NEET UG 2009 — Chemistry Physical Chemistry

Given, $\mathrm{K}_{\mathrm{a}}\left(\mathrm{NH}_4 \mathrm{OH}\right)=1.77 \times 10^{-5}$Ka​(NH4​OH)=1.77×10−5 $$\begin{aligned} & \mathrm{NH}_4 \mathrm{OH} \rightleftharpoons \mathrm{NH}_4^{+}+\mathrm{OH}^{-} \\ & \mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_4 \mathrm{OH}\right]}=1.77 \times 10^{-5} \end{aligned}$$​NH4​OH⇌NH4+​+OH−Ka​=[NH4​OH][NH4+​][OH−]​=1.77×10−5​ Hydrolysis of $\mathrm{NH}_4 \mathrm{Cl}$NH4​Cl takes place as, $$\mathrm{NH}_4 \mathrm{Cl}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{NH}_4 \mathrm{OH}+\mathrm{HCl}$$NH4​Cl+H2​O⟶NH4​OH+HCl or $\mathrm{NH}_4^{+}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{NH}_4 \mathrm{OH}+\mathrm{H}^{+}$NH4+​+H2​O⟶NH4​OH+H+ Hydrolysis constant $$, \mathrm{K}_{\mathrm{h}}=\frac{\left[\mathrm{NH}_4 \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{NH}_4^{+}\right]} \ldots$$,Kh​=[NH4+​][NH4​OH][H+]​… or $$\mathrm{K}_{\mathrm{h}}=\frac{\left[\mathrm{NH}_4 \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{OH}^{-}\right]}$$Kh​=[NH4+​][OH−][NH4​OH][H+][OH−]​ From Eqs (i) and (iii) $$\begin{aligned} \mathrm{K}_{\mathrm{h}} & =\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}}} \quad\left[\because\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=\mathrm{K}_{\mathrm{w}}\right] \\ & =\frac{10^{-14}}{1.77 \times 10^{-5}} \\ & =5.65 \times 10^{-10} \end{aligned}$$Kh​​=Ka​Kw​​[∵[H+][OH−]=Kw​]=1.77×10−510−14​=5.65×10−10​