AP Chemistry Chapter 14 Answers Zumdahl 14.41 ph = log[h + ]; poh = log[oh ]; At 25 o C, ph + poh = 14.00; For Exercise 13.37: ph = log[h + ] = log(1.0 x 10 7 ) = 7.00; poh = 14.00 ph = 14.00 7.00 = 7.00 ph = log(8.3 x 10 16 ) = 15.08; poh = 14.00 15.08 = 1.08 ph = log(12) = 1.08; poh = 14.08 (1.08) = 15.08 ph = log(5.4 x 10 5 ) = 4.27; poh = 14.00 4.27 = 9.73 Note that ph is less than zero when [H + ] is greater than 1.0 M (an extremely acidic solution). For Exercise13.38: poh = log[oh ] = log(1.5) = 0.18; ph = 14.00 poh = 14.00 (0.18) = 14.18 poh = log(3.6 x 10 15 ) = 14.44; ph = 14.00 14.44 = 0.44 poh = log(1.0 x 10 7 ) = 7.00; ph = 14.00 7.00 = 7.00 poh = log(7.3 x 10 4 ) = 3.14; ph = 14.00 3.14 = 10.86 Note that ph is greater than 14.00 when [OH ] is greater than 1.0 M (an extremely basic solution). 14.43 poh = 14.00 6.88 = 7.12; [H + ] = 10 6.88 = 1.3 x 10 7 M [OH ] = 10 7.12 = 7.6 x 10 8 M; acidic [H + ] = 1.0 x 10 14 / 8.4 x 10 14 = 0.12 M; ph = log(0.12) = 0.92 poh = 14.00 0.92 = 13.08; acidic ph = 14.00 3.11 = 10.89; [H + ] = 10 10.89 = 1.3 x 10 11 M [OH ] = 10 3.11 = 7.8 x 10 4 M; basic ph = =log(1.0 x 10 7 ) = 7.00; poh = 14.00 7.00 = 7.00
[OH ] = 10 7.00 = 1.0 x 10 7 M; neutral 14.45 poh = 14.0 ph = 14.0 2.1 = 11.9; [H + ] = 10 ph = 10 2.1 = 8 x 10 3 M (1 sig fig) [OH ] = K w / [H + ] = 1.0 x 10 14 / 8.0 x 10 3 = 1 x 10 12 M or [OH ] = 10 poh = 10 11.9 = 1 x 10 12 M The sample of gastric juice is acidic since the ph is less than 7.00 at 25 o C. 14.47 All the acids in this problem are strong acids that are always assumed to completely dissociate in water. The general dissociation reaction for a strong acid is: HA (aq) H + (aq) + A (aq) where A is the conjugate base of the strong acid HA. For 0.250 M solutions of these strong acids, 0.250 M H + and 0.250 M A are present when the acids completely dissociate. The amount of H + donated from water will be insignificant in this problem since H 2 O is a very weak aci Major species present after dissociation = H +, ClO 4, and H 2 O; ph = log[h + ] = log(0.250) = 0.602 Major species = H +, NO 3, and H 2 O; ph = 0.602 14.49 Both are strong acids. 0.0500 L x 0.050 mol/l = 2.5 x 10 3 mol HCl = 2.5 x 10 3 mol H + + 2.5 x 10 3 mol Cl 0.1500 L x 0.10 mol/l = 1.5 x 10 2 mol HNO 3 = 1.5 x 10 2 mol H + + 1.5 x 10 2 mol NO 3 [H + ] = (2.5 x 10 3 + 1.5 x 10 2 ) mol / 0.2000 L = 0.088 M; [OH ] = K w / [H + ] = 1.1 x 10 13 M [Cl ] = 2.5 x 10 3 mol/0.2000 L = 0.013 M; [NO 3 ] = 1.5 x 10 2 mol/0.2000 L = 0.075 M 14.51 [H + ] = 10 1.50 = 3.16 x 10 2 M (carrying one extra sig fig); M 1 V 1 = M 2 V 2 V 1 = M 2 V 2 / M 1 = 3.16 x 10 2 mol/l x 1.6 L / 12 mol/l = 4.2 x 10 3 L To 4.2 ml of 12 M HCl, add enough water to make 1600 ml of solution. The resulting solution will have [H + ] = 3.2 x 10 2 M and ph = 1.50. 14.53 HNO 2 (K a = 4.0 x 10 4 ) and H 2 O (K a = K w = 1.0 x 10 14 ) are the major species. HNO 2 is a much stronger acid than H 2 O so it is the major source of H +. However, HNO 2 is a weak acid (K a < 1) so it only partially dissociates in water. We must solve an equilibrium problem to determine [H + ]. In the Solutions Guide, we will summarize the initial, change, and equilibrium concentration into one table called the ICE table. Solving the weak acid problem:
