Question

The normality of the excess ions $\left(H^{+}/O{H}^{-}\right)$ in a solution obtained by mixing $100\text{mL}$ of a $0.2\text{M}{H}_{2}S{O}_4$ solution with $100\text{mL}$ of a $0.2\text{M}NaOH$ solution is:

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

Answer 1

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

Normality = n-factor $\times$ Molarity
Normality of $0.2\text{M}{H}_{2}S{O}_4$ solution $=2\times 0.2=0.4\text{N}$
Normality of $0.2\text{M}NaOH$ solution $=1\times 0.2=0.2N$
Now, $N_1{V}_1-N_2{V}_2=N_f{V}_f$ (1 is the acid and 2 is the base)
$(0.4\times 100)–(0.2\times 100)=N_f\times 200$
$N_f=0.1\text{N}$
Since, there are more equivalents of the acid than the base, the final solution will be acidic and it will contain an excess of $H^{+}$ ions
Normality of $H^{+}$ ions = $0.1\text{N}$