Chemistry - Ionic Equilibrium Question with Solution | TestHub

First there will be an acid-base reaction between$\mathrm{K}{\mathrm{H}}_2\mathrm{B}{\mathrm{O}}_3$and$\mathrm{H}\mathrm{C}\mathrm{l}$.

conc. will become half after reaction of an equal volume of${\mathrm{K}\mathrm{H}}_2{\mathrm{B}\mathrm{O}}_3$and$\mathrm{H}\mathrm{C}\mathrm{l}$
So$\frac{0.1}{5}=0.05$
Now,

$\left[{\mathrm{H}}^{+}\right]=\sqrt{{\mathrm{K}}_{\mathrm{a}}\mathrm{C}}=\sqrt{12.8\times {10}^{-10}\times 0.05}=8\times 10^{-6}$
$\mathrm{p}\mathrm{H}=-\log \left(8\times {10}^{-6}\right)\approx 5.0$