Relative Strengths of Acids and Bases

Transcription

1 OpenStax-CNX module: m Relative Strengths of Acids and Bases OpenStax This work is produced by OpenStax-CNX and licensed under the Creative Commons Attribution License 4.0 By the end of this section, you will be able to: Abstract Assess the relative strengths of acids and bases according to their ionization constants Rationalize trends in acidbase strength in relation to molecular structure Carry out equilibrium calculations for weak acidbase systems We can rank the strengths of acids by the extent to which they ionize in aqueous solution. The reaction of an acid with water is given by the general expression: HA (aq) + H 2 O (l) H 3 O + (aq) + A (aq) (1) Water is the base that reacts with the acid HA, A is the conjugate base of the acid HA, and the hydronium ion is the conjugate acid of water. A strong acid yields 100% (or very nearly so) of H 3 O + and A when the acid ionizes in water; Figure 1 lists several strong acids. A weak acid gives small amounts of H 3 O + and A. Version 1.10: Apr 13, :55 pm

2 OpenStax-CNX module: m Figure 1: Some of the common strong acids and bases are listed here. The relative strengths of acids may be determined by measuring their equilibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is called the acid-ionization constant, Ka. For the reaction of an acid HA: HA (aq) + H 2 O (l) H 3 O + (aq) + A (aq), (2) we write the equation for the ionization constant as: H3 O +] A ] K a = HA] where the concentrations are those at equilibrium. Although water is a reactant in the reaction, it is the solvent as well, so we do not include H 2 O] in the equation. The larger the K a of an acid, the larger the concentration of H 3 O + and A relative to the concentration of the nonionized acid, HA. Thus a stronger acid has a larger ionization constant than does a weaker acid. The ionization constants increase as the strengths of the acids increase. (A table of ionization constants of weak acids appears in Appendix H 1, with a partial listing in Table 1.) The following data on acid-ionization constants indicate the order of acid strength CH 3 CO 2 H < HNO 2 < HSO 4 : CH 3 CO 2 H (aq) + H 2 O (l) H 3 O + (aq) + CH 3 CO 2 (aq) K a = (4) (3) HNO 2 (aq) + H 2 O (l) H 3 O + (aq) + NO 2 (aq) K a = (5) HSO 4 (aq) + H 2 O (aq) H 3 O + (aq) + SO 4 2 (aq) K a = (6) 1 "Ionization Constants of Weak Acids" <

3 OpenStax-CNX module: m Another measure of the strength of an acid is its percent ionization. The percent ionization of a weak acid is the ratio of the concentration of the ionized acid to the initial acid concentration, times 100: % ionization = H3 O +] eq HA] 0 100(7) Because the ratio includes the initial concentration, the percent ionization for a solution of a given weak acid varies depending on the original concentration of the acid, and actually decreases with increasing acid concentration. Example 1 Calculation of Percent Ionization from ph Calculate the percent ionization of a M solution of nitrous acid (a weak acid), with a ph of Solution The percent ionization for an acid is: H3 O +] eq HNO 2 ] (8) The chemical equation for the dissociation of the nitrous acid is: HNO 2 (aq) + H 2 O (l) NO 2 (aq) + H 3 O + (aq). Since 10 ph = H 3 O +], we nd that = M, so that percent ionization is: = 6.5% (9) Remember, the logarithm 2.09 indicates a hydronium ion concentration with only two signicant gures. Check Your Learning Calculate the percent ionization of a 0.10-M solution of acetic acid with a ph of note: 1.3% ionized We can rank the strengths of bases by their tendency to form hydroxide ions in aqueous solution. reaction of a Brønsted-Lowry base with water is given by: The B (aq) + H 2 O (l) HB + (aq) + OH (aq) (10) Water is the acid that reacts with the base, HB + is the conjugate acid of the base B, and the hydroxide ion is the conjugate base of water. A strong base yields 100% (or very nearly so) of OH and HB + when it reacts with water; Figure 1 lists several strong bases. A weak base yields a small proportion of hydroxide ions. Soluble ionic hydroxides such as NaOH are considered strong bases because they dissociate completely when dissolved in water. note: View the simulation 2 of strong and weak acids and bases at the molecular level. 2

4 OpenStax-CNX module: m As we did with acids, we can measure the relative strengths of bases by measuring their base-ionization constant (Kb) in aqueous solutions. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base has a larger ionization constant than does a weaker base. For the reaction of a base, B: B (aq) + H 2 O (l) HB + (aq) + OH (aq), (11) we write the equation for the ionization constant as: ] HB + OH ] K b = B] where the concentrations are those at equilibrium. Again, we do not include H 2 O] in the equation because water is the solvent. The chemical reactions and ionization constants of the three bases shown are: NO 2 (aq) + H 2 O (l) HNO 2 (aq) + OH (aq) K b = (13) (12) CH 3 CO 2 (aq) + H 2 O (l) CH 3 CO 2 H (aq) + OH (aq) K b = (14) NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH (aq) K b = (15) A table of ionization constants of weak bases appears in Appendix I 3 (with a partial list in Table 2). As with acids, percent ionization can be measured for basic solutions, but will vary depending on the base ionization constant and the initial concentration of the solution. Consider the ionization reactions for a conjugate acid-base pair, HA A : HA (aq) + H 2 O (l) H 3 O + (aq) + A H3 O +] A ] (aq) K a = (16) HA] A (aq) + H 2 O (l) OH (aq) + HA (aq) K b = HA] OH] A ] (17) Adding these two chemical equations yields the equation for the autoionization for water: )HA(aq + H 2 O (l) + )A (aq) + H 2 O (l) H 3 O + (aq) + )A (aq) + OH (aq) + )HA(aq (18) 2H 2 O (l) H 3 O + (aq) + OH (aq) (19) As shown in the previous chapter on equilibrium, the K expression for a chemical equation derived from adding two or more other equations is the mathematical product of the other equations' K expressions. Multiplying the mass-action expressions together and cancelling common terms, we see that: K a K b = H3 O +] A ] HA] HA] OH ] A ] = H 3 O +] OH ] = K w (20) For example, the acid ionization constant of acetic acid (CH 3 COOH) is , and the base ionization constant of its conjugate base, acetate ion ( CH 3 COO ), is The product of these two constants is indeed equal to K w : K a K b = ( ) ( ) = = K w (21) 3 "Ionization Constants of Weak Bases" <

5 OpenStax-CNX module: m The extent to which an acid, HA, donates protons to water molecules depends on the strength of the conjugate base, A, of the acid. If A is a strong base, any protons that are donated to water molecules are recaptured by A. Thus there is relatively little A and H 3 O + in solution, and the acid, HA, is weak. If A is a weak base, water binds the protons more strongly, and the solution contains primarily A and H 3 O + the acid is strong. Strong acids form very weak conjugate bases, and weak acids form stronger conjugate bases (Figure 2). Figure 2: This diagram shows the relative strengths of conjugate acid-base pairs, as indicated by their ionization constants in aqueous solution. Figure 3 lists a series of acids and bases in order of the decreasing strengths of the acids and the corresponding increasing strengths of the bases. The acid and base in a given row are conjugate to each other.

