Question 22.8: GRADED Calculate equilibrium constants for the two acid-base...
GRADED
Calculate equilibrium constants for the two acid-base reactions above.
ⓐ CH_{3}NH_{2}(aq) + H^{+}(aq) → CH_{3}NH_{3}^{+}(aq)
ⓑ CH_{3}NH_{2}(aq) + CH_{3}COOH(aq) \rightleftharpoons CH_{3}NH_{3}^{+}(aq) + CH_{3}COO^{-}(aq)
ⓐ
| ANALYSIS | |
| reaction (CH_{3}NH_{2}(aq) + H^{+}(aq) → CH_{3}NH_{3}^{+}(aq)) | Information given: |
| K_{a} for CH_{3}NH_{3}^{+} | Information implied |
| K | Asked for |
STRATEGY
1. Note that the given equation is the reverse of the equation for the dissociation of a weak acid.
2. Find K_{a} for CH_{3}NH_{3}^{+} in Appendix 1 and use the reciprocal rule (Chapter 12).
ⓑ
| ANALYSIS | |
| reaction (CH_{3}NH_{2}(aq) + CH_{3}COOH(aq) → CH_{3}NH_{3}^{+}(aq) + CH_{3}COO^{-}(aq)) From (a): K (4.2 × 10^{10}) for the reaction CH_{3}NH_{2}(aq) + H^{+}(aq) \rightleftharpoons CH_{3}NH_{3}^{+}(aq) |
Information given: |
| K_{a} for CH_{3}COOH | Information implied |
| K | Asked for |
STRATEGY
1. Break up the given equation into two reactions
CH_{3}NH_{2} \rightleftharpoons CH_{3}NH_{3}^{+} (1)
CH_{3}COOH \rightleftharpoons CH_{3}COO^{-} (2)
2. Find K for each reaction
3. Combine the two equations and their Ks. Apply the Rule of Multiple Equilibria (Chapter 12).
ⓐ
| CH_{3}NH_{3}^{+}(aq) \rightleftharpoons CH_{3}NH_{2}(aq) + H^{+}(aq) K_{a} 5= 2.4 × 10^{-11} | 1. K_{a} expression |
| CH_{3}NH_{2}(aq) + H^{+}(aq) \rightleftharpoons CH_{3}NH_{3}^{+}(aq) K = 1/K_{a} = 1/2.4 × 10^{-11} = 4.2 × 10^{10} | 2. K |
ⓑ
| From (a): K_{1} = 4.2 × 10^{10} | K for equation (1) |
| From Appendix 1: K_{2} = K_{a} = 1.8 × 10^{-5} | K for equation (2) |
| CH_{3}NH_{2}(aq) + H^{+}(aq) \rightleftharpoons CH_{3}NH_{3}^{+}(aq) K_{1} = 4.2 × 10^{10}
\frac{CH_{3}COOH(aq) \rightleftharpoons CH_{3}COO^{-}(aq) + H^{+}(aq) K_{2} = 1.8 × 10^{-5}}{CH_{3}NH_{2}(aq) + CH_{3}COOH(aq) \rightleftharpoons CH_{3}NH_{3}^{+}(aq) + CH_{3}COO^{-}(aq)} |
Combine both equations |
| K = K_{1} × K_{2} = (4.2 × 10^{10})(1.8 × 10^{-5}) = 7.5 × 10^{5} | K |