Question

The equivalent weight of phosphoric acid $\left(H_{3}P{O}_4\right)$ in the reaction,
$NaOH+H_{3}P{O}_4⟶Na{H}_{2}P{O}_4+H_{2}O$, is

• A. 59
• B. 49
• C. 25
• D. 98

Answer 1

Answer: (D)

Here $1$ mole of $NaOH$ reacts with $1$ mole of $H_{3}P{O}_4$. Since substances react in the ratio of their gram equivalents; thus, $1g$ equivalent weight of $NaOH$ reacts with $1g$ equivalent weight of $H_{3}P{O}_4$. Therefore, equivalent weight of $H_{3}P{O}_4$ in this reaction is $98$.