Question

Equilibrium constant for the reaction, $N{H}_{4}OH+H^{+}\rightleftharpoons N{H}_4^{+}+H_{2}O$ is $1.8\times {10}^9$. Hence, equilibrium constant for $N{H}_3\left(aq\right)+H_{2}O\rightleftharpoons N{H}_4^{+}+O{H}^{-}$ is:

• A. $1.8\times {10}^{-5}$
• B. $1.8\times {10}^5$
• C. $1.8\times {10}^{-9}$
• D. $5.59\times {10}^{-10}$

Answer 1

The correct option is A $1.8\times {10}^{-5}$

$N{H}_{4}OH+H^{+}\rightleftharpoons N{H}_4^{+}+H_{2}O$

$\therefore \frac{\left[N{H}_4^{+}\right]\left[H_{2}O\right]}{\left[N{H}_{4}OH\right]\left[H^{+}\right]}=1.8\times {10}^9$

Again,
$N{H}_3+H_{2}O\to N{H}_{4}OH\rightleftharpoons N{H}_4^{+}+O{H}^{-}$

$\therefore K^{\prime }=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{\left[N{H}_{4}OH\right]}$

Now $\frac{K^{\prime }}{K}=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{\left[N{H}_{4}OH\right]}\times \frac{\left[N{H}_{4}OH\right]\left[H^{+}\right]}{\left[N{H}_4^{+}\right]\left[H_{2}O\right]}$

$=\left[O{H}^{-}\right]\left[H^{+}\right](\because H_{2}O$ is in excess)

$=K_w=1\times {10}^{-14}$

$\therefore K^{\prime }=K\times 1\times {10}^{-14}$

$=1.8\times {10}^9\times {10}^{-14}=1.8\times {10}^{-5}$