Question

The volume of 0.25 M $H_{3}P{O}_3$ required to neutralize 25 ml of 0.03 m $Ca(OH{)}_2$ is ?

• A. 1.32 l
• B. 3 mL
• C. 26.4 mL
• D. 2.0 mL

Answer 1

The correct option is B 3 mL

Let, the required volume of $H_3{PO}_3$ be $VmL$.

Equivalent of acid reacted = Equivalent of base reacted.

$\Rightarrow [{\mathrm{Vol}}^m\times$ molarity $x$ n-factor of acid $]=\left[\begin{array}{c}{V_c|}^m\times \text{molarity}x\text{-n}\text{fact of}\\ \text{base}\end{array}\right]$

$\begin{array}{c}\Rightarrow (V\times 0.25\times 2)=(25\times 0.03\times 2)\\ \therefore V=\frac{25\times 0.03}{0.25}=3mL\text{ans}\end{array}$

Note that, n-factor of acid is its besiaity and that of base

is its acidity.

se, correct option is(B)