Chemical and Ionic Equilibrium 2 Question 60
60. The ionisation constant of $\mathrm{NH}{4}^{+}$in water is $5.6 \times 10^{-10}$ at $25^{\circ} \mathrm{C}$. The rate constant for the reaction of $\mathrm{NH}{4}^{+}$and $\mathrm{OH}^{-}$to form $\mathrm{NH}{3}$ and $\mathrm{H}{2} \mathrm{O}$ at $25^{\circ} \mathrm{C}$ is $3.4 \times 10^{10} \mathrm{~L} / \mathrm{mol} / \mathrm{s}$. Calculate the rate constant per proton transfer from water to $\mathrm{NH}_{3}$.
(1996, 3M)
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Solution:
- $K_{a}\left(\mathrm{NH}_{4}^{+}\right)=5.6 \times 10^{-10}$
$$ \begin{aligned} & \qquad K_{b}\left(\mathrm{NH}{3}\right)=K{w} / K_{a}=\frac{10^{-14}}{5.6 \times 10^{-10}}=1.8 \times 10^{-5} \ & \text { i.e. } \mathrm{NH}{3}+\mathrm{H}{2} \mathrm{O} \stackrel{k_{1}}{\underset{k_{2}}{\rightleftharpoons}} \mathrm{NH}{4}^{+}+\mathrm{OH}^{-} \ & K=\frac{k{1}}{k_{2}}=1.8 \times 10^{-5} \ & k_{1}=K k_{2}=1.8 \times 10^{-5} \times 3.4 \times 10^{10}=6.12 \times 10^{5} \end{aligned} $$
