ISM_chapter16_part2

ISM_chapter16_part2 - Chapter 16 Acids and Bases 742 Assume...

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Unformatted text preview: Chapter 16: Acids and Bases 742 Assume x is very small and: 0.20 – x ≅ 0.20. 1.8 × 10 –5 = x 2 (0.20) x 2 = (1.8 × 10 –5 )(0.20) x = 1.9 × 10 –3 M = [H 3 O + ] = [CH 3 COO – ] [CH 3 COOH] = 0.20 M – x = 0.20 M – 1.9 × 10 –3 M = 0.20 M (as assumed) 58 . Answer: [H 3 O + ] = 1.3 × 10 –5 ; [A – ] = 1.3 × 10 –5 ; [HA] = 0.040 Strategy and Explanation: Combine the methods described in Questions 33 and 54 with those described in Section 16.6 and in the solution to Question 14.45. The equation for the equilibrium and the equilibrium expression are: HA(aq) + H 2 O( l ) H 3 O + (aq) + A – (aq) K a = [H 3 O + ][A " ] [HA] As the reactants decompose, the concentrations of the products increase stoichiometrically, until they reach equilibrium concentrations. HA(aq) H 3 O + (aq) A – (aq) initial conc. (M) 0.040 1.0 × 10 –7* 0 change as reaction occurs (M) – x + x + x equilibrium conc. (M) 0.040 – x x x * from dissociation of pure water. Assume this concentration will be small compared to the acid added. At equilibrium K a = (x)(x) (0.040 " x) = 4.0 × 10 –9 Assume x is very small and: 0.040 – x ≅ 0.040. 4.0 × 10 –9 = x 2 (0.040) x 2 = (4.0 × 10 –9 )(0.040) x = 1.3 × 10 –5 M = [H 3 O + ] = [A – ] [HA] = 0.040 M – x = 0.040 M – 1.3 × 10 –5 M = 0.040 M (as assumed) 59. Answer: (a) 2.74 (b) 3.6% Strategy and Explanation: Use the method described in the answers to Questions 33 and 58. The equation for the equilibrium and the equilibrium expression are: C 6 H 5 COOH(aq) + H 2 O( l ) H 3 O + (aq) + C 6 H 5 COO – (aq) K a = [H 3 O + ][C 6 H 5 COO " ] [C 6 H 5 COOH] As the reactants decompose, the concentrations of the products increase stoichiometrically, until they reach equilibrium concentrations. C 6 H 5 COOH(aq) H 3 O + (aq) C 6 H 5 COO – (aq) initial conc. (M) 0.050 1.0 × 10 –7* 0 change as reaction occurs (M) – x + x + x equilibrium conc. (M) 0.050 – x x x * from dissociation of pure water. Assume this concentration will be small compared to the acid added. At equilibrium K a = (x)(x) (0.050 " x) = 6.3 × 10 –5 Assume x is very small and: 0.050 – x ≅ 0.050. Chapter 16: Acids and Bases 743 6.3 × 10 –5 = x 2 (0.050) x 2 = (6.3 × 10 –5 )(0.050) x = 1.8 × 10 –3 M = [H 3 O + ] = [CH 3 COO – ] (a) pH = – log[H 3 O + ] = –log(1.8 × 10 –3 ) = 2.74 (b) Percent ionization is defined as the equilibrium concentration of H 3 O + from the ionization compared to the initial concentration of the acid (0.050 M): % acid ionized = x 0.050 " 100% = 1.8 " 10 # 3 0.050 " 100% = 3.6% 60 . Answer: 1.4 × 10 –5 Strategy and Explanation: Use the method described in the solution to Question 56. The equation for the equilibrium and the equilibrium expression are: CH 3 CH 2 COOH(aq) + H 2 O( l ) H 3 O + (aq) + CH 3 CH 2 COO – (aq) K a = [H 3 O + ][CH 3 CH 2 COO " ] [CH 3 CH 2 COOH] At equilibrium, pH = 2.93, so [H 3 O + ] = 10 –pH = 10 –2.93 = 1.2 × 10...
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ISM_chapter16_part2 - Chapter 16 Acids and Bases 742 Assume...

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