6 OpenStax-CNX module: m Figure 3: The chart shows the relative strengths of conjugate acid-base pairs. The rst six acids in Figure 3 are the most common strong acids. These acids are completely dissociated in aqueous solution. The conjugate bases of these acids are weaker bases than water. When one of these acids dissolves in water, their protons are completely transferred to water, the stronger base. Those acids that lie between the hydronium ion and water in Figure 3 form conjugate bases that can compete with water for possession of a proton. Both hydronium ions and nonionized acid molecules are present in equilibrium in a solution of one of these acids. Compounds that are weaker acids than water (those found below water in the column of acids) in Figure 3 exhibit no observable acidic behavior when dissolved in water. Their conjugate bases are stronger than the hydroxide ion, and if any conjugate base were formed, it would react with water to re-form the acid. The extent to which a base forms hydroxide ion in aqueous solution depends on the strength of the base relative to that of the hydroxide ion, as shown in the last column in Figure 3. A strong base, such as one

7 OpenStax-CNX module: m of those lying below hydroxide ion, accepts protons from water to yield 100% of the conjugate acid and hydroxide ion. Those bases lying between water and hydroxide ion accept protons from water, but a mixture of the hydroxide ion and the base results. Bases that are weaker than water (those that lie above water in the column of bases) show no observable basic behavior in aqueous solution. Example 2 The Product Ka Kb = Kw Use the K b for the nitrite ion, NO 2, to calculate the K a for its conjugate acid. Solution K b for NO 2 is given in this section as The conjugate acid of NO 2 is HNO 2 ; K a for HNO 2 can be calculated using the relationship: Solving for K a, we get: K a K b = = K w (22) K a = K w K b = = (23) This answer can be veried by nding the K a for HNO 2 in Appendix H 4. Check Your Learning We can determine the relative acid strengths of NH 4+ and HCN by comparing their ionization constants. The ionization constant of HCN is given in Appendix H 5 as The ionization constant of NH 4+ is not listed, but the ionization constant of its conjugate base, NH 3, is listed as Determine the ionization constant of NH 4 +, and decide which is the stronger acid, HCN or NH 4 +. note: NH 4+ is the slightly stronger acid (K a for NH 4+ = ). 1 The Ionization of Weak Acids and Weak Bases Many acids and bases are weak; that is, they do not ionize fully in aqueous solution. A solution of a weak acid in water is a mixture of the nonionized acid, hydronium ion, and the conjugate base of the acid, with the nonionized acid present in the greatest concentration. Thus, a weak acid increases the hydronium ion concentration in an aqueous solution (but not as much as the same amount of a strong acid). Acetic acid, CH 3 CO 2 H, is a weak acid. When we add acetic acid to water, it ionizes to a small extent according to the equation: CH 3 CO 2 H (aq) + H 2 O (l) H 3 O + (aq) + CH 3 CO 2 (aq), (24) giving an equilibrium mixture with most of the acid present in the nonionized (molecular) form. This equilibrium, like other equilibria, is dynamic; acetic acid molecules donate hydrogen ions to water molecules and form hydronium ions and acetate ions at the same rate that hydronium ions donate hydrogen ions to acetate ions to reform acetic acid molecules and water molecules. We can tell by measuring the ph of an aqueous solution of known concentration that only a fraction of the weak acid is ionized at any moment (Figure 4). The remaining weak acid is present in the nonionized form. For acetic acid, at equilibrium: H3 O +] CH 3 CO 2 ] K a = = (25) CH 3 CO 2 H] 4 "Ionization Constants of Weak Acids" < 5 "Ionization Constants of Weak Acids" <

8 OpenStax-CNX module: m Figure 4: ph indicates that a 0.l-M solution of HCl (beaker on left) has a ph of 1. ˆ paper + ionized and = M 2 H3 O = 0.1 M. A 0.1- M ) because the weak acid CH CO H is only partially ionized. CH CO H]. (credit: modi cation of work by Sahar Atwa) The acid is fully ˆ solution of CH CO H (beaker on right) is a ph of 3 ( H3 O ˆ In this solution, H3 O + + <

9 OpenStax-CNX module: m Ionization Constants of Some Weak Acids Ionization Reaction Ka at 25 C HSO 4 + H 2 O H 3 O + + SO 4 2 HF + H 2 O H 3 O + + F HNO 2 + H 2 O H 3 O + + NO 2 HCNO + H 2 O H 3 O + + NCO HCO 2 H + H 2 O H 3 O + + HCO 2 CH 3 CO 2 H + H 2 O H 3 O + + CH 3 CO 2 HCIO + H 2 O H 3 O + + CIO HBrO + H 2 O H 3 O + + BrO HCN + H 2 O H 3 O + + CN Table Table 1 gives the ionization constants for several weak acids; additional ionization constants can be found in Appendix H 6. At equilibrium, a solution of a weak base in water is a mixture of the nonionized base, the conjugate acid of the weak base, and hydroxide ion with the nonionized base present in the greatest concentration. Thus, a weak base increases the hydroxide ion concentration in an aqueous solution (but not as much as the same amount of a strong base). For example, a solution of the weak base trimethylamine, (CH 3 ) 3 N, in water reacts according to the equation: (CH 3 ) 3 N (aq) + H 2 O (l) (CH 3 ) 3 NH + (aq) + OH (aq), (26) giving an equilibrium mixture with most of the base present as the nonionized amine. This equilibrium is analogous to that described for weak acids. We can conrm by measuring the ph of an aqueous solution of a weak base of known concentration that only a fraction of the base reacts with water (Figure 5). The remaining weak base is present as the unreacted form. The equilibrium constant for the ionization of a weak base, K b, is called the ionization constant of the weak base, and is equal to the reaction quotient when the reaction is at equilibrium. For trimethylamine, at equilibrium: (CH3 ) K b = 3 NH +] OH ] (27) (CH 3 ) 3 N] 6 "Ionization Constants of Weak Acids" <

