Chemistry - Ionic Equilibrium Question with Solution | TestHub

Let us consider the formation of a salt of a weak acid and a strong base. $$\begin{array}{rl}{\mathrm{In}}^{-}+{\mathrm{H}}_2\mathrm{O}& \rightleftharpoons \mathrm{HIn}+{\mathrm{OH}}^{-}\\ {\mathrm{K}}_h& =\frac{[\mathrm{HIn}]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{In}}^{-}\right]}\end{array}$$ $$\begin{array}{rl}& \mathrm{HIn}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{In}}^{-}\\ & {\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{OH}}^{-}\\ & {\mathrm{K}}_{\ln }=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{In}}^{-}\right]}{[\mathrm{HIn}]}\\ & {\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]\end{array}$$ From (ii) and (iii), $$\begin{array}{rl}& \frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{In}}}=\frac{[\mathrm{HIn}]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{In}}^{-}\right]}\\ & \left[{\mathrm{OH}}^{-}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{In}}}\frac{\left[{\mathrm{In}}^{-}\right]}{[\mathrm{HIn}]}\\ & \log \left[{\mathrm{OH}}^{-}\right]={\mathrm{logK}}_{\mathrm{w}}-{\log }_{\ln }+\log \\ & \frac{\left[{\mathrm{In}}^{-}\right]}{[\mathrm{HIn}]}\end{array}$$ $$\begin{array}{rl}& -\mathrm{pOH}=-{\mathrm{pK}}_w+{\mathrm{pK}}_{\ln }+\log \\ & \frac{\left[{\mathrm{In}}^{-}\right]}{[\mathrm{HIn}]}\\ & {\mathrm{pK}}_w-\mathrm{pOH}={\mathrm{pK}}_{\ln }+\log \frac{\left[{\mathrm{In}}^{-}\right]}{[\mathrm{HIn}]}\\ & \text{ or, }\mathrm{pH}={\mathrm{pK}}_{\ln }+\log \frac{\left[{\mathrm{In}}^{-}\right]}{[\mathrm{HIn}]}\\ & \log \frac{\left[{\mathrm{In}}^{-}\right]}{[\mathrm{HIn}]}=\mathrm{pH}-{\mathrm{pK}}_{\ln }\\ & \end{array}$$