the ph of a 2.0 10 4 M solution of HBr. The H + concentration is 2.0 10 4 and the log of 2.0 10 4 is -3.70. (Recall that in a logarithm only the digits after the decimal are significant.) Since ph is -log[h + ], the ph of the solution is 3.70. As an example of a strong base, suppose we have a solution which is 2.0 10 4 M in Ca(OH) 2. This leads to a hydroxide concentration of 4.0 10 4. The log of 4.0 10 4 is -3.40. However since this is based on hydroxide ion and not hydrogen ion, the log corresponds to something we call the poh and not the ph. Here the poh will be 3.40. Because [H + ][OH ] = 1.00 10 14 we know that the log of H + plus the log of OH add up to 14 (i.e. ph + poh = 14.0). Thus, ph = 14.0 - poh = 14.00-3.40 = 10.60 2. Weak Acids and Weak Bases Acids which are not listed among the strong acids (i.e. acids which are not H 2 SO 4, HNO 3, HCl, HBr, or HI) are classed as weak acids. Bases which are not strong (i.e. bases other than the hydroxides of Group IA and Group IIA metals) are weak. In contrast to a strong acid, which is completely dissociated in aqueous solution, a weak acid dissociates only partly. That is, the dissociation of a weak acid or base is an equilibrium reaction. In order to calculate the results of that equilibrium, we need to use the equilibrium constant. Let us consider the dissociation of a 0.50 M HF, a weak acid which reacts as below. HF H + + F The dissociation has an equilibrium constant, referred to as "K a ", equal to 7.2 10 4. -1-
The reaction starts with [HF] = 0.50 M and [H + ] = [F ] = 0. If the amount of HF which dissociates is x, then [HF] = (0.50 - x) and [H + ] = [F ] = x. Therefore: Clearing the fraction and combining like terms leads to a quadratic equation. Generally, however, it is possible to make an approximation which allows us to avoid using the quadratic formula. If K is small, x should be small. In fact it will probably be so small that (0.50 - x) will be equal to 0.50, and the x on the bottom can be ignored. Neglecting x gives the following equation. This is easily solved to give x = 0.019, which means that: [H + ] = [F ] = 0.019 M and [HF] = 0.50-0.019 0.50 M As before, we use the "5% rule" to determine whether neglecting x was valid. Since 0.019 is less than 5% of 0.50, the assumption is valid. Of course in some problems you know the value of x and are using it to calculate K. In such a case you don't neglect x even if it is less than 5%. Weak base calculations typically involve ammonia or a similar compound (e.g. trimethylamine, N(CH 3 ) 3 ). Let us calculate the ph of a 0.10 M ammonia solution, which reacts with water: NH 3 + H 2 O NH 4 + + OH The equilibrium constant, K b, is equal to 1.8 10 5. The equilibrium expression is, therefore: Notice that water is in the reaction but not in the mass action expression. Substituting the concentrations gives: Neglecting the addition of x to 0.10, and multiplying gives: x 2 = 1.8 10 6 M and x = [OH ] = 1.34 10 3 M poh = 2.87 and ph = 11.13-2-
3. Polyprotic Acids A polyprotic acid has more than one removable proton. An example would be hydrogen sulfide: H 2 S H + + HS K a1 = 1.1 10 7 HS H + + S 2 K a2 = 1.2 10 13 Why are the equilibrium constants for the removal of the two protons different? This is because the first proton is removed from a neutral molecule while the second is removed from an anion. As an example, we will calculate the ion concentration in 0.100 M H 2 S solution. K A1 = [H+ ][HS - ] [H 2 S] After neglecting the x in the denominator, we solve to get: x = [H + ] = [HS - ] = 1.0 10 4 M Additional hydrogen ions will be formed in the second ionization. However, the H + formed in later steps is ALWAYS neglected. Let us now determine the sulfide concentration: K A2 = [H+ ][S -2 ] [HS - ] Notice that we used "y" instead of "x" in order to lessen confusion. After neglecting the additive y's, the problem becomes enormously easy, and we get: y = [S -2 ] = 1.2 10 13 M Neglecting the y was clearly justified. Also note that "y" is equal to the H + which comes from the second dissociation. As stated above, y is always small and can be ignored in calculating H +. 4. Hydrolysis Hydrolysis occurs when the conjugate base of a weak acid or the conjugate acid of a weak base is dissolved in water. To calculate the ph of a sodium acetate solution, one calculates the hydroxide concentration caused by the hydrolysis of the acetate. C 2 H 3 O 2 + H 2 O HC 2 H 3 O 2 + OH You will notice that in this reaction acetate functions as a base. Therefore to solve the problem you must use K b for the acetate ion which is K w divided by K a. Thus for 0.15 M acetate you calculate: Setting x = [OH ] = [HC 2 H 3 O 2 ] -3-
