Equivalent weight of H₃PO₄ in the following reaction is ${\mathrm{H}}_3{\mathrm{PO}}_4+\mathrm{NaOH}\to {\mathrm{NaH}}_2{\mathrm{PO}}_4+{\mathrm{H}}_2\mathrm{O}$

${\mathrm{H}}_3{\mathrm{PO}}_4+\mathrm{NaOH}\to {\mathrm{NaH}}_2{\mathrm{PO}}_4+{\mathrm{H}}_2\mathrm{O}$

This is a neutralization reaction; 

In the above reaction, only one of the H-atom of H₃PO₄ is replaced.

$$\begin{gathered} \therefore \mathrm{EW} \mathrm{of} {\mathrm{H}}_3{\mathrm{PO}}_4=\frac{\mathrm{M}}{1} \\ \mathrm{EW} \mathrm{of} {\mathrm{H}}_3{\mathrm{PO}}_4=\frac{98}{1}=98 \end{gathered}$$