${N H_{4} O H \rightleftharpoons N H_{4}^{+}+O H^{-}}$

${K_{a}=\frac{\left[N H_{4}^{+}\right]\left[O H^{-}\right]}{\left[N H_{4} O H\right]}}$${=1.77 \times 10^{-5} \ldots .(i)}$

Hydrolysis of $N H_{4} C l$ takes place as,

$N H_{4} C l+H_{2} O \rightarrow N H_{4} O H+H C l$

$\quad N H_{4}^{+}+H_{2} O \rightarrow N H_{4} O H+H^{+}$

Hydrolysis constant, $K_{h}=\frac{\left[N H_{0} O H\right]\left[H^{+}\right]}{\left[N H_{4}^{+}\right]} \ldots \ldots(i i)$

$ \quad K_{h}=\frac{\left[N H_{4} O H\right]\left[H^{+}\right]\left[O H^{-}\right]}{\left[N H_{4}^{+}\right][O H]} \ldots \ldots(i i i)$

From Eqs. $(i)$ and $(iii)$

${K_{h}=\frac{K_{w}}{K_{a}}}$ ${\left[\therefore\left[H^{+}\right]\left[O H^{-}\right]=K_{w}\right]}$

${=\frac{10^{-14}}{1.77 \times 10^{-5}}=5.65 \times 10^{-10}}$