Neglecting the addition of x to 0.15 gives: x 2 = 8.32 10 11 and x = [OH ] = 9.12 10 6 M poh = 5.04 and ph = 8.96 In hydrolysis only the conjugate base (or conjugate acid) is present. If the conjugate base is accompanied by the acid if acetate is accompanied by acetic acid the solution is a buffer. The following solutions should be treated as hydrolysis problems: Sodium carbonate Ammonium chloride Equal moles of acetic acid and sodium hydroxide A titration at its equivalence point The following solutions should not be treated as hydrolysis problems: Acetic acid Ammonia Unequal numbers of moles of acetic acid and sodium hydroxide A titration anyplace except at its equivalence point For qualitative questions that is whether a given compound forms an acidic, basic or neutral solution you must know the following: The salt of a strong acid and a strong base is neutral; the salt of a strong acid and a weak base is acidic; the salt of a weak acid and a strong base is basic; the salt of a weak acid and a weak base is a nasty question which no one will ask you. Using this logic, you should predict that ammonium nitrate is acidic, calcium chloride is neutral, potassium phosphate is basic and aluminum sulfate is acidic. 5. Buffers A buffer is an aqueous solution which contains roughly equal concentrations of a weak acid and its conjugate base or a weak base and its conjugate acid. A buffer resists ph change because no matter whether H + or OH is added; there is something in the solution which will react with the added ion. Consider a buffer made from the weak acid HF and its conjugate base F. If an acid is added, it reacts with the fluoride: H + + F HF If a base is added, it will react with the hydrofluoric acid: OH + HF H 2 O + F The ph of a buffer is easy to calculate. Consider a buffer based on HF, which has a K A = 7.2 10 4. In a solution where concentrations of both the weak acid and its conjugate base are equal to 0.10 M, the ph can be calculated from the mass action expression. -4-
[H + ] = 7.2 10 4 M ph = 3.14 If H + is added to the buffer, it all reacts with F to form HF. (This is important!) So if we add (without dilution) 0.01 M H +, we end up with 0.11 M HF and 0.90 M F. Therefore: Remember that adding a strong base to a mixture of HA and A - both decreases the HA and increases the A -. [H + ] = 8.8 10 4 M and ph = 3.05 Now let us try another buffer calculation, based on hypochlorous acid, HOCl, for which K a = 3.5 10 8. Suppose we start with 1 liter of 0.50 M HOCl and add, without diluting the solution, 0.20 moles of KOH. This is typical of how buffer problems start; so it is important you understand it. The strong base (KOH) reacts completely with the weak acid (HOCl) to form H 2 O and OCl. HOCl + OH OCl + H 2 O This can be summarized in the ICE chart on the right: The H + is formed by "bounce back" and its concentration is determined by the equilibrium expression: K A = [H+ ][OCl - ] [HOCl] HOCl OH - OCl - Initial 0.50 0.20 0.00 Change -0.20-0.20 +0.20 Equilibrium 0.30 0.00 0.20 Therefor [H + ] = 5.25 10 8 M and ph = 7.28 Now consider a buffer made by mixing 70.0 ml of 0.100 M ammonia with 30.0 ml of 0.100 M HCl. We will assume the volumes are additive. After dilution, the concentrations are: [NH 3 ] = 0.070 M [HCl] = 0.030 M The HCl (actually the H + from the HCl) reacts with the NH 3 as shown below: H + + NH 3 NH 4 + This gives: [NH 4 + ] = 0.030 M [NH 3 ] = 0.040 M Plugging this in to the mass action expression gives us: poh = 4.62 and ph = 9.38 then [OH ] = 2.4 10 5 M -5-
