Chemical and Ionic Equilibrium - Result Question 116
####60. 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 $NH _3$.
(1996, 3M)
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Solution:
- $K _a\left(NH _4^{+}\right)=5.6 \times 10^{-10}$
$$ \begin{aligned} & \qquad K _b\left(NH _3\right)=K _w / K _a=\frac{10^{-14}}{5.6 \times 10^{-10}}=1.8 \times 10^{-5} \ & \text { i.e. } NH _3+H _2 O \stackrel{k _1}{\underset{k _2}{\rightleftharpoons}} NH _4^{+}+OH^{-} \ & K=\frac{k _1}{k _2}=1.8 \times 10^{-5} \ & k _1=K k _2=1.8 \times 10^{-5} \times 3.4 \times 10^{10}=6.12 \times 10^{5} \end{aligned} $$
