Question

The volume of $0.1M{H}_{2}S{O}_4$ required to neutralize 30 mL of $2.0MNaOH$ is:

• A. 100 mL
• B. 300 mL
• C. 400 mL
• D. 200 mL

Answer 1

The correct option is B 300 mL

Solution:- (B) $300mL$

Let $V$ be the volume of $H_{2}S{O}_4$ required.

Given:-

$M_{H_{2}S{O}_4}=0.1M$

$M_{NaOH}=2M$

$V_{NaOH}=30mL$

Now, as we know that

$N_1{V}_1=N_2{V}_2$

$\because N=n\times M$

Whereas $n$ is the valence factor

Therefore,

$n_1{M}_1{V}_1=n_2{M}_2{V}_2$

For $H_{2}S{O}_4;n=2$

For $NaOH;n=1$

Therefore,

$2\times 0.1\times V=1\times 2\times 30$

$\Rightarrow V=\frac{60}{0.2}$

$\Rightarrow V=300mL$

Hence $300mL$ of $H_{2}S{O}_4$ is required to neutralize $30mL$ of $2MNaOH$.