Question

The ionisation constant of $NH^+_4$ in water is $ 5.6 \times 10^{- 10} $ at $ 25^{\circ} C$. The rate constant for the reaction of $NH_4 $ and $ OH^- $ to form $ NH_3$ and $ H_2 O$ at $ 25^{\circ} C$ is $3.4 \times 10^{ 10} L/mol/s$. Calculate the rate constant per proton transfer from water to $NH_3 $

Answer 1

$ K_a = ( NH_4^- ) = 5.6 \times 10^{ - 10 } $
$ K_b = ( NH_3) = K_w / K_a = \frac{ 10^{ - 14 }}{ 5.6 \times 10^{ - 10 }} = 1.8 \times 10^{ - 5 } $
i,e., $ NH_3 + H_2O $ $ \xrightleftharpoons[ k_2] {K_1} $ $ NH_4^+ \, + OH^- $
$ K = \frac{ k_1 }{ k_2 } = 1.8 \times 10^{ - 5 } $
$ k_1 = Kk_2 = .8 \times 10^{ - 5 } \times 3.4 \times 10^{ 10} $
$ = 6.12 \times 10^ 5 $