How many ml of 1 (M) ${\mathrm{H}}_2{\mathrm{SO}}_4$ is required to neutralize 10 ml of 1 (M) NaOH solution ?

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

$\begin{array}{l}\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)\frac{{\mathrm{M}}_1{\mathrm{V}}_1}{{\mathrm{n}}_1}=\frac{{\mathrm{M}}_2{\mathrm{V}}_2}{{\mathrm{n}}_2}\left(\mathrm{NaOH}\right)\\ {\mathrm{n}}_1 , {\mathrm{n}}_2 \mathrm{are} \mathrm{no}.\mathrm{of} \mathrm{moles} \mathrm{from} \mathrm{equation}\\ \frac{1\times {\mathrm{V}}_1}{1}=\frac{1\times 10}{2}\\ \hspace{0.17em}{\mathrm{V}}_1=5\mathrm{ml}.\end{array}$