The ionisation constant of ammonium hydroxide is $$1.77 \times {10^{ - 5}}$$   at $$298\,K.$$  Hydrolysis constant of ammonium chloride is

Solution :
$$\eqalign{ & {\text{Given,}}\,{K_a}\left( {N{H_4}OH} \right) = 1.77 \times {10^{ - 5}} \cr & N{H_4}OH \rightleftharpoons NH_4^ + + O{H^ - } \cr & {K_a} = \frac{{\left[ {NH_4^ + } \right]\left[ {O{H^ - }} \right]}}{{\left[ {N{H_4}OH} \right]}} \cr & \,\,\,\,\,\,\,\, = 1.77 \times {10^{ - 5}}\,\,\,\,...{\text{(i)}} \cr & {\text{Hydrolysis of}}\,N{H_4}Cl\,{\text{takes place as,}} \cr & N{H_4}Cl + {H_2}O \to N{H_4}OH + HCl \cr & {\text{or}}\,\,NH_4^ + + {H_2}O \to N{H_4}OH + {H^ + } \cr & {\text{Hydrolysis constant,}} \cr & \,\,\,\,\,\,{K_h} = \frac{{\left[ {N{H_4}OH} \right]\left[ {{H^ + }} \right]}}{{\left[ {NH_4^ + } \right]}}...{\text{(ii)}} \cr & {\text{or}}\,{{\text{K}}_h} = \frac{{\left[ {N{H_4}OH} \right]\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]}}{{\left[ {NH_4^ + } \right]\left[ {O{H^ - }} \right]}}...{\text{(iii)}} \cr & {\text{From Eqs}}{\text{. (i), (ii) and (iii)}} \cr & {{\text{K}}_h} = \frac{{{K_w}}}{{{K_a}}}\,\,\,\left[ {\because \,\,\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] = {K_w}} \right] \cr & \,\,\,\,\,\,\,\, = \frac{{{{10}^{ - 14}}}}{{1.77 \times {{10}^{ - 5}}}} \cr & \,\,\,\,\,\,\,\, = 5.65 \times {10^{ - 10}} \cr} $$