Question

The equivalent weight of phosphoric acid $\left(H_{3}P{O}_4\right)$ in the reaction :
$NaOH+H_{3}P{O}_4\to Na{H}_{2}P{O}_4+H_{2}O$
(correct answer +1, wrong answer -0.25)

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

Answer 1

The correct option is D $98$

n-factor of $H_{3}P{O}_4$ is one, since only one $H$ is replaced.

Equivalent weight $=\frac{\text{molecular weight}}{1}=\frac{98}{1}=98$