In the reaction $2\mathrm{NaOH}+{\mathrm{H}}_3{\mathrm{PO}}_4\to {\mathrm{Na}}_2{\mathrm{HPO}}_4+2{\mathrm{H}}_2\mathrm{O}$, the Equivalent weight of the acid is

We are required to calculate the Equivalent weight of ${\mathrm{H}}_3{\mathrm{PO}}_4$.

$2\mathrm{NaOH}+{\mathrm{H}}_3{\mathrm{PO}}_4\to {\mathrm{Na}}_2{\mathrm{HPO}}_4+2{\mathrm{H}}_2\mathrm{O}$

In the reaction

The number of hydrogen ions replaced by the base of acid is 2.

The molar mass of the acid= 98 g

Now we know that equivalent weight$=\frac{\mathrm{Molecular} \mathrm{Weight}}{\mathrm{Net} \mathrm{positive} \mathrm{valence}}$

Therefore, the Equivalent weight of ${\mathrm{H}}_3{\mathrm{PO}}_4$$=\frac{98}{2}=49$

Hence, option(a) is correct.