SATHEE: Chemical and Ionic Equilibrium

Chemical and Ionic Equilibrium - Result Question 116

####60. 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 $3.4 \times 10^{10} L / m o l / s$ . Calculate the rate constant per proton transfer from water to $N H_{3}$ .

(1996, 3M)

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Solution:

$K_{a} (N H_{4}^{+}) = 5.6 \times 10^{- 10}$

$\begin{aligned} K_{b} (N H_{3}) = K_{w} / K_{a} = \frac{10^{- 14}}{5.6 \times 10^{- 10}} = 1.8 \times 10^{- 5} & i.e. N H_{3} + H_{2} O \overset{k_{1}}{\underset{k_{2}}{⇌}} N H_{4}^{+} + O H^{-} & K = \frac{k_{1}}{k_{2}} = 1.8 \times 10^{- 5} & k_{1} = K k_{2} = 1.8 \times 10^{- 5} \times 3.4 \times 10^{10} = 6.12 \times 10^{5} \end{aligned}$

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Chemical and Ionic Equilibrium - Result Question 116

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