100 ml of 0.1 M $\mathrm{HCl}$, 200 ml 0.1 M ${\mathrm{H}}_2{\mathrm{SO}}_4$ and 200 ml of 0.2 M $\mathrm{NaOH}$ are mixed. ${\mathrm{p}}^{\mathrm{H}}$ of the resultant mixture is

$\begin{array}{l}{\mathrm{V}}_{\mathrm{a}}{\mathrm{N}}_{\mathrm{a}}\left(\mathrm{HCl}\right)=100\left(0.1\right)1=10\mathrm{milli} \mathrm{equivalents}\\ {\mathrm{V}}_{\mathrm{a}}{\mathrm{N}}_{\mathrm{a}}\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)=200\left(0.1\right)2=40\mathrm{milli} \mathrm{equivalents}\\ {\mathrm{V}}_{\mathrm{b}}{\mathrm{N}}_{\mathrm{b}}\left(\mathrm{NaOH}\right)=200\left(0.2\right)1=40\mathrm{milli} \mathrm{equivalents}\\ {\mathrm{V}}_{\mathrm{a}}{\mathrm{N}}_{\mathrm{a}}=50\mathrm{milli} \mathrm{equivalents}\\ {\mathrm{V}}_{\mathrm{b}}{\mathrm{N}}_{\mathrm{b}}=40\mathrm{milli} \mathrm{equivalents}\\ \left[{\mathrm{H}}^{+}\right]=\frac{{\mathrm{V}}_{\mathrm{a}}{\mathrm{N}}_{\mathrm{a}}-{\mathrm{V}}_{\mathrm{b}}{\mathrm{N}}_{\mathrm{b}}}{{\mathrm{V}}_{\mathrm{total}}}\\ \left[{\mathrm{H}}^{+}\right]=\frac{10}{500}\\ \left[{\mathrm{H}}^{+}\right]=2\times {10}^{-2}\mathrm{M}\\ {\mathrm{p}}^{\mathrm{H}}=-\log \left(2\times {10}^{-2}\mathrm{M}\right)\\ {\mathrm{p}}^{\mathrm{H}}=1.699\end{array}$