Many textbooks explain buffers in terms of the Henderson-Hasselbalch Equation: ph = pk A + log [base] [acid] This is equivalent to what we have been doing, and many students find Henderson-Hasselbalch makes simple problems even simpler. However, it makes hard problems even more difficult. So we are not using it here. 6. Indicators How do people determine the ph of a solution? An electrochemical device called a ph meter can be used. However, these are delicate and expensive and they need frequent calibration. A simpler method involves the use of an organic acid or base which has different colors in its acidic and basic forms. Such a material, an example of which is phenolphthalein, is called an indicator. Phenolphthalein is colorless in its acid form (Figure 1) but becomes pink when it loses a proton. The equilibrium constant for the loss of a proton by this indicator (HIn H + + In ) is K a = 1 10 9 which gives a pk A of 9. The pk A of an acid is significant since it equals the ph at which half the material is in the acid form and half is in the basic form. Thus the pk roughly equals the ph at which color change occurs. Of course one can see the pink color of the basic form long before the indicator is half converted. Consider what happens at ph 8. Figure 1. The acidic form of phenolphthalein Dropping the ph by one unit causes the fraction of indicator in the basic form to decrease from 50% to about 9%. However, a tinge of pink in an otherwise colorless solution can probably be seen at even less than 9%. For most indicators the color changes over a range of about one ph unit on either side of the pk. Fortunately, this is narrow enough for most purposes. Indicator ph Range pka Acid Form Base Form Methyl violet 0.0-1.6 0.8 yellow blue Methyl yellow 2.9-4.0 3.3 red yellow Methyl orange 3.1-4.4 4.2 red yellow Methyl red 4.2-6.2 5.0 red yellow Chlorophenol red 4.8-6.4 6.0 yellow red Bromothymol blue 6.0-7.6 7.1 yellow blue Phenol red 6.4-8.0 7.4 yellow red Cresol purple 7.4-9.0 8.3 yellow purple Thymol blue 8.0-9.6 8.9 yellow blue Phenolphthalein 8.0-9.8 9.0 colorless red Figure 2. Properties of Acid-Base Indicators at 25 C The strips of ph paper which most of you have used for measuring ph are based on indicators. Typically a mixture of three indicators, which change color at different ph values, is used to give a -6-
solution which changes color continuously over a wide ph range. Litmus is also an indicator, one which turns red in acidic solution and blue in basic solution. (Remember bbblue for bbbasic.) Indicators are widely used to determine the end point in a titration. What you need to know, how to choose which indicator to use in a given titration, will be discussed in a later section. 7. Titrations Titrations are used to determine the amount of acid or base in a sample. If the sample to be analyzed is an acid, a base solution of known concentration (the titrant) is added. The point at which the moles of acid and the moles of base are equal is known as the "end point" or "equivalence point." Calculations are based on the volume of titrant needed to reach the end point. Thus one can calculate the moles of acid in a sample: Moles A = M B V B If the acid is in a solution of known volume, its molarity can be calculated from: M A V A = M B V B Suppose a sample of organic acid weighing 0.255 g was found to require 35.6 ml of 0.0110 M KOH to reach the equivalence point. What is the molecular weight of the unknown acid? We know the molecular weight = grams/moles and that at the end point moles A = moles B. So: What is the molarity of an acid solution if 5.0 ml are titrated by 28.8 ml of 0.0110 M KOH? M A V A = M B V B M A 5.0 = 0.0110 28.8 M A = 0.0634 M Since volume is stated in ml on both sides of the equation, the units cancel. All this assumes a monoprotic acid. If the acid is diprotic, and assuming that both protons are titrated, it is necessary to multiply the acid molarity by two. Thus: 2 M A V A = M B V B 8. Titration Curves To understand what happens during a titration it is helpful to graph the way ph changes as the titration proceeds. This graph is known as a titration curve. -7-
First look at what happens when a strong acid is titrated with a strong base (Figure 3). The ph starts low. As the titration proceeds (as base is added) the ph increases slowly. Suddenly, as the titration nears the equivalence point, the slope increases and the ph changes more rapidly. The equivalence point is, in fact, the point at which the titration curve has its greatest slope. This is why titrations work so well. At the equivalence point, the very point you want to determine, a small amount of titrant will cause a large change in ph. A single drop could cause the ph to change from 4 to 10, for example. This is fortunate, since indicators change color gradually over a range of several ph units. This