Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\mathrm{H}\mathrm{A}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{A}}^{-}$
${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{H}\mathrm{A}\right]}$…(i)
Also,$\mathrm{H}\mathrm{A}+{\mathrm{B}}^{+}+\mathrm{O}{\mathrm{H}}^{-}\to {\mathrm{B}}^{+}+{\mathrm{A}}^{-}+{\mathrm{H}}_2\mathrm{O}$
${\mathrm{K}}_{\mathrm{e}\mathrm{q}}=\frac{\left[{\mathrm{H}}_2\mathrm{O}\right]\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{H}\mathrm{A}\right]\left[\mathrm{O}{\mathrm{H}}^{-}\right]}$…(ii)
By Eqs. (i) and (ii),
$\frac{{\mathrm{K}}_{\mathrm{e}\mathrm{q}}}{{\mathrm{K}}_{\mathrm{a}}}=\frac{1}{\left[{\mathrm{H}}^{+}\right]\left[\mathrm{O}{\mathrm{H}}^{-}\right]}=\frac{1}{{\mathrm{K}}_{\mathrm{w}}}$
So${\mathrm{K}}_{\mathrm{e}\mathrm{q}}=\frac{{\mathrm{K}}_{\mathrm{a}}}{{\mathrm{K}}_{\mathrm{w}}}=\frac{10^{-4}}{10^{-14}}=10^{10}$.