Question

Hydrolysis constant of $N{H}_4^{+}$ is $5.55\times {10}^{-10}.$ Calculate the ionisation constant of $N{H}_4^{+}$.

• A. $5.55\times {10}^{-10}$
• B. $5.55\times {10}^{-5}$
• C. $1.88\times {10}^{-10}$
• D. $1.88\times {10}^{-4}$

Answer 1

The correct option is A $5.55\times {10}^{-10}$

$N{H}_4^{+}+H_{2}O\rightleftharpoons N{H}_{4}OH+H^{+}$
$K_h=\frac{\left[N{H}_{4}OH\right]\left[H^{+}\right]}{\left[N{H}_4^{+}\right]}$
In the above reaction , at equilibrium ,
$\left[N{H}_{4}OH\right]=\left[H^{+}\right].....\left(1\right)$

Now,

$N{H}_4^{+}\rightleftharpoons N{H}_3+H^{+}$
If $K_i$ is ionisation constant, then
$K_i=\frac{\left[N{H}_3\right]\left[H^{+}\right]}{\left[N{H}_4^{+}\right]}$
Also from the above reaction , at equilibrium,
$\left[N{H}_3\right]=\left[H^{+}\right].....\left(2\right)$

$\text{From 1 and 2}{K}_h=K_i=\frac{[H^{+}{]}^2}{\left[N{H}_4^{+}\right]}$
Since, $K_h=K_i\therefore K_i=5.55\times {10}^{-10}$