titration is often done using phenolphthalein (pk A = 9) as an indicator. Figure 3. The titration of a strong acid with a strong base. The titration of a weak acid with a strong base gives a similar curve (Figure 4), but there are differences. The initial ph is higher, since the acid is weaker, and the ph at the equivalence point will be higher as well. What this means is that the vertical portion of the titration curve, the segment near the inflection point, is smaller than with a strong acid. This means that determining the end point is more difficult with a weak acid than with a strong acid, and choice of indicator is more critical. This is a good time to address a point which some students find Figure 4. The titration of a confusing. The simple comparison between strong and weak weak acid with a strong base. acids given in the previous paragraph works only if the concentrations are the same. A concentrated solution of a weak acid can be more acidic than a dilute solution of strong acid. You must know what species are present at various points on the titration curve. Consider the titration of an HF solution with a solution of KOH. At first the solution contains only HF. As the reaction proceeds, the HF reacts with OH to form H 2 O and F. Thus, the principal species are K +, HF and F. At the equivalence point, all the HF has been converted to F and the principal species present are K + and F. What happens when the volume of titrant is half of that needed to reach the equivalence point? (This is called the half equivalence point). Here the concentrations of HF and F are equal. If [HF] = [F - ], we get the following: K A = [H+ ][F - ] [HF] and K A = [H + ] Therefore the ph is equal to the pk. This is not a calculation which a practicing chemist does often, but it occurs frequently on the AP test. Watch for the half-equivalence point! To calculate the ph at various points on the curve you should remember that the initial solution is a weak acid, a calculation which you know well. The solution at the half-equivalence point (or at any other point part way through the titration) is a buffer. Finally, at the equivalence point all the weak acid and strong base have reacted. All the HF has been converted to F and therefore the [H + ] and [OH ] concentrations are based on the reaction: -8-
F + H 2 O HF + OH The titration of a weak base with a strong acid is simply the reverse of this. It starts at a high ph (but not as high a ph as it would if it were a strong base) and has its equivalence point below 7. Similarly the titration of a strong base with a strong acid gives a curve which is the same as the titration of a strong acid with a strong base. They both have their equivalence points at 7, but in one the ph increases and in the other it decreases. Students worry about what happens when you titrate a weak acid with a weak base. Don't! The experimenter gets to choose the titrant and no one would choose to titrate with a weak base. The titration of a polyprotic acid is more complex. Consider a diprotic acid with: K A1 = 10 3 and K a2 = 10 10. Since the acid has two K's, the titration curve has two inflection points. When the K's of a polyprotic acid are close, one deprotonation starts before the other has finished and determining the end point is difficult. 9. Mixing Different Types of Acid/Base Problems Acid/base problems are not difficult. Unfortunately students often don't know what category a problem falls into and try to solve the wrong type. There are four types of acid/base problems. We talk about acids, but everything here is equally applicable to bases. Strong Acid -- nothing present except for a strong acid Figure 5. Titration Curve for a diprotic acid. Weak Acid -- nothing present except for the weak acid and its dissociation products. Finding the ph of an acetic acid solution is a weak acid problem. Buffer -- a weak acid and its conjugate base. Acetic acid and sodium acetate would make a buffer. Acetic acid and sodium hydroxide would also make a buffer, because the hydroxide would react with the acetic acid to form acetate. However hydrochloric acid and sodium chloride would not form a buffer, because hydrochloric acid is strong. Hydrolysis -- only the conjugate base of a weak acid. Thus, sodium acetate solution is a hydrolysis problem. However, if you add an acid to this solution, it becomes a buffer. Some examples: 1.0 M SODIUM FLUORIDE. Forms sodium ions and fluoride ions in solution. The fluoride ions react: F + H 2 O HF + OH, which is hydrolysis. 