The volume of a$0.5 \mathrm{M} {\mathrm{H}}_2{\mathrm{SO}}_4$ required to react completely with $25 \mathrm{mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}$ will be:

$2\mathrm{NaOH} + {\mathrm{H}}_2{\mathrm{SO}}_4 ⟶ {\mathrm{Na}}_2{\mathrm{SO}}_4 + {\mathrm{H}}_2\mathrm{O}$

$\begin{array}{cc}2\mathrm{mol}& 1\mathrm{mol}\\ 2\mathrm{m}\cdot \mathrm{mol}& 1\mathrm{m}\cdot \mathrm{mol}\end{array}$

     $\begin{array}{r}\frac{M_1{V}_1\left(\mathrm{NaOH}\right)}{M_2{V}_2\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)}=\frac{\text{ no. of mol of }\mathrm{NaOH}}{\text{ no. of mol of }{\mathrm{H}}_2{\mathrm{SO}}_4};\\ \frac{0.1\times 2.5}{0.5\times V_2}=\frac{2}{1};V_2=\frac{0.1\times 25}{0.5\times 2}=2.5\mathrm{mL}\end{array}$