Acid-Base Equilibria: pH Changes with Addition of Conjugate Acids and Bases, Assignments of Chemistry

Download Acid-Base Equilibria: pH Changes with Addition of Conjugate Acids and Bases and more Assignments Chemistry in PDF only on Docsity! 1 Chapter 17 17.4 (a) pH increases (adding conjugate base of weak acid will shift the equilibrium back toward reactants) (b) pH decreases (adding conjugate acid of weak base will shift the equilibrium back toward reactants) (c) pH increases (adding weak base to a strong acid will remove some of the strong acid from solution) (d) pH increases (adding conjugate base of weak acid will shift equilbrium to reactants) (e) remain the same (NaClO4 will not change the pH) 17.6 (a) HCHO2 + H2O ↔ H3O + + CHO2 – Initial 0.100 --- 0 0.090 Change –x --- +x +x Equilibrium 0.100 –x --- x 0.090+x Ka = [H3O + ][CHO2 – ] [HCHO2 ] = x(0.090+x) (0.100– x) = 1.8 × 10–4 ≈ x(0.090) (0.100) x = 2.0 × 10–4 = [H3O +] pH = –log(2.0 × 10–4) = 3.70 (b) C5H5N + H2O ↔ OH – + C5H5NH + Initial 0.0750 --- 0 0.0850 Change –x --- +x +x Equilibrium 0.0750 –x --- x 0.0850+x Kb = [OH– ][C 5H5NH + ] [C 5H5N] = x(0.0850+x) (0.0750 – x) = 1.7 × 10–9 ≈ x(0.0850) (0.0750) x = 1.6 × 10–9 = [OH–] pH = 14–log(1.6 × 10–9) = 5.18 17.10 (a) pH of a buffer is determined by the Ka of the weak acid and the ratio of weak base to weak acid. (b) The buffer capacity of a buffer is determined by that acid/base ratio. 2 17.12 (a) HCO3 – + H2O ↔ H3O + + CO3 2– Initial 0.100 --- 0 0.125 Change –x --- +x +x Equilibrium 0.100 –x --- x 0.125+x Ka = [H3O + ][HCO3 – ] [CO3 2– ] = x(0.125+x) (0.100– x) = 5.6 × 10–11 ≈ x(0.125) (0.100) x = 4.5 × 10–11 = [H3O +] pH = –log(4.5 × 10–11) = 10.35 (b) [HCO3 –] = (0.055 L)(0.20 M) / 0.120 L = 0.092 M [CO3 2–] = (0.065 L)(0.15 M) / 0.120 L = 0.081 M HCO3 – + H2O ↔ H3O + + CO3 2– Initial 0.092 --- 0 0.081 Change –x --- +x +x Equilibrium 0.092 –x --- x 0.081+x Ka = [H3O + ][HCO3 – ] [CO3 2– ] = x(0.081+x) (0.092– x) = 5.6 × 10–11 ≈ x(0.081) (0.092) x = 6.4 × 10–11 = [H3O +] pH = –log(6.4 × 10–11) = 10.20 17.14 mol NH3 = (5.0 g)(1 mol / 17.0304 g) = 0.2936 mol [NH3] = 0.2936 mol / 2.50 L = 0.1174 M mol NH4Cl = (20.0 g)(1 mol / 53.4913 g) = 0.3739 mol [NH4Cl] = 0.3739 mol / 2.50 L = 0.1496 M