1.0 M SODIUM FLUORIDE + 0.5 M HYDROFLUORIC ACID. These dissolve to form F and HF. They don't react; they form a buffered equilibrium. 1.0 M HYDROFLUORIC ACID + 1.0 M SODIUM HYDROXIDE. They form HF and OH -, which react according to HF + OH F + H 2 O. Because these are present in equal concentrations, you end up with 1.0 M fluoride which then hydrolyzes. 1.0 M SODIUM FLUORIDE + 1.0 M HYDROCHLORIC ACID. The solution contains 1.0 M F and 1.0 M H +. They react to form 1.0 M HF which then dissociates in a typical weak acid reaction. -9-
0.5 M SODIUM FLUORIDE + 1.0 M HYDROCHLORIC ACID. These form H + and F ions. All the fluoride is protonated by 0.5 M of the H + and the remaining 0.5 M H + is left to act as a strong acid. 1.0 M HYDROFLUORIC ACID + 0.5 M SODIUM HYDROXIDE. These form HF and OH -. All of the hydroxide reacts with 0.5 M HF to form 0.5 M F -. This mixture, which now contains 0.5 M fluoride and 0.5 M HF, acts as a buffer. 1.0 M SODIUM FLUORIDE AND 0.5 M HYDROCHLORIC ACID. These dissolve to give a solution containing H + and F ions. All the H + from the hydrochloric acid is used to protonate 0.5 M of the fluoride forming 0.5 M HF. This leaves a mixture containing 0.5 M HF and the remaining 0.5 M F -, which is a buffer. 10. Brønsted and Lewis Models After all we have learned about acids and bases and their properties, it is strange to realize that the definition of acids and bases which we have been using is only one of several. The definition which we have used, that an acid supplies H + ions and a base gives OH ions, is the original definition as put forth by the Swedish chemist Svante Arrhenius. It is, therefore, called the Arrhenius model or the Arrhenius definition. It is very useful, but there are others. The Brønsted Model, or more properly the Brønsted-Lowry Model, defines an acid as a proton donor and a base as a proton acceptor. Consider the reaction of hydrofluoric acid with ammonia HF + NH 3 F + NH 4 + The hydrofluoric acid donates a proton to the ammonia, making the hydrofluoric acid an acid and the ammonia a base. Fluoride, which results from the loss of a proton by hydrofluoric acid, is considered a base since in the reverse reaction it gains a proton. It is said to be the "conjugate base" of hydrofluoric acid. Similarly the ammonium ion, which loses a proton in the reverse reaction, is said to be the "conjugate acid" of ammonia. Although this is less obvious, the dissolving of hydrogen chloride in water can also be considered an acid base reaction. HCl + H 2 O H 3 O + + Cl Here HCl is the acid, water is the base, hydronium ion (H 3 O +, essentially another way of writing H + ) is the conjugate acid and chloride is the conjugate base. Water is a base. How strange! Now look at what happens when ammonia is added to water: NH 3 + H 2 O NH 4 + + OH Ammonia is a base, water is an acid, ammonium ion is the conjugate acid and hydroxide ion is the conjugate base. So water is either an acid or base, depending on the circumstances. In the Brønsted model acid/base reactions can be considered to be a competition between the base and the conjugate as to which one wants the proton more. -10-
Imagine a creature on some distant planet where lakes are made of ammonia instead of water. Acid/base chemistry in ammonia would be based on the reaction: NH 3 + NH 3 NH 4 + + NH 2 Chemistry students on this planet might think it strange that acid/base neutralization reactions could occur in liquid water. Yet they could easily understand aqueous chemistry in terms of the Brønsted model. Another way of defining acids and bases is seen in the Lewis model. Lewis acids are electron pair acceptors and Lewis bases are electron pair donors. This is consistent with the Arrhenius model, since a proton is itself an excellent acceptor of electrons while hydroxide is a good electron donor. Thus a neutralization reaction is water is still an acid base reaction. H + + OH H 2 O However, other reactions which we do not think of as acid/base reactions are acid/base reactions according to the Lewis model. Consider the reaction of trimethylamine, which we know to have an unshared pair of electrons and to thus be a Lewis base. Suppose it reacts