10 OpenStax-CNX module: m Figure 5: ph paper indicates that a 0.1-M solution of NH 3 (left) is weakly basic. The solution has a poh of 3 (OH ] = M) because the weak base NH 3 only partially reacts with water. A 0.1-M solution of NaOH (right) has a poh of 1 because NaOH is a strong base. (credit: modication of work by Sahar Atwa) The ionization constants of several weak bases are given in Table 2 and in Appendix I 7. Ionization Constants of Some Weak Bases Ionization Reaction Kb at 25 C (CH 3 ) 2 NH + H 2 O (CH 3 ) 2 NH OH CH 3 NH 2 + H 2 O CH 3 NH OH (CH 3 ) 3 N + H 2 O (CH 3 ) 3 NH + + OH NH 3 + H 2 O NH OH C 6 H 5 NH 2 + H 2 O C 6 N 5 NH OH Table Example 3 Determination of Ka from Equilibrium Concentrations Acetic acid is the principal ingredient in vinegar (Figure 6); that's why it tastes sour. At equilibrium, a solution contains CH 3 CO 2 H] = M and H 3 O +] = CH 3 CO 2 ] = M. What is the value of K a for acetic acid? 7 "Ionization Constants of Weak Bases" <

11 OpenStax-CNX module: m Figure 6: Vinegar is a solution of acetic acid, a weak acid. (credit: modication of work by HomeSpot HQ/Flickr) Solution We are asked to calculate an equilibrium constant from equilibrium concentrations. At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction: CH 3 CO 2 H (aq) + H 2 O (l) H 3 O + (aq) + CH 3 CO 2 (aq) (28) K a = H3 O +] CH 3 CO 2 ] CH 3 CO 2 H] = ( ) ( ) = (29)

12 OpenStax-CNX module: m Check Your Learning What is the equilibrium constant for the ionization of the HSO 4 ion, the weak acid used in some household cleansers: HSO 4 (aq) + H 2 O (l) H 3 O + (aq) + SO 4 2 (aq) (30) In one mixture of NaHSO 4 and Na 2 SO 4 at equilibrium, H 3 O +] = M; HSO 4 ] = 0.29 M; and SO 4 2 ] = 0.13 M. note: K a for HSO 4 = Example 4 Determination of Kb from Equilibrium Concentrations Caeine, C 8 H 10 N 4 O 2 is a weak base. What is the value of K b for caeine if a solution at equilibrium has C 8 H 10 N 4 O 2 ] = M, C 8 H 10 N 4 O 2 H +] = M, and OH ] = M? Solution At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction: C 8 H 10 N 4 O 2 (aq) + H 2 O (l) C 8 H 10 N 4 O 2 H + (aq) + OH (aq) (31) C8 H 10 N 4 O 2 H +] OH ] ( ) ( ) K b = = = (32) C 8 H 10 N 4 O 2 ] Check Your Learning What is the equilibrium constant for the ionization of the HPO 2 4 ion, a weak base: HPO 4 2 (aq) + H 2 O (l) H 2 PO 4 (aq) + OH (aq) (33) In a solution containing a mixture of NaH 2 PO 4 and Na 2 HPO 4 at equilibrium, OH ] = M; H 2 PO 4 ] = M; and HPO 4 2 ] = M. note: K b for HPO 4 2 = Example 5 Determination of Ka or Kb from ph The ph of a M solution of nitrous acid, HNO 2, is What is its K a? HNO 2 (aq) + H 2 O (l) H 3 O + (aq) + NO 2 (aq) (34) Solution We determine an equilibrium constant starting with the initial concentrations of HNO 2, H 3 O +, and NO 2 as well as one of the nal concentrations, the concentration of hydronium ion at equilibrium. (Remember that ph is simply another way to express the concentration of hydronium ion.) We can solve this problem with the following steps in which x is a change in concentration of a species in the reaction:

13 OpenStax-CNX module: m We can summarize the various concentrations and changes as shown here (the concentration of water does not appear in the expression for the equilibrium constant, so we do not need to consider its concentration): To get the various values in the ICE (Initial, Change, Equilibrium) table, we rst calculate H 3 O +], the equilibrium concentration of H 3 O +, from the ph: H3 O +] = = M (35) The change in concentration of H 3 O +,x H3O + ], is the dierence between the equilibrium concentration of H 3 O +, which we determined from the ph, and the initial concentration, H 3 O +]. The i initial concentration of H 3 O + is its concentration in pure water, which is so much less than the nal concentration that we approximate it as zero ( 0). The change in concentration of NO 2 is equal to the change in concentration of H 3 O +]. For each 1 mol of H 3 O + that forms, 1 mol of NO 2 forms. The equilibrium concentration of HNO 2 is equal to its initial concentration plus the change in its concentration. Now we can ll in the ICE table with the concentrations at equilibrium, as shown here:

14 OpenStax-CNX module: m Finally, we calculate the value of the equilibrium constant using the data in the table: K a = H3 O +] NO 2 ] (0.0046) (0.0046) = HNO 2 ] (0.0470) = (36) Check Your Learning. The ph of a solution of household ammonia, a M solution of NH 3, is What is K b for NH 3. note: K b = Example 6 Equilibrium Concentrations in a Solution of a Weak Acid Formic acid, HCO 2 H, is the irritant that causes the body's reaction to ant stings (Figure 7). Figure 7: The pain of an ant's sting is caused by formic acid. (credit: John Tann)

