Question

The equivalent weight of phosphoric acid $\left(H_{3}P{O}_4\right)$ in the reaction,

$NaOH+H_{3}P{O}_4\to Na{H}_{2}P{O}_4+H_{2}O$ is:

• A. $25$
• B. $49$
• C. $59$
• D. $98$

Answer 1

The correct option is D $98$

We know that,

Equivalent weight of a compound = Molecular Weight / change in valency

For the given reaction,

$NaOH+H_{3}P{O}_4\to Na{H}_{2}P{O}_4+H_{2}O$

Molecular weight of $H_{3}P{O}_4=98u$

Change in valency of $H_{3}P{O}_4$ can be given by the $H^{+}$ lost or transferred which is $1$ .

Substituting these values in the formula for equivalent weight, we get

$Eq.Wt.=\frac{98}{1}=98$

Hence the correct option is option D.