with BF 3, a molecule which lacks an octet and is, therefore, capable of acting as an electron acceptor. The molecule which results, BF 3 -N(CH 3 ) 3, is referred to as a "Lewis acid/lewis base adduct." Figure 6. A Lewis acid Lewis base reaction 11. Relation of Structure to Acidity A compound which contains a hydrogen atom will often give it up when dissolved, forming an acidic solution. However many compounds, for example organic compounds other than carboxylic acids, do not lose protons in water. The question of why some proton-containing compounds ionize easily (strong acids), others ionize with difficulty (weak acids) and still others don't ionize at all (non-acids) has to do with bond strength. Why bonds strengths vary as they do is complex. The simplest example of a structure/acidity relationship occurs in the hydrogen halides HF, HCl, HBr and HI. Since HF contains the most electronegative, and therefore the most non-metallic, of the halogens, it seems that it should form the most acidic hydride. However that's not the way it works. Being the most electronegative element means it will form the strongest bond with hydrogen, which means it will form the weakest acid. Acid Bond Strength (kj/mol) Acid strength (in Water) HF 565 weak HCl 427 strong HBr 363 strong HI 295 strong -11-
The next trend is among the oxy-acids, for example HClO 4, HClO 3, HClO 2 and HClO. The trend is that the more oxygens there are, the stronger the acid. Thus the trend in acid strength is: HClO 4 > HClO 3 > HClO 2 > HClO. To understand this you must first realize that the formulas written above give a misleading picture of the structures. In almost all oxy-acids the hydrogen is bound to an oxygen and not to the central atom. Thus, HClO 2 is actually H-O-Cl-O. (That's why HCl, a strong acid, doesn't fit the trend. The hydrogen in HCl is bound directly to chlorine.) Why do additional oxygen atoms make an oxy-acid stronger? The easiest way to view this (and there are several legitimate ways) is to say that oxygen, being very electronegative, withdraws electron density from the chlorine. Since losing a proton (a hydrogen cation) will leave behind a negative charge, something which withdraws negative charge will make the ion more stable. The acidity of other oxy-acids follows the same trend. Thus: H 2 SO 4 > H 2 SO 3 and HNO 3 > HNO 2. Another trend you should know is that metal hydroxides are basic while non-metal hydroxides are acidic. Thus ClOH is an acid (usually written as HOCl) and KOH is a base. Similarly metal oxides generally react with water to form basic solutions, for example: CaO + H 2 O Ca 2+ + 2 OH Non-metal oxides, on the other hand, react with water to form acidic solutions, for example: SO 2 + H 2 O H 2 SO 3 Why is this? The answer is that an electronegative element, such as chlorine, forms a covalent bond with oxygen. This bond, like other covalent bonds, does not break easily in water. Thus the only bond which can break is the H-O bond. In metal hydroxides, however, the central atom is electropositive and thus forms an ionic bond. The ionic bond between the metal ion and the hydroxide ion is strong, but unlike the covalent bond it comes apart easily in a polar solvent such as water. (As you may recall, polar water molecules surround the ions and stabilize the charges.) The intermediate case, in which the central atom is neither strongly electronegative nor strongly electropositive, leads to a situation where both the X-O bond and the O-H bond are capable of breaking. This leads to a phenomenon known as "amphoterism," in which the oxide or hydroxide is capable of acting as either an acid or a base. It is difficult to predict whether a transition metal oxide will be acidic, basic or amphoteric. In general you will not have to do that. However, when we get to the reaction section of this course, you will learn that aluminum and zinc form amphoteric oxides. But that's for another day. 12. "How many acids" problems One good way of testing your knowledge of acid-base chemistry is to ask how many compounds in a list form acidic (or basic) solutions in water. Consider the following list of compounds. CaBr 2, NH 4 Cl, SO 3, AlCl 3, KCN If you answered that "two form acidic solutions," you are wrong. The correct answer is "three," specifically NH 4 Cl, SO 3, and AlCl 3. Make sure you understand this. -12-