Question

The normality of solution obtained by mixing $100\text{mL}$ of $0.2M$ $H_{2}S{O}_4$ with $100\text{mL}$ of $0.2M$ $NaOH$ is:

• A. $0.1\text{N}$
• B. $0.2\text{N}$
• C. $0.5\text{N}$
• D. $0.3\text{N}$

Answer 1

The correct option is A $0.1\text{N}$

Number of eq. of $H_{2}S{O}_4=N_1{V}_1$
$=0.2\times 2\times \frac{100}{1000}$
$=0.04$

Number of eq. of $NaOH=N_2{V}_2$
$=0.2\times 1\times \frac{100}{1000}$
$=0.02$

Resulting equivalent $=N_1{V}_1-N_2{V}_2$
$=0.04-0.02$
$=0.02$
$N_{mix}=\frac{\text{Resulting eq.}}{\text{V(L)}}=\frac{0.02}{0.2}=0.1\text{N}$