15 OpenStax-CNX module: m What is the concentration of hydronium ion and the ph in a M solution of formic acid? Solution HCO 2 H (aq) + H 2 O (l) H 3 O + (aq) + HCO 2 (aq) K a = (37) Step 1. Determine x and equilibrium concentrations. The equilibrium expression is: HCO 2 H (aq) + H 2 O (l) H 3 O + (aq) + HCO 2 (aq) (38) The concentration of water does not appear in the expression for the equilibrium constant, so we do not need to consider its change in concentration when setting up the ICE table. The table shows initial concentrations (concentrations before the acid ionizes), changes in concentration, and equilibrium concentrations follows (the data given in the problem appear in color): Step 2. Solve for x and the equilibrium concentrations. At equilibrium: K a = = H3 O +] HCO 2 ] HCO 2 H] (39) = (x) (x) x = (40) Now solve for x. Because the initial concentration of acid is reasonably large and K a is very small, we assume that x 0.534, which permits us to simplify the denominator term as (0.534 x) = This gives: Solve for x as follows: K a = = x (41) x 2+ = ( ) = (42) x = (43) = (44) To check the assumption that x is small compared to 0.534, we calculate: x = = (1.8% of 0.534) (45)

16 OpenStax-CNX module: m x is less than 5% of the initial concentration; the assumption is valid. We nd the equilibrium concentration of hydronium ion in this formic acid solution from its initial concentration and the change in that concentration as indicated in the last line of the table: H3 O +] = 0 + x = M. (46) = M (47) The ph of the solution can be found by taking the negative log of the H 3 O +], so: log ( ) = 2.01 (48) Check Your Learning Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of acetic acid in a M solution of acetic acid, CH 3 CO 2 H? CH 3 CO 2 H (aq) + H 2 O (l) H 3 O + (aq) + CH 3 CO 2 (aq) K a = (49) (Hint: Determine CH 3 CO 2 ] at equilibrium.) Recall that the percent ionization is the fraction of CH 3CO 2 ] acetic acid that is ionized 100, or CH 3CO 2H] 100. initial note: percent ionization = 1.3% The following example shows that the concentration of products produced by the ionization of a weak base can be determined by the same series of steps used with a weak acid. Example 7 Equilibrium Concentrations in a Solution of a Weak Base Find the concentration of hydroxide ion in a 0.25-M solution of trimethylamine, a weak base: (CH 3 ) 3 N (aq) + H 2 O (l) (CH 3 ) 3 NH + (aq) + OH (aq) K b = (50) Solution This problem requires that we calculate an equilibrium concentration by determining concentration changes as the ionization of a base goes to equilibrium. The solution is approached in the same way as that for the ionization of formic acid in Example 6. The reactants and products will be dierent and the numbers will be dierent, but the logic will be the same:

17 OpenStax-CNX module: m Step 1. Determine x and equilibrium concentrations. The table shows the changes and concentrations: Step 2. Solve for x and the equilibrium concentrations. At equilibrium: (CH3 ) K b = 3 NH +] OH ] (x) (x) = (CH 3 ) 3 N] 0.25 x = (51) If we assume that x is small relative to 0.25, then we can replace (0.25 x) in the preceding equation with Solving the simplied equation gives: x = (52) This change is less than 5% of the initial concentration (0.25), so the assumption is justied. Recall that, for this computation, x is equal to the equilibrium concentration of hydroxide ion in the solution (see earlier tabulation): OH ] = 0 + x = x = M (53) = M (54) Then calculate poh as follows: poh = log ( ) = 2.40 (55) Using the relation introduced in the previous section of this chapter: ph + poh = pk w = (56) permits the computation of ph: ph = poh = = (57) Step 3. Check the work. A check of our arithmetic shows that K b = Check Your Learning (a) Show that the calculation in Step 2 of this example gives an x of and the calculation in Step 3 shows K b = (b) Find the concentration of hydroxide ion in a M solution of ammonia, a weak base with a K b of Calculate the percent ionization of ammonia, the fraction ionized 100, or NH4+ ] NH 3] 100

18 OpenStax-CNX module: m note: M, 2.33% Some weak acids and weak bases ionize to such an extent that the simplifying assumption that x is small relative to the initial concentration of the acid or base is inappropriate. As we solve for the equilibrium concentrations in such cases, we will see that we cannot neglect the change in the initial concentration of the acid or base, and we must solve the equilibrium equations by using the quadratic equation. Example 8 Equilibrium Concentrations in a Solution of a Weak Acid Sodium bisulfate, NaHSO 4, is used in some household cleansers because it contains the HSO 4 ion, a weak acid. What is the ph of a 0.50-M solution of HSO 4? HSO 4 (aq) + H 2 O (l) H 3 O + (aq) + SO 4 2 (aq) K a = (58) Solution We need to determine the equilibrium concentration of the hydronium ion that results from the ionization of HSO 4 so that we can use H 3 O +] to determine the ph. As in the previous examples, we can approach the solution by the following steps: Step 1. Determine x and equilibrium concentrations. This table shows the changes and concentrations: Step 2. Solve for x and the concentrations. As we begin solving for x, we will nd this is more complicated than in previous examples. As we discuss these complications we should not lose track of the fact that it is still the purpose of this step to determine the value of x. At equilibrium: K a = H3 O +] SO ] 2 4 (x) (x) = = (59) HSO 4 ] 0.50 x If we assume that x is small and approximate (0.50 x) as 0.50, we nd: x = (60)

19 OpenStax-CNX module: m When we check the assumption, we calculate: x HSO 4 ] i (61) x 0.50 = = 0.15 (15%) (62) The value of x is not less than 5% of 0.50, so the assumption is not valid. quadratic formula to nd x. The equation: We need the K a = = (x) (x) 0.50 x (63) gives x = x 2+ (64) or x x = 0 (65) This equation can be solved using the quadratic formula. For an equation of the form ax 2+ + bx + c = 0, (66) x is given by the equation: x = b ± b 2+ 4ac 2a (67) In this problem, a = 1, b = , and c = Solving for x gives a negative root (which cannot be correct since concentration cannot be negative) and a positive root: x = (68) Now determine the hydronium ion concentration and the ph: H3 O +] = 0 + x = M (69) = M (70) The ph of this solution is: ph = log H 3 O +] = log = 1.14 (71)