HNO 2 H + + NO 2 Initial 0.250 M ~0 0 x mol/l HNO 2 dissociates to reach equilibrium Equil. 0.250 x x x K a = [H + ][NO 2 ] / [HNO 2 ] = 4.0 x 10 4 = x 2 / 0.250 x; If we assume x << 0.250, then: 4.0 x 10 4 x 2 / 0.250, x = (4.0 x 10 4 (0.250)) 1/2 = 0.010 M We must check the assumption: (x / 0.250) x 100 = (0.010 / 0.250) x 100 = 4.0% All the assumptions are goo The H + contribution from water (10 7 M) is negligible, and x is small compared to 0.250 (percent error = 4.0%). If the percent error is less than 5% for an assumption, we will consider it a valid assumption (called the 5% rule). Finishing the problem: x = 0.010 M = [H + ]; ph = log(0.010) = 2.00 CH 3 CO 2 H (K a = 1.8 x 10 5 ) and H 2 O (K a = K w = 1.0 x 10 14 ) are the major species. CH 3 CO 2 H is the major source of H +. Solving the weak acid problem: CH 3 CO 2 H H + + CH 3 CO 2 Initial 0.250 M ~0 0 x mol/l CH 3 CO 2 H dissociates to reach equilibrium Equil. 0.250 x x x K a = [H + ][CH 3 CO 2 ] / [CH 3 CO 2 H] = 1.8 x 10 5 = x 2 / 0.250 x x 2 / 0.250 (assuming x<<0.250) x = 2.1 x 10 3 M; Checking assumption: (2.1 x 10 3 / 0.250) x 100 = 0.84%. Assumptions goo [H + ] = x = 2.1 x 10 3 M; ph = log(2.1 x 10 3 ) = 2.68 14.55 [CH 3 COOH] 0 = [HC 2 H 3 O 2 ] 0 = 0.0560 g HC 2 H 3 O 2 x (1 mol HC 2 H 3 O 2 /60.05 g) / 0.05000 L = 1.87 x 10 2 M HC 2 H 3 O 2 H + + C 2 H 3 O 2 K a = 1.8 x 10 5 Initial 0.0187 M ~0 0 x mol/l HC 2 H 3 O 2 dissociates to reach equilibrium Equil. 0.0187 x x x K a = 1.8 x 10 5 = [H + ] [C 2 H 3 O 2 ] / [HC 2 H 3 O 2 ] = x 2 / 0.0187 x x 2 / 0.0187 x = [H + ] = 5.8 x 10 4 M; ph = 3.24 Assumptions good (x is 3.1% of 0.0187) [H + ] = [C 2 H 3 O 2 ] = [CH 3 COO ] = 5.8 x 10 4 M; [CH 3 COOH] = 0.0187 5.8 x 10 4 = 0.0181 M
14.57 This is a weak acid in water. Solving the weak acid problem: HF H + + F K a = 7.2 x 10 4 Initial 0.020 M 0 0 x mol/l HF dissociates to reach equilibrium Equil. 0.020 x x x K a = 7.2 x 10 4 = [H + ][F ] / [HF] = x 2 / 0.020 x x 2 / 0.020 (assuming x<<0.020) x = [H + ] = 3.8 x 10 3 M; Check assumptions: (x / 0.020) x 100 = 3.8 x 10 3 / 0.020 = 19% The assumption x<<0.020 is not good (x is more than 5% of 0.020). We must solve x 2 / (0.020 x) = 7.2 x 10 4 exactly by using either the quadratic formula or by the method of successive approximations (see Appendix 1.4 of text). Using successive approximations, we let 0.016 M be a new approximation for [HF]. That is, in the denominator, try x = 0.0038 (the value of x we calculated making