Question 15.13: (a) Kb for trimethylamine is 6.5 × 10^-5. Calculate Ka for t......
(a) K_{b} for trimethylamine is 6.5\times 10^{-5}. Calculate K_{a} for the trimethylammonium ion, (CH_{3})_{3}NH^{+}.
(b) K_{a} for HCN is 4.9\times 10^{-10}. Calculate K_{b} for CN^{-}.
(c) Pyridine (C_{5}H_{5}N), an organic solvent, has pK_{b} = 8.74. What is the value of pK_{a} for the pyridinium ion, C_{5}H_{5}NH^{+}?
STRATEGY
To calculate K_{a} from K_{b} (or vice versa), use the equation K_{a} = K_{w}/K_{b} or K_{b} = K_{w}/K_{a}. To calculate pK_{a} from pK_{b}, use the equation pK_{a} = 14.00 – pK_{b}.
(a) K_{a} for (CH_{3})_{3}NH^{+} is the equilibrium constant for the acid-dissociation reaction
(CH_{3})_{3}NH^{+}(aq)+H_{2}O(l)\xrightleftharpoons{}H_{3}O^{+}(aq)+(CH_{3})_{3}N(aq)Because K_{a} = K_{w}/K_{b}, we can find K_{a} for (CH_{3})_{3}NH^{+} from K_{b} for its conjugate base (CH_{3})_{3}N:
K_{a}=\frac{K_{w}}{K_{b}}=\frac{1.0\times 10^{-14}}{6.5\times 10^{-5}}=1.5\times 10^{-10}(b) K_{b} for CN^{-} is the equilibrium constant for the base-protonation reaction
CN^{-}(aq)+H_{2}O(l)\xrightleftharpoons{}HCN(aq)+OH^{-}(aq)Because K_{b} = K_{w}/K_{a}, we can find K_{b} for CN^{-} from K_{a} for its conjugate acid HCN:
K_{b}=\frac{K_{w}}{K_{a}}=\frac{1.0\times 10^{-14}}{4.9\times 10^{-10}}=2.0\times 10^{-5}(c) We can find pK_{a} for C_{5}H_{5}NH^{+} from pK_{b} for C_{5}H_{5}N:
pK_{a}=14.00 – pK_{b} = 14.00 – 8.74 = 5.26