20 OpenStax-CNX module: m Check Your Learning (a) Show that the quadratic formula gives x = (b) Calculate the ph in a M solution of caeine, a weak base: C 8 H 10 N 4 O 2 (aq) + H 2 O (l) C 8 H 10 N 4 O 2 H + (aq) + OH (aq) K b = (72) (Hint: It will be necessary to convert OH ] to H 3 O +] or poh to ph toward the end of the calculation.) note: ph The Relative Strengths of Strong Acids and Bases Strong acids, such as HCl, HBr, and HI, all exhibit the same strength in water. The water molecule is such a strong base compared to the conjugate bases Cl, Br, and I that ionization of these strong acids is essentially complete in aqueous solutions. In solvents less basic than water, we nd HCl, HBr, and HI dier markedly in their tendency to give up a proton to the solvent. For example, when dissolved in ethanol (a weaker base than water), the extent of ionization increases in the order HCl < HBr < HI, and so HI is demonstrated to be the strongest of these acids. The inability to discern dierences in strength among strong acids dissolved in water is known as the leveling eect of water. Water also exerts a leveling eect on the strengths of strong bases. For example, the oxide ion, O 2, and the amide ion, NH 2, are such strong bases that they react completely with water: O 2 (aq) + H 2 O (l) U+27F6] OH (aq) + OH (aq) (73) NH 2 (aq) + H 2 O (l) U+27F6] NH 3 (aq) + OH (aq) (74) Thus, O 2 and NH 2 appear to have the same base strength in water; they both give a 100% yield of hydroxide ion. 3 Eect of Molecular Structure on Acid-Base Strength In the absence of any leveling eect, the acid strength of binary compounds of hydrogen with nonmetals (A) increases as the H-A bond strength decreases down a group in the periodic table. For group 7A, the order of increasing acidity is HF < HCl < HBr < HI. Likewise, for group 6A, the order of increasing acid strength is H 2 O < H 2 S < H 2 Se < H 2 Te. Across a row in the periodic table, the acid strength of binary hydrogen compounds increases with increasing electronegativity of the nonmetal atom because the polarity of the H-A bond increases. Thus, the order of increasing acidity (for removal of one proton) across the second row is CH 4 < NH 3 < H 2 O < HF; across the third row, it is SiH 4 < PH 3 < H 2 S < HCl (see Figure 8).

21 OpenStax-CNX module: m Figure 8: As you move from left to right and down the periodic table, the acid strength increases. As you move from right to left and up, the base strength increases. Compounds containing oxygen and one or more hydroxyl (OH) groups can be acidic, basic, or amphoteric, depending on the position in the periodic table of the central atom E, the atom bonded to the hydroxyl group. Such compounds have the general formula O n E(OH) m, and include sulfuric acid, O 2 S(OH) 2, sulfurous acid, OS(OH) 2, nitric acid, O 2 NOH, perchloric acid, O 3 ClOH, aluminum hydroxide, Al(OH) 3, calcium hydroxide, Ca(OH) 2, and potassium hydroxide, KOH: If the central atom, E, has a low electronegativity, its attraction for electrons is low. Little tendency exists for the central atom to form a strong covalent bond with the oxygen atom, and bond a between the element and oxygen is more readily broken than bond b between oxygen and hydrogen. Hence bond a is ionic, hydroxide ions are released

22 OpenStax-CNX module: m to the solution, and the material behaves as a basethis is the case with Ca(OH) 2 and KOH. Lower electronegativity is characteristic of the more metallic elements; hence, the metallic elements form ionic hydroxides that are by denition basic compounds. If, on the other hand, the atom E has a relatively high electronegativity, it strongly attracts the electrons it shares with the oxygen atom, making bond a relatively strongly covalent. The oxygen-hydrogen bond, bond b, is thereby weakened because electrons are displaced toward E. Bond b is polar and readily releases hydrogen ions to the solution, so the material behaves as an acid. High electronegativities are characteristic of the more nonmetallic elements. Thus, nonmetallic elements form covalent compounds containing acidic OH groups that are called oxyacids. Increasing the oxidation number of the central atom E also increases the acidity of an oxyacid because this increases the attraction of E for the electrons it shares with oxygen and thereby weakens the O-H bond. Sulfuric acid, H 2 SO 4, or O 2 S(OH) 2 (with a sulfur oxidation number of +6), is more acidic than sulfurous acid, H 2 SO 3, or OS(OH) 2 (with a sulfur oxidation number of +4). Likewise nitric acid, HNO 3, or O 2 NOH (N oxidation number = +5), is more acidic than nitrous acid, HNO 2, or ONOH (N oxidation number = +3). In each of these pairs, the oxidation number of the central atom is larger for the stronger acid (Figure 9). Figure 9: As the oxidation number of the central atom E increases, the acidity also increases. Hydroxy compounds of elements with intermediate electronegativities and relatively high oxidation numbers (for example, elements near the diagonal line separating the metals from the nonmetals in the periodic table) are usually amphoteric. This means that the hydroxy compounds act as acids when they react with strong bases and as bases when they react with strong acids. The amphoterism of aluminum hydroxide, which commonly exists as the hydrate Al(H 2 O) 3 (OH) 3, is reected in its solubility in both strong acids and strong bases. In strong bases, the relatively insoluble hydrated aluminum hydroxide, Al(H 2 O) 3 (OH) 3, is