the normal assumption), so 0.020 0.0038 = 0.016, then solve for a new value of x in the numerator. x 2 = 0.020 x x 2 / 0.016 = 7.2 x 10 4, x = 3.4 x 10 3 We use this new value of x to further refine our estimate of [HF], i.e., 0.020 x = 0.020 0.0034 = 0.0166 (carry extra significant figure). x 2 = 0.020 x x 2 / 0.0166 = 7.2 x 10 4, x = 3.5 x 10 3 We repeat until we get an answer that repeats itself. This would be the same answer we would get solving exactly using the quadratic equation. In this case it is: x = 3.5 x 10 3 So: [H + ] = [F ] = x = 3.5 x 10 3 M; [OH ] = K w / [H + ] = 2.9 x 10 12 M [HF] = 0.020 x = 0.020 0.0035 = 0.017 M; ph = 2.46 Note: When the 5% assumption fails, use whichever method you are most comfortable with to solve exactly. The method of successive approximations is probably fastest when the percent error is less than ~25% (unless you have a calculator that can solve quadratic equations). 14.59 Major species: HC 2 H 2 ClO 2 (K a = 1.35 x 10 3 ) and H 2 O; Major source of H + : HC 2 H 2 ClO 2 HC 2 H 2 ClO 2 H + + C 2 H 2 ClO 2 Initial 0.10 M 0 0 x mol/l HC 2 H 2 ClO 2 dissociates to reach equilibrium Equil. 0.010 x x x K a = 1.35 x 10 3 = x 2 / 0.10 x x 2 / 0.10, x = 1.2 x 10 2 M Checking the assumptions finds that x is 12% of 0.10 which fails the 5% rule. We must solve 1.35 x 10 3 = x 2 / (0.10 x) exactly using either the method of successive approximations or the
quadratic equation. Using either method gives x = [H + ] = 1.1 x 10 2 M. ph = log[h + ] = log(1.1 x 10 2 ) = 1.96 14.61 HCl is a strong aci It will produce 0.10 M H +. HOCl is a weak aci Let s consider the equilibrium: HOCl H + + OCl K a = 3.5 x 10 8 Initial 0.10 M 0.10 M 0 x mol/l HOCl dissociates to reach equilibrium Equil. 0.10 x x x K a = 3.5 x 10 8 = [H + ][OCl ] / [HOCl] = (0.10 + x)(x) / (0.10 x) x, x = 3.5 x 10 8 M Assumptions are great (x is 3.5 x 10 5 % of 0.10). We are really assuming that HCl is the only important source of H +, which it is. The [H + ] contribution from HOCl, x, is negligible. Therefore, [H + ] = 0.10 M; ph = 1.00 HNO 3 is a strong acid, giving an initial concentration of H + equal to 0.050 M. Consider the equilibrium: HC 2 H 3 O 2 H + + C 2 H 3 O 2 K a = 1.8 x 10 5 Initial 0.50 M 0 0 x mol/l HC 2 H 3 O 2 dissociates to reach equilibrium Equil. 0.50 x x x K a = 1.8 x 10 5 = [H + ][C 2 H 3 O 2 ] / [HC 2 H 3 O 2 ] = (0.050 + x)(x) / 0.50 x = 0.050 x / 0.50 x = 1.8 x 10 4 ; Assumptions are good (well within the 5% rule). [H + ] = 0.050 + x = 0.050 M and ph = 1.30 14.63 In all parts of this problem, acetic acid (HC 2 H 3 O 2 ) is the best weak acid present. We must solve a weak acid problem. HC 2 H 3 O 2 H + + C 2 H 3 O 2 Initial 0.50 M 0 0 x mol/l HC 2 H 3 O 2 dissociates to reach equilibrium Equil. 0.50 x x x K a = 1.8 x 10 5 = [H + ][C 2 H 3 O 2 ] / [HC 2 H 3 O 2 ] = x 2 / 0.50 x x 2 / 0.50 x = [H + ] = [C 2 H 3 O 2 ] = 3.0 x 10 3 M Assumptions goo
