$0.5 ext{ g}$ of fuming $ ext{H}_2 ext{SO}_4$ (oleum) is diluted with water. This solution is co

Question

$0.5 	ext{ g}$ of fuming $	ext{H}_2	ext{SO}_4$ (oleum) is diluted with water. This solution is completely neutralized by $26.7 	ext{ mL}$ of $0.4 	ext{ N NaOH}$. Find the percentage of free $	ext{SO}_3$ in the sample of oleum. (Report your answer to the nearest integer).

Professor PC Thomas Classes · Answer

Total equivalents of acid = equivalents of base = $26.7 	imes 0.4 	imes 10^{-3} = 10.68 	imes 10^{-3}$ eq. Let the mass of free $	ext{SO}_3$ be $x$ g, making the mass of $	ext{H}_2	ext{SO}_4$ equal to $(0.5-x)$ g. Equivalent weight of $	ext{SO}_3 = 80/2 = 40$ and $	ext{H}_2	ext{SO}_4 = 98/2 = 49$. 
$\frac{x}{40} + \frac{0.5-x}{49} = 10.68 	imes 10^{-3}$. Multiplying by 1960 gives $49x + 20 - 40x = 20.9328 \implies 9x = 0.9328 \implies x \approx 0.1036$ g. 
$\% 	ext{ free SO}_3 = \frac{0.1036}{0.5} 	imes 100 = 20.72\% \approx 21\%$.