If 200 mL of a 0.031 molar solution of H₂SO₄ is added to 84 mL of a 0.150 M KOH solution, The value of pH of the resulting solution is

$$\begin{gathered} \mathrm{mmol}\text{ of }{\mathrm{H}}^{+}\text{(initial) }=200\times 0.031\times 2=12.4 \\ \mathrm{mmol}\text{ of }{\mathrm{OH}}^{-}\text{(initial) }=84\times 0.15=12.6 \\ \mathrm{mmol}\text{ of }{\mathrm{OH}}^{-}\text{(left) after neutralisation }=12.6-12.4=0.2 \\ {\left[{\mathrm{OH}}^{-}\right]}_{\text{final }}=\frac{0.2}{284}=7\times {10}^{-4}\mathrm{M} \\ \mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right]=-\log \left(7\times {10}^{-4}\right) \\ \mathrm{pOH}=3.15\text{ and }\mathrm{pH}=10.85 \\ \therefore \mathrm{pH}=14-\mathrm{pOH}=14-3.15=10.85\approx 10.9 \end{gathered}$$