23 OpenStax-CNX module: m converted into the soluble ion, Al(H 2 O) 2 (OH) 4 ], by reaction with hydroxide ion: Al(H 2 O) 3 (OH) 3 (aq) + OH (aq) H 2 O (l) + Al(H 2 O) 2 (OH) 4 ] (aq) (75) In this reaction, a proton is transferred from one of the aluminum-bound H 2 O molecules to a hydroxide ion in solution. The Al(H 2 O) 3 (OH) 3 compound thus acts as an acid under these conditions. On the other hand, when dissolved in strong acids, it is converted to the soluble ion Al(H 2 O) 6 ] 3+ by reaction with hydronium ion: 3H 3 O + (aq) + Al(H 2 O) 3 (OH) 3 (aq) Al(H 2 O) 6 3+ (aq) + 3H 2 O (l) (76) In this case, protons are transferred from hydronium ions in solution to Al(H 2 O) 3 (OH) 3, and the compound functions as a base. 4 Key Concepts and Summary The strengths of Brønsted-Lowry acids and bases in aqueous solutions can be determined by their acid or base ionization constants. Stronger acids form weaker conjugate bases, and weaker acids form stronger conjugate bases. Thus strong acids are completely ionized in aqueous solution because their conjugate bases are weaker bases than water. Weak acids are only partially ionized because their conjugate bases are strong enough to compete successfully with water for possession of protons. Strong bases react with water to quantitatively form hydroxide ions. Weak bases give only small amounts of hydroxide ion. The strengths of the binary acids increase from left to right across a period of the periodic table (CH 4 < NH 3 < H 2 O < HF), and they increase down a group (HF < HCl < HBr < HI). The strengths of oxyacids that contain the same central element increase as the oxidation number of the element increases (H 2 SO 3 < H 2 SO 4 ). The strengths of oxyacids also increase as the electronegativity of the central element increases H 2 SeO 4 < H 2 SO 4 ]. 5 Key Equations ˆ K a = H3O+ ]A ] HA] ˆ K b = HB+ ]OH ] B] ˆ K a K b = = K w ˆ Percent ionization = H3O+ ] eq HA] Chemistry End of Chapter Exercises Exercise 1 Explain why the neutralization reaction of a strong acid and a weak base gives a weakly acidic solution. Exercise 2 (Solution on p. 30.) Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basic solution. Exercise 3 Use this list of important industrial compounds (and Figure 3) to answer the following questions regarding: CaO, Ca(OH) 2, CH 3 CO 2 H, CO 2, HCl, H 2 CO 3, HF, HNO 2, HNO 3, H 3 PO 4, H 2 SO 4, NH 3, NaOH, Na 2 CO 3. (a) Identify the strong Brønsted-Lowry acids and strong Brønsted-Lowry bases. (b) List those compounds in (a) that can behave as Brønsted-Lowry acids with strengths lying between those of H 3 O + and H 2 O.

24 OpenStax-CNX module: m (c) List those compounds in (a) that can behave as Brønsted-Lowry bases with strengths lying between those of H 2 O and OH. Exercise 4 (Solution on p. 30.) The odor of vinegar is due to the presence of acetic acid, CH 3 CO 2 H, a weak acid. List, in order of descending concentration, all of the ionic and molecular species present in a 1- M aqueous solution of this acid. Exercise 5 Household ammonia is a solution of the weak base NH 3 in water. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this base. Exercise 6 (Solution on p. 30.) Explain why the ionization constant, K a, for H 2 SO 4 is larger than the ionization constant for H 2 SO 3. Exercise 7 Explain why the ionization constant, K a, for HI is larger than the ionization constant for HF. Exercise 8 (Solution on p. 30.) Gastric juice, the digestive uid produced in the stomach, contains hydrochloric acid, HCl. Milk of Magnesia, a suspension of solid Mg(OH) 2 in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction, and identify the conjugate acid-base pairs. Exercise 9 Nitric acid reacts with insoluble copper(ii) oxide to form soluble copper(ii) nitrate, Cu(NO 3 ) 2, a compound that has been used to prevent the growth of algae in swimming pools. Write the balanced chemical equation for the reaction of an aqueous solution of HNO 3 with CuO. Exercise 10 (Solution on p. 30.) What is the ionization constant at 25 C for the weak acid CH 3 NH + 3, the conjugate acid of the weak base CH 3 NH 2, K b = Exercise 11 What is the ionization constant at 25 C for the weak acid (CH 3 ) 2 NH + 2, the conjugate acid of the weak base (CH 3 ) 2 NH, K b = ? Exercise 12 (Solution on p. 30.) Which base, CH 3 NH 2 or (CH 3 ) 2 NH, is the stronger base? Which conjugate acid, (CH 3 ) 2 NH 2+ or (CH 3 ) 2 NH, is the stronger acid? Exercise 13 Which is the stronger acid, NH 4+ or HBrO? Exercise 14 (Solution on p. 30.) Which is the stronger base, (CH 3 ) 3 N or H 2 BO 3? Exercise 15 Predict which acid in each of the following pairs is the stronger and explain your reasoning for each. (a) H 2 O or HF (b) B(OH) 3 or Al(OH) 3 (c) HSO 3 or HSO 4 (d) NH 3 or H 2 S (e) H 2 O or H 2 Te Exercise 16 (Solution on p. 30.) Predict which compound in each of the following pairs of compounds is more acidic and explain your reasoning for each. (a) HSO 4 or HSeO 4

25 OpenStax-CNX module: m (b) NH 3 or H 2 O (c) PH 3 or HI (d) NH 3 or PH 3 (e) H 2 S or HBr Exercise 17 Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign. (a) acidity: HCl, HBr, HI (b) basicity: H 2 O, OH, H, Cl (c) basicity: Mg(OH) 2, Si(OH) 4, ClO 3 (OH) (Hint: Formula could also be written as HClO 4 ). (d) acidity: HF, H 2 O, NH 3, CH 4 Exercise 18 (Solution on p. 30.) Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign. (a) acidity: NaHSO 3, NaHSeO 3, NaHSO 4 (b) basicity: BrO 2,ClO 2,IO 2 (c) acidity: HOCl, HOBr, HOI (d) acidity: HOCl, HOClO, HOClO 2, HOClO 3 (e) basicity: NH 2, HS, HTe, PH 2 (f) basicity: BrO, BrO 2,BrO 3,BrO 4 Exercise 19 Both HF and HCN ionize in water to a limited extent. Which of the conjugate bases, F or CN, is the stronger base? See Table 2. Exercise 20 (Solution on p. 30.) The active ingredient formed by aspirin in the body is salicylic acid, C 6 H 4 OH(CO 2 H). The carboxyl group ( CO 2 H) acts as a weak acid. The phenol group (an OH group bonded to an aromatic ring) also acts as an acid but a much weaker acid. List, in order of descending concentration, all of the ionic and molecular species present in a M aqueous solution of C 6 H 4 OH(CO 2 H). Exercise 21 What do we represent when we write: CH 3 CO 2 H (aq) + H 2 O (l) H 3 O + (aq) + CH 3 CO 2 (aq)? Exercise 22 (Solution on p. 30.) Explain why equilibrium calculations are not necessary to determine ionic concentrations in solutions of certain strong electrolytes such as NaOH and HCl. Under what conditions are equilibrium calculations necessary as part of the determination of the concentrations of all ions of some other strong electrolytes in solution? Exercise 23 Are the concentrations of hydronium ion and hydroxide ion in a solution of an acid or a base in water directly proportional or inversely proportional? Explain your answer. Exercise 24 (Solution on p. 30.) What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak acid? Exercise 25 What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak base? Exercise 26 (Solution on p. 31.) Which of the following will increase the percent of NH 3 that is converted to the ammonium ion in water (Hint: Use LeChâtelier's principle.)?

