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Q.
The equivalent weight of phosphoric acid $ ({{H}_{3}}P{{O}_{4}}) $ in the reaction: $ NaOH+{{H}_{2}}P{{O}_{4}}\xrightarrow{{}}Na{{H}_{2}}P{{O}_{4}}+{{H}_{2}}O $ is:
BHUBHU 2005
Solution:
$ \underset{\begin{smallmatrix} 23+16+1 \\ =40 \\ 1g.eq \end{smallmatrix}}{\mathop{NaOH}}\,+\underset{\begin{smallmatrix} 1\times 3+31+16\times 4 \\ =98 \\ 1g-eq \end{smallmatrix}}{\mathop{{{H}_{3}}P{{O}_{4}}}}\,\xrightarrow[{}]{{}}Na{{H}_{2}}P{{O}_{4}}+{{H}_{2}}O $
$ \therefore $ equivalent weight of $ {{H}_{3}}P{{O}_{4}} $ is 98 g In the above reaction no change in oxidation state of phosphorus takes place. So, equivalent weight will by equal to its molecular weight.