Question

Equivalent weight of $H_{3}P{O}_4$ in each of the reaction will be respectively:
$H_{3}P{O}_4+O{H}^{-}\to H_{2}P{O}_4^{-}+H_{2}O$
$H_{3}P{O}_4+2O{H}^{-}\to HP{O}_4^{2-}+2H_{2}O$
$H_{3}P{O}_4+3O{H}^{-}\to P{O}_4^{3-}+3H_{2}O$

• A. 98, 49, 32.67
• B. 49, 98, 32.67
• C. 98, 32.67, 49
• D. 32.67, 49, 98

Answer 1

The correct option is A 98, 49, 32.67

$\text{Equivalent weight}=\frac{\text{Molecular weight}}{\text{n-factor}}$
In these reactions, the n-factor is the number of $H^{+}$ released.
In first reaction, one $H^{+}$ is released so n-factor is one.
Therefore, $\text{Equivalent weight = Molecular weight}$
Molecular weight of $H_{3}P{O}_4$ = 98
$\therefore$ Equivalent weight of $H_{3}P{O}_4$ = 98

In second reaction, n-factor is 2
$\therefore$ Equivalent weight of $H_{3}P{O}_4=\frac{98}{2}=49$

In third reaction, n-factor is 3
$\therefore$ Equivalent weight of $H_{3}P{O}_4=\frac{98}{3}=32.67$
Hence, option (a) is correct.