Chemistry - Ionic Equilibrium Question with Solution | TestHub

Given, ${\mathrm{K}}_{\mathrm{a}}\left({\mathrm{NH}}_4\mathrm{OH}\right)=1.77\times 10^{-5}$ $$\begin{array}{rl}& {\mathrm{NH}}_4\mathrm{OH}\rightleftharpoons {\mathrm{NH}}_4^{+}+{\mathrm{OH}}^{-}\\ & {\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{NH}}_4^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{NH}}_4\mathrm{OH}\right]}=1.77\times 10^{-5}\end{array}$$ Hydrolysis of ${\mathrm{NH}}_4\mathrm{Cl}$ takes place as, $${\mathrm{NH}}_4\mathrm{Cl}+{\mathrm{H}}_2\mathrm{O}⟶{\mathrm{NH}}_4\mathrm{OH}+\mathrm{HCl}$$ or ${\mathrm{NH}}_4^{+}+{\mathrm{H}}_2\mathrm{O}⟶{\mathrm{NH}}_4\mathrm{OH}+{\mathrm{H}}^{+}$ Hydrolysis constant $$,{\mathrm{K}}_{\mathrm{h}}=\frac{\left[{\mathrm{NH}}_4\mathrm{OH}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{NH}}_4^{+}\right]}\dots$$ or $${\mathrm{K}}_{\mathrm{h}}=\frac{\left[{\mathrm{NH}}_4\mathrm{OH}\right]\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{NH}}_4^{+}\right]\left[{\mathrm{OH}}^{-}\right]}$$ From Eqs (i) and (iii) $$\begin{array}{rl}{\mathrm{K}}_{\mathrm{h}}& =\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}}\left[\because \left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]={\mathrm{K}}_{\mathrm{w}}\right]\\ & =\frac{10^{-14}}{1.77\times 10^{-5}}\\ & =5.65\times 10^{-10}\end{array}$$