The equivalent weight of phosphoric acid $\left({\mathrm{H}}_3{\mathrm{PO}}_4\right)$ in the reaction, $\mathrm{NaOH}+{\mathrm{H}}_3{\mathrm{PO}}_4\to {\mathrm{NaH}}_2{\mathrm{PO}}_4+{\mathrm{H}}_2\mathrm{O}$ is

$$\begin{gathered} \begin{array}{l}\mathrm{NaOH}+{\mathrm{H}}_3{\mathrm{PO}}_4\to {\mathrm{NaH}}_2{\mathrm{PO}}_4+{\mathrm{H}}_2\mathrm{O} \\ \begin{array}{l}40\mathrm{g} 98\mathrm{g}\\ \mathrm{lmol} 1\mathrm{mol} \\ 1\mathrm{g} \mathrm{eq}.\end{array}\\ \\ \end{array} \end{gathered}$$
Here 1 mole of NaOH reacts with 1 mole of ${\mathrm{H}}_3{\mathrm{PO}}_4$ Since substances react in the
ratio of their gram equivalents; thus, 1 g equivalent weight of NaOH reacts with 1 g
equivalent weight of  ${\mathrm{H}}_3{\mathrm{PO}}_4$ . Therefore, equivalent weight of ${\mathrm{H}}_3{\mathrm{PO}}_4$ in this reaction is 98.