Question

In the reaction $NaOH+H_{3}P{O}_4\to N{a}_{2}HP{O}_4+H_{2}O$, the equivalent weight of $H_{3}P{O}_4$ (phosphoric acid) is:

• A. 98
• B. 49
• C. 32.66
• D. 24.5

Answer 1

The correct option is B 49

n-factor of $H_{3}P{O}_4$ for this particular reaction = 2
(As only 2 hydrogens are getting replaced)
Equivalent weight of $H_{3}P{O}_4$ = Molar wt/2 $=\frac{98}{2}=49$