Question

$5g$ sample contain only $N{a}_{2}C{O}_3$ and $N{a}_{2}S{O}_4$. This sample is dissolved and the volume made up to $500mL.25mL$ of this solution neutralizes $20mL$ of $0.1M{H}_{2}S{O}_4$. Calculate the percentage of $N{a}_{2}S{O}_4$ in the sample.

Answer 1

Answer: 15.2

Only $N{a}_{2}C{O}_3$ reacts with $H_{2}S{O}_4$;
$N{a}_{2}C{O}_3+H_{2}S{O}_4\to N{a}_{2}S{O}_4+H_{2}C{O}_3$
$m$-moles of $N{a}_{2}C{O}_3=m$-moles of $H_{2}S{O}_4$
$=20\times 0.1=2$
$m$-moles of $N{a}_{2}C{O}_3$ in $500mL$ solution
$=\frac{500}{25}\times 2=40$
wt. of $N{a}_{2}C{O}_3=40\times 106\times 10^{-3}=4.24g$
$\%$ of $N{a}_{2}C{O}_3=\frac{4.23}{5}\times 100=84.8$
$\therefore \%$ of $N{a}_{2}S{O}_4=100-84.4=15.2$