Chemistry - Ionic Equilibrium Question with Solution | TestHub

${\text{X}}^{-}\text{+}{\text{H}}_{\text{2}}\text{O}\rightleftharpoons {\text{OH}}^{-}\text{+}\text{HX}$
${\mathrm{K}}_{\mathrm{b}}=\frac{\left[\mathrm{O}{\mathrm{H}}^{-}\right]\mathrm{}\mathrm{}\left[\mathrm{H}\mathrm{X}\right]}{\left[{\mathrm{X}}^{-}\right]}$
$\text{HX}\rightleftharpoons {\text{H}}^{\text{+}}\text{+}{\text{X}}^{-}$
${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\mathrm{}\mathrm{}\left[{\mathrm{X}}^{-}\right]}{\left[\mathrm{H}\mathrm{X}\right]}$
$\therefore \mathrm{}\mathrm{}{\mathrm{K}}_{\mathrm{a}}\times {\mathrm{K}}_{\mathrm{b}}=\left[{\mathrm{H}}^{+}\right]\mathrm{}\mathrm{}\left[\mathrm{O}{\mathrm{H}}^{-}\right]={\mathrm{K}}_{\mathrm{w}}=10^{-14}$
Hence${\mathrm{K}}_{\mathrm{a}}=10^{-4}$
Now as$\left[{\mathrm{X}}^{-}\right]=\left[\mathrm{H}\mathrm{X}\right],\mathrm{}\mathrm{}\mathrm{p}\mathrm{H}=\mathrm{p}{\mathrm{K}}_{\mathrm{a}}=4$.