Chemistry - Ionic Equilibrium Question with Solution | TestHub

${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$is a diprotic and in the first stage of ionization behaves as a strong acid, but in the second stage of ionization it behaves as weak acid.
${\text{HSO}}_{\text{4}}^{-}\stackrel{{\text{H}}_{\text{2}}\text{O}}{\to }{\text{H}}^{\text{+}}\text{+}{\text{SO}}_{\text{4}}^{\text{2}-}$
Thus, we also get some${\mathrm{H}}^{+}$in the second stage.
Hence,$\left[{\mathrm{H}}^{+}\right]>\left[\mathrm{H}\mathrm{S}{\mathrm{O}}_4^{-}\right]$