26 OpenStax-CNX module: m (a) addition of NaOH (b) addition of HCl (c) addition of NH 4 Cl Exercise 27 Which of the following will increase the percent of HF that is converted to the uoride ion in water? (a) addition of NaOH (b) addition of HCl (c) addition of NaF Exercise 28 (Solution on p. 31.) What is the eect on the concentrations of NO 2, HNO 2, and OH when the following are added to a solution of KNO 2 in water: (a) HCl (b) HNO 2 (c) NaOH (d) NaCl (e) KNO The equation for the equilibrium is: NO 2 (aq) + H 2 O (l) HNO 2 (aq) + OH (aq) Exercise 29 What is the eect on the concentration of hydrouoric acid, hydronium ion, and uoride ion when the following are added to separate solutions of hydrouoric acid? (a) HCl (b) KF (c) NaCl (d) KOH (e) HF The equation for the equilibrium is: HF (aq) + H 2 O (l) H 3 O + (aq) + F (aq) Exercise 30 (Solution on p. 31.) Why is the hydronium ion concentration in a solution that is 0.10 M in HCl and 0.10 M in HCOOH determined by the concentration of HCl? Exercise 31 From the equilibrium concentrations given, calculate K a for each of the weak acids and K b for each of the weak bases. (a) CH 3 CO 2 H: H 3 O +] = M; CH 3 CO 2 ] = M; CH 3 CO 2 H] = M; (b) ClO : OH ] = M; HClO] = M; ClO ] = M; (c) HCO 2 H: HCO 2 H] = M; H3 O +] = M; HCO 2 ] = M; (d) C 6 H 5 NH 3 + :C 6 H 5 NH 3 + ] = M; C 6 H 5 NH 2 ] = M; H3 O +] = M

27 OpenStax-CNX module: m Exercise 32 (Solution on p. 31.) From the equilibrium concentrations given, calculate K a for each of the weak acids and K b for each of the weak bases. (a) NH 3 : OH ] = M; NH 4 + ] = M; NH 3 ] = M; (b) HNO 2 : H 3 O +] = M; NO 2 ] = M; HNO 2 ] = 1.07 M; (c) (CH 3 ) 3 N: (CH 3 ) 3 N] = 0.25 M; (CH 3 ) 3 NH + ] = M; OH ] = M; (d) NH 4 + :NH 4 + ] = M; NH 3 ] = M; H 3 O + ] = M Exercise 33 Determine K b for the nitrite ion, NO 2. In a 0.10-M solution this base is % ionized. Exercise 34 (Solution on p. 31.) Determine K a for hydrogen sulfate ion, HSO 4. In a 0.10-M solution the acid is 29% ionized. Exercise 35 Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid: (a) F (b) NH 4 + (c) AsO 4 3 (d) (CH 3 ) 2 NH 2 + (e) NO 2 (f) HC 2 O 4 (as a base) Exercise 36 (Solution on p. 31.) Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid: (a) HTe (as a base) (b) (CH 3 ) 3 NH + (c) HAsO 4 3 (as a base) (d) HO 2 (as a base) (e) C 6 H 5 NH 3 + (f) HSO 3 (as a base) Exercise 37 For which of the following solutions must we consider the ionization of water when calculating the ph or poh? (a) M HNO 3 (b) 0.10 g HCl in 1.0 L of solution (c) g NaOH in 0.50 L of solution (d) M Ca(OH) 2 (e) M KNO 3 Exercise 38 (Solution on p. 31.) Even though both NH 3 and C 6 H 5 NH 2 are weak bases, NH 3 is a much stronger acid than C 6 H 5 NH 2. Which of the following is correct at equilibrium for a solution that is initially 0.10 M in NH 3 and 0.10 M in C 6 H 5 NH 2?

28 OpenStax-CNX module: m (a) OH ] = NH 4 + ] (b) NH 4 + ] = C 6 H 5 NH 3 + ] (c) OH ] = C 6 H 5 NH 3 + ] (d) NH 3 ] = C 6 H 5 NH 2 ] (e) both a and b are correct Exercise 39 Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.25 M in HCO 2 H and 0.10 M in HClO. Exercise 40 (Solution on p. 31.) Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is M in HNO 2 and M in HBrO. Exercise 41 Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.25 M in CH 3 NH 2 and 0.10 M in C 5 H 5 N (K b = ). Exercise 42 (Solution on p. 31.) Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is M in NH 3 and M in C 6 H 5 NH 2. Exercise 43 Using the K a value of , place Al(H 2 O) 6 3+ in the correct location in Figure 3. Exercise 44 (Solution on p. 31.) Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected. Ionization constants can be found in Appendix H 8 and Appendix I 9. (a) M HClO, a weak acid (b) M C 6 H 5 NH 2, a weak base (c) M HCN, a weak acid (d) 0.11 M (CH 3 ) 3 N, a weak base (e) MFe(H 2 O) 6 2+ a weak acid, K a = Exercise 45 Propionic acid, C 2 H 5 CO 2 H (K a = ), is used in the manufacture of calcium propionate, a food preservative. What is the hydronium ion concentration in a M solution of C 2 H 5 CO 2 H? Exercise 46 (Solution on p. 32.) White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white vinegar is g/cm 3, what is the ph? Exercise 47 The ionization constant of lactic acid, CH 3 CH(OH)CO 2 H, an acid found in the blood after strenuous exercise, is If 20.0 g of lactic acid is used to make a solution with a volume of 1.00 L, what is the concentration of hydronium ion in the solution? Exercise 48 (Solution on p. 32.) Nicotine, C 10 H 14 N 2, is a base that will accept two protons (K 1 = , K 2 = ). What is the concentration of each species present in a M solution of nicotine? Exercise 49 The ph of a 0.20-M solution of HF is Determine K a for HF from these data. Exercise 50 (Solution on p. 32.) The ph of a 0.15-M solution of HSO 4 is Determine K a for HSO 4 from these data. 8 "Ionization Constants of Weak Acids" < 9 "Ionization Constants of Weak Bases" <

29 OpenStax-CNX module: m Exercise 51 The ph of a 0.10-M solution of caeine is Determine K b for caeine from these data: C 8 H 10 N 4 O 2 (aq) + H 2 O (l) C 8 H 10 N 4 O 2 H + (aq) + OH (aq) Exercise 52 (Solution on p. 32.) The ph of a solution of household ammonia, a M solution of NH 3, is Determine K b for NH 3 from these data.

