Question

$12.5$gm of fuming $H_{2}S{O}_4$ (labelled as $112\%$) is mixed with $100$lit water. Molar concentration of $H^{+}$ in resultant solution is:

• A. $\frac{14}{5}$
• B. $\frac{2}{350}$
• C. $\frac{3}{350}$
• D. $\frac{14}{70}$

Answer 1

The correct option is A $\frac{14}{5}$

$H_2{S}_2{O}_7+H_{2}O⟶2H_{2}S{O}_4$

$178g$ $H_{2}S{O}_2{O}_7$ combines with $18g$ $H_{2}O$ to give $2\times 98=196g$ $H_{2}S{O}_4$.

$112g$ will give $\frac{196}{178}\times 112=123.32g$ $H_{2}S{O}_4$.

$\therefore 12.5g$ will form $\frac{196}{178}\times 12.5=13.76g$ $H_{2}S{O}_4$.

Concentration $=\frac{13.7}{98}\times \frac{1000}{100}=1.4M=\frac{7}{5}M$

$\therefore$ Concentration of $H^{+}=2\times \frac{7}{5}=\frac{14}{5}M$