Question

The ionization constant of $N{H}_4^{+}$ in water is $2.8\times {10}^{-10}$ 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 $1.4Lmo{l}^{-1}{s}^{-1}$ ,the rate constant per proton transfer from water to $N{H}_3$ will be

• A. $6\times {10}^5$
• B. $8\times {10}^5$
• C. $2\times {10}^5$
• D. $5\times {10}^5$

Answer 1

The correct option is D $5\times {10}^5$

$N{H}_3+H_{2}O\underset{K_b}{\stackrel{k_f}{\rightleftharpoons }}+O{H}^{-};K_b=1.4\times {10}^{10}$
$N{H}_4^{+}+H_{2}O\rightleftharpoons N{H}_{4}OH+H^{+};K_a=2.8\times {10}^{-10}$
$K_{Base}=\frac{K_f}{K_b}=\frac{K_W}{K_{acid}}$
$K_{acid}\times K_{base}=K_W={10}^{-14}$
$\frac{K_f}{1.4\times {10}^{10}}=\frac{{10}^{-14}}{2.8\times {10}^{-10}}$
$K_f=5\times {10}^5$