Unit 7, Lesson 05: Answers to Working with Ka, Kb and Kw

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1 Unit 7, Lesson 05: Answers to Working with Ka, Kb and Kw 4. For each of the following acids: write the ionization reaction in water and the Ka expression write the ionization reaction for its conjugate base and Kb expression calculate the value of Kb for the conjugate base a) benzoic acid C 6 H 5 COOH (aq) + H 2 O (l) C 6 H 5 COO 1 Ka = [C 6 H 5 COO 1 ] [H 3 O 1+ ] [C 6 H 5 COOH] = 6.3 x 10-5 (from page 597) C 6 H 5 COO (aq) + H 2 O (l) C 6 H 5 COOH (aq) + OH (aq) Kb = [C 6 H 5 COOH] [OH 1 ] [C 6 H 5 COO ] = 1.0 x / 6.3 x 10-5 = 1.6 x b) hydrofluoric acid HF (aq) + H 2 O (l) F 1 Ka = [F 1 ] [H 3 O 1+ ] [HF] = 6.3 x 10-4 (from page 597) F (aq) + H 2 O (l) HF (aq) + OH (aq) Kb = [HF] [OH 1 ] [F ] = 1.0 x / 6.3 x 10-4 = 1.6 x 10-11

2 c) iodic acid HIO 3 HIO 3 (aq) + H 2 O (l) IO 3 1 Ka = [IO 3 1 ] [H 3 O 1+ ] [HIO 3 ] = 1.7 x 10-1 (from page 597) IO 3 (aq) + H 2 O (l) HIO 3 (aq) + OH (aq) Kb = [HIO 3 ] [OH 1 ] [IO 3 ] = 1.0 x / 1.7 x 10-1 = 5.9 x d) H 2 BO 3 (aq) H 2 BO 3 (aq) + H 2 O (l) HBO 3 Ka = [HBO 3 ] [H 3 O 1+ ] [H 2 BO 3 1 ] < 1.0 x (from table E.10 on page 597) HBO 3 (aq) + H 2 O (l) H 2 BO 3 (aq) + OH (aq) Kb = [H 2 BO 3 ] [OH 1 ] [HBO 3 ] = 1.0 x / < 1.0 x > 1 (so HBO 3 2- is a very strong base)

3 e) HCO 3 (aq) HCO 3 (aq) + H 2 O (l) CO 3 Ka = [CO 3 ] [H 3 O 1+ ] [HCO 3 1 ] = 4.7 x (from table E.10 on page 597) CO 3 (aq) + H 2 O (l) HCO 3 (aq) + OH (aq) Kb = [HCO 3 ] [OH 1 ] [CO 3 ] = 1.0 x / 4.7 x = 2.1 x For each of the following bases: write the ionization reaction in water and the Kb expression write the ionization reaction for its conjugate acid and Ka expression calculate the value of Ka for the conjugate acid a) hydroxlamine NH 2 OH (aq) + H 2 O (l) NH 3 OH + (aq) + OH (aq) Kb = [NH 3 OH + ] [OH 1 ] [NH 2 OH] = 8.8 x 10-9 (from table E.11 on page 597) NH 3 OH + (aq) + H 2 O (l) NH 2 OH Ka = [NH 2 OH] [H 3 O 1+ ] [NH 3 OH + ] = Kw / Kb = 1.0 x / 8.8 x 10-9 = 1.1 x 10-6

4 b) ethanamine C 2 H 5 NH 2 (aq) + H 2 O (l) C 2 H 5 NH 3 + (aq) + OH (aq) Kb = [C 2 H 5 NH 3 + ] [OH 1 ] [C 2 H 5 NH 2 ] = 4.5 x 10-4 (from table E.11 on page 597) C 2 H 5 NH 3 + (aq) + H 2 O (l) C 2 H 5 NH 2 Ka = [C 2 H 5 NH 2 ] [H 3 O 1+ ] [C 2 H 5 NH 3 + ] = Kw / Kb = 1.0 x / 4.5 x 10-4 = 2.2 x c) urea NH 2 CONH 2 (aq) + H 2 O (l) NH 2 CONH 3 + (aq) + OH (aq) Kb = [NH 2 CONH 3 + ] [OH 1 ] [NH 2 CONH 2 ] = 1.3 x (from table E.11 on page 597) NH 2 CONH 3 + (aq) + H 2 O (l) NH 2 CONH 2 Ka = [NH 2 CONH 2 ] [H 3 O 1+ ] [NH 2 CONH 3 + ] = Kw / Kb = 1.0 x / 1.3 x = 0.77 (quite a strong acid)

5 d) H 2 BO 3 (aq) H 2 BO 3 (aq) + H 2 O (l) H 3 BO 3 (aq) + OH (aq) Kb = [H 3 BO 3 ] [OH 1 ] [H 2 BO 3 ] We only know the Ka for H 3 BO 3 (aq) = 5.4 x Kb using this value: (from table E.10 on page 597), so we can calculate Kb = 1.0 x / 5.4 x = 1.85 x 10-5 or 1.9 x 10-5 (2 sd) H 3 BO 3 (aq) + H 2 O (l) H 2 BO 3 Ka = [H 2 BO 3 ] [H 3 O 1+ ] [H 3 BO 3 ] = 5.4 x (from table E.10 on page 597) e) HCO 3 (aq) HCO 3 (aq) + H 2 O (l) H 2 CO 3 ( (aq) + OH (aq) Kb = [H 2 CO 3 ] [OH 1 ] [HCO 3 ] We only know the Ka for H 2 CO 3 (aq) = 4.5 x 10-7 Kb using this value: (from table E.10 on page 597), so we can calculate Kb = 1.0 x / 4.5 x 10-7 = 2.2 x 10-8 H 2 CO 3 (aq) + H 2 O (l) HCO 3 Ka = [HCO 3 ] [H 3 O 1+ ] [H 2 CO 3 ] = 4.5 x 10-7 (from table E.10 on page 597)

6 6. Given the relationship K a K b = K w, complete the following statements: a) If the K a for an acid is large, the K b for its conjugate base is: small b) If the K a for an acid is small, the K b for its conjugate base is: large c) If the K b for a base is large, the K a for its conjugate acid is: small d) If the K b for a base is small, the K a for its conjugate acid is: large