How many millilitres of 0.1 N ${\mathrm{H}}_2{\mathrm{SO}}_4$ solution will be required for complete reaction with a solution containing 0.125 g of pure ${\mathrm{Na}}_2{\mathrm{CO}}_3?$

$\begin{array}{l}\mathrm{Meq}\mathrm{of}{\mathrm{H}}_2{\mathrm{SO}}_4=\mathrm{Meq}\mathrm{of}{\mathrm{Na}}_2{\mathrm{CO}}_3\\ {\left(\frac{\mathrm{NV}\left(\mathrm{ml}\right)}{1000}\right)}_{{\mathrm{H}}_2{\mathrm{SO}}_4}={\left(\frac{\mathrm{W}}{\mathrm{E}}\right)}_{{\mathrm{Na}}_2{\mathrm{CO}}_3}\\ 0.1\times \frac{\mathrm{V}}{1000}=\frac{0.125}{106}\times 2\\ \mathrm{V}=23.6\mathrm{mL}\end{array}$

Note:  Meq=no.of milli equivalents

              $\mathrm{Meq}=\frac{\mathrm{W}}{\mathrm{E}}=\frac{\mathrm{NV}\left(\mathrm{ml}\right)}{1000}$