Percent dissociation = [H + ] / [HC 2 H 3 O 2 ] 0 x 100% = (3.0 x 10 3 / 0.50) x 100 = 0.60% The setups for solutions b and c are similar to solution a except the final equation is slightly different, reflecting the new concentration of HC 2 H 3 O 2. K a = 1.8 x 10 5 = [H + ][C 2 H 3 O 2 ] / [HC 2 H 3 O 2 ] = x 2 / 0.050 x x 2 / 0.050 x = [H + ] = [C 2 H 3 O 2 ] = 9.5 x 10 4 M Assumptions goo % dissociation = (9.5 x 10 4 / 0.050) x 100 = 1.9% K a = 1.8 x 10 5 = [H + ][C 2 H 3 O 2 ] / [HC 2 H 3 O 2 ] = x 2 / 0.0050 x x 2 / 0.0050 x = [H + ] = [C 2 H 3 O 2 ] = 3.0 x 10 4 M; Check assumptions. Assumption that x is negligible is borderline (6.0% error). We should solve exactly. Using the method of successive approximations (see Appendix 1.4 of text0: 1.8 x 10 5 = x 2 / 0.0050 3.0 x 10 4 = x 2 = 0.0047, x = 2.9 x 10 4 Next trial also gives x = 2.9 x 10 4. % dissociation = (2.9 x 10 4 /5.0 x 10 3 ) x 100 = 5.8% As we dilute a solution, all concentrations decrease. Dilution will shift the equilibrium to the side with the greater number of particles. For example, suppose we double the volume of an equilibrium mixture of a weak acid by adding water, then: Q = [([H + ] eq / 2)([X ] eq /2)] / ([HX] eq / 2) = 1/2 K a Q < K a, so the equilibrium shifts to the right or towards a greater percent dissociation. e. [H + ] depends on the initial concentration of a weak acid and on how much weak acid dissociates. For solutions ac the initial concentration of acid decreases more rapidly than the percent dissociations increases. Thus, [H + ] decreases. 14.65 Let HX symbolize the weak aci Setup the problem like a typical weak acid equilibrium problem. HX H + + X Initial 0.15 M 0 0 x mol/l HX dissociates to reach equilibrium Equil. 0.15 x x x If the acid is 3.0% dissociated, then x = [H + ] is 3.0% of 0.15: x = 0.030 x (0.15 M) = 4.5 x 10 3 M. Now that we know the value of x, we can solve for K a :
K a = [H + ][X ] / [HX] = x 2 / 0.15 x = (4.5 x 10 3 ) 2 / 0.15 4.5 x 10 3 = 1.4 x 10 4 M 14.67 Setup the problem using the K a equilibrium reaction for HOCN. HOCN H + + OCN Initial 0.0100 M 0 0 x mol/l HOCN dissociates to reach equilibrium Equil. 0.0100 x x x K a = [H + ][OCN ] / [HOCN] = x 2 / 0.0100 x; ph = 2.77: x = [H + ] = 10 ph = 10 2.77 = 1.7 x 10 3 M K a = (1.7 x 10 3 ) 2 / (0.0100 1.7 x 10 3 ) = 3.5 x 10 4 14.69 Major species: HCOOH and H 2 O; Major source of H + ; HCOOH HCOOH H + + HCOO Initial C 0 0 where C = [HCOOH] 0 x mol/l HCOOH dissociates to reach equilibrium Equil. C x x x K a = 1.8 x 10 4 = [H + ][HCOO ] / [HCOOH] = x 2 / C x where x = [H + ] 1.8 x 10 4 = [H + ] 2 / C [H + ] ; ph = 2.70, so: [H + ] = 10 2.70 = 2.0 x 10 3 M 1.8 x 10 4 = (2.0 x 10 3 ) 2 / C (2.0 x 10 3 ), C (2.0 x 10 3 ) = 4.0 x 10 6 / 1.8 x 10 4, C = 2.4 x 10 2 M A 0.024 M formic acid solution will have ph = 2.70