30 OpenStax-CNX module: m Solutions to Exercises in this Module Solution to Exercise (p. 23) The salt ionizes in solution, but the anion slightly reacts with water to form the weak acid. This reaction also forms OH, which causes the solution to be basic. Solution to Exercise (p. 24) H 2 O] > CH 3 CO 2 H] > H 3 O +] CH 3 CO 2 ] > OH ] Solution to Exercise (p. 24) The oxidation state of the sulfur in H 2 SO 4 is greater than the oxidation state of the sulfur in H 2 SO 3. Solution to Exercise (p. 24) Mg(OH) 2 (s) + 2HCl (aq) U+27F6] Mg 2+ (aq) + 2Cl (aq) + 2H 2 O (l) BB BA CB CA Solution to Exercise (p. 24) K a = Solution to Exercise (p. 24) The stronger base or stronger acid is the one with the larger K b or K a, respectively. In these two examples, they are (CH 3 ) 2 NH and CH 3 NH + 3. Solution to Exercise (p. 24) triethylamine. Solution to Exercise (p. 24) (a) HSO 4 ; higher electronegativity of the central ion. (b) H 2 O; NH 3 is a base and water is neutral, or decide on the basis of K a values. (c) HI; PH 3 is weaker than HCl; HCl is weaker than HI. Thus, PH 3 is weaker than HI. (d) PH 3 ; in binary compounds of hydrogen with nonmetals, the acidity increases for the element lower in a group. (e) HBr; in a period, the acidity increases from left to right; in a group, it increases from top to bottom. Br is to the left and below S, so HBr is the stronger acid. Solution to Exercise (p. 25) (a) NaHSeO 3 < NaHSO 3 < NaHSO 4 ; in polyoxy acids, the more electronegative central elements, in this caseforms the stronger acid. The larger number of oxygen atoms on the central atom (giving it a higher oxidation state) also creates a greater release of hydrogen atoms, resulting in a stronger acid. As a salt, the acidity increases in the same manner. (b) ClO 2 < BrO 2 < IO 2 ; the basicity of the anions in a series of acids will be the opposite of the acidity in their oxyacids. The acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three. (c) HOI < HOBr < HOCl; in a series of the same form of oxyacids, the acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three. (d) HOCl < HOClO < HOClO 2 < HOClO 3 ; in a series of oxyacids of the same central element, the acidity increases as the number of oxygen atoms increases (or as the oxidation state of the central atom increases). (e) HTe < HS PH 2 < NH 2 ;PH 2 and NH 2 are anions of weak bases, so they act as strong bases toward H +. HTe and HS are anions of weak acids, so they have less basic character. In a periodic group, the more electronegative element has the more basic anion. (f) BrO 4 < BrO 3 < BrO 2 < BrO ; with a larger number of oxygen atoms (that is, as the oxidation state of the central ion increases), the corresponding acid becomes more acidic and the anion consequently less basic. Solution to Exercise (p. 25) H 2 O] > C 6 H 4 OH (CO 2 H)] > H + ] 0 > C 6 H 4 OH(CO 2 ) C 6 H 4 O(CO 2 H) ] > OH ] Solution to Exercise (p. 25) Strong electrolytes are 100% ionized, and, as long as the component ions are neither weak acids nor weak bases, the ionic species present result from the dissociation of the strong electrolyte. Equilibrium calculations are necessary when one (or more) of the ions is a weak acid or a weak base.

31 OpenStax-CNX module: m Solution to Exercise (p. 25) 1. Assume that the change in initial concentration of the acid as the equilibrium is established can be neglected, so this concentration can be assumed constant and equal to the initial value of the total acid concentration. 2. Assume we can neglect the contribution of water to the equilibrium concentration of H 3 O +. Solution to Exercise (p. 25) (b) The addition of HCl Solution to Exercise (p. 26) (a) Adding HCl will add H 3 O + ions, which will then react with the OH ions, lowering their concentration. The equilibrium will shift to the right, increasing the concentration of HNO 2, and decreasing the concentration of NO 2 ions. (b) Adding HNO 2 increases the concentration of HNO 2 and shifts the equilibrium to the left, increasing the concentration of NO 2 ions and decreasing the concentration of OH ions. (c) Adding NaOH adds OH ions, which shifts the equilibrium to the left, increasing the concentration of NO 2 ions and decreasing the concentrations of HNO 2. (d) Adding NaCl has no eect on the concentrations of the ions. (e) Adding KNO 2 adds NO 2 ions and shifts the equilibrium to the right, increasing the HNO 2 and OH ion concentrations. Solution to Exercise (p. 26) This is a case in which the solution contains a mixture of acids of dierent ionization strengths. In solution, the HCO 2 H exists primarily as HCO 2 H molecules because the ionization of the weak acid is suppressed by the strong acid. Therefore, the HCO 2 H contributes a negligible amount of hydronium ions to the solution. The stronger acid, HCl, is the dominant producer of hydronium ions because it is completely ionized. In such a solution, the stronger acid determines the concentration of hydronium ions, and the ionization of the weaker acid is xed by the H 3 O + ] produced by the stronger acid. Solution to Exercise (p. 26) (a) K b = ; (b) K a = ; (c) K b = ; (d) K a = Solution to Exercise (p. 27) K a = Solution to Exercise (p. 27) (a) K b = ; (b) K a = ; (c) K b = ; (d) K b = ; (e) K b = ; (f) K b = Solution to Exercise (p. 27) (a) is the correct statement. Solution to Exercise (p. 28) H 3 O + ] = M HNO 2 ] = OH ] = M BrO ] = M HBrO] = M Solution to Exercise (p. 28) OH ] = NO 4 + ] = M NH 3 ] = M H 3 O + ] = M C 6 H 5 NH 3 + ] = M C 6 H 5 NH 2 ] = M