Question

1 g of fuming sulphuric acid $(H_{2}S{O}_4+S{O}_3)$, called oleum, is diluted with water. This solution is completely neutralised 26.7 ml of 0.4 M NaOH. Find the percentage of free $S{O}_3$ in the sample solution.

• A. $20.73$%
• B. $43.80$%
• C. $79.27$%
• D. $60.74$%

Answer 1

The correct option is A $20.73$%

$H_2{S}_2{O}_7+H_{2}O\to 2H_{2}S{O}_4$
$S{O}_3$ of $H_2{S}_2{O}_7$ is converted into $H_{2}S{O}_4$, hence $S{O}_3$ acts also as a dibasic acid.

Eq. wt. $\left(S{O}_3\right)=\frac{M}{2}=40$

Let $H_{2}S{O}_4$ in fuming $H_{2}S{O}_4=xg$

$S{O}_3=(1-x)g$

Equivalent of $H_{2}S{O}_4=\frac{x}{49}$

$S{O}_3=\frac{1-x}{40}$

Equivalent of $NaOH$ used $=\frac{26.7\times 0.8}{1000}$
$=0.02136$

$\frac{x}{49}+\frac{1-x}{40}=0.02136$

$x=0.7927g{H}_{2}S{O}_4$ in $1g$ oleum.

Percentage of $H_{2}S{O}_4=79.27$%

$S{O}_3=20.73$%.