Question

The ionisation constant of $N{H}_4^{+}$ in water is $5.6\times 10^{-10}$ at $25^{\circ }C$. The rate constant for the reaction of $N{H}_4$ and $O{H}^{-}$ to form $N{H}_3$ 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 $N{H}_3$

Answer 1

$K_a=(N{H}_4^{-})=5.6\times 10^{-10}$
$K_b=(N{H}_3)=K_w/K_a=\frac{10^{-14}}{5.6\times 10^{-10}}=1.8\times 10^{-5}$
i,e., $N{H}_3+H_{2}O$ $\underset{k_2}{\overset{K_1}{\rightleftharpoons }}$ $N{H}_4^{+}+O{H}^{-}$
$K=\frac{k_1}{k_2}=1.8\times 10^{-5}$
$k_1=K{k}_2=.8\times 10^{-5}\times 3.4\times 10^{10}$
$=6.12\times 10^5$