Question

$0.5g$ of fuming $H_{2}S{O}_4$ (oleum) is diluted with water. This solution is completely neutralised by $26.7mL$ of $0.4NNaOH$. Find the percentage of free $S{O}_3$ in the sample solution.

Answer 1

Oleum consists of $S{O}_3$ and $H_{2}S{O}_4$.

Let the mass of $S{O}_3$ in the given sample of oleum be $=xg$

Mass of $H_{2}S{O}_4$ in the given sample of oleum $=(0.5-x)g$

Eq. mass of $S{O}_3=\frac{80}{2}=40$

No. of $g$ equivalents of $S{O}_3=\frac{x}{40}$

$[2NaOH+S{O}_3\to N{a}_{2}S{O}_4+H_{2}O$

$2NaOH+H_{2}S{O}_4\to N{a}_{2}S{O}_4+2H_{2}O$]

Eq. mass of $H_{2}S{O}_4=\frac{98}{2}=49$

No. of $g$ equivalents of $H_{2}S{O}_4=\frac{(0.5-x)}{49}$

Total no. of $g$ equivalents $=\frac{x}{40}+\frac{(0.5-x)}{49}$

$26.7mL$ of $0.4NNaOH$ contain no. of equivalents of $NaOH$

$=\frac{0.4}{1000}\times 26.7$

At equivalence point,
No. of $g$ equivalents of $NaOH=\frac{x}{40}+\frac{(0.5-x)}{49}$

So, $\frac{0.4\times 26.7}{1000}=\frac{49x+(40\times 0.5-40x)}{40\times 49}$

$x=\frac{0.9328}{9}=0.1036$

Hence, % of free $S{O}_3=\frac{0.1036}{0.5}\times 100$

$=20.72$%