Chemical And Ionic Equilibrium

Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise

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 $\mathrm{NH}_3$.

(1996,3 M)

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Solution:

$K_a\left(\mathrm{NH}_4^{+}\right)=5.6 \times 10^{-10}$

$K_b\left(\mathrm{NH}_3\right)=K_w / K_a=\frac{10^{-14}}{5.6 \times 10^{-10}}=1.8 \times 10^{-5}$

$\begin{aligned} & \text { i.e. } \mathrm{NH}_3+\mathrm{H}_2 \mathrm{O} \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} \mathrm{NH}_4^{+}+\mathrm{OH}^{-} \\ & K=\frac{k_1}{k_2}=1.8 \times 10^{-5} \\ & k_1=K k_2=1.8 \times 10^{-5} \times 3.4 \times 10^{10}=6.12 \times 10^5 \\ \end{aligned}$

Chemical And Ionic Equilibrium

Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise

Solution:

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