Chemistry - Ionic Equilibrium Question with Solution | TestHub

Given, for HCl,$\mathrm{p}\mathrm{H}=3$
$\left[{\mathrm{H}}^{+}\right]=10^{-3}={\mathrm{M}}_1$
${\mathrm{V}}_1=2\mathrm{m}\mathrm{l}$
For NaOH,$\mathrm{p}\mathrm{H}=\mathrm{}10,\left[{\mathrm{H}}^{+}\right]=10^{-10}={\mathrm{M}}_2$
${\mathrm{V}}_2=3\mathrm{m}\mathrm{l}$
${\mathrm{M}}_1{\mathrm{V}}_1-{\mathrm{M}}_2{\mathrm{V}}_2={\mathrm{M}}_{\mathrm{R}}({\mathrm{V}}_1+{\mathrm{V}}_2)$
$10^{-3}\times 2-10^{-10}\times 3={\mathrm{M}}_{\mathrm{R}}(2+3)$
$2\times 10^{-3}=5\times {\mathrm{M}}_{\mathrm{R}}$
${\mathrm{M}}_{\mathrm{R}}=4\times 10^{-4}=\left[{\mathrm{H}}^{+}\right]$
$\mathrm{p}\mathrm{H}=\mathrm{l}\mathrm{o}\mathrm{g}(4\times 10^{-4})=4-0.6020=3.39\approx 3.5$