Question

$5 \,g$ sample contain only $Na_2CO_3$ and $Na_2SO_4$. This sample is dissolved and the volume made up to $500 \,mL. 25 \,mL$ of this solution neutralizes $20 \,mL$ of $0.1 \,M \,H_2SO_4$. Calculate the percentage of $Na_2SO_4$ in the sample.

Answer 1

Only $Na_2CO_3$ reacts with $H_2SO_4$;
$Na_2CO_3 + H_2SO_4 \to Na_2SO_4+ H_2CO_3$
$m$-moles of $Na_2CO_3 = m$-moles of $H_2SO_4$
$= 20 \times 0.1 = 2 $
$m$-moles of $Na_2CO_3$ in $500 \,mL$ solution
$ = \frac{500}{25} \times 2 = 40$
wt. of $Na_2CO_3 = 40 \times 106 \times 10^{-3} = 4.24 \,g$
$\%$ of $Na_2CO_3 = \frac{4.23}{5} \times 100 = 84.8$
$\therefore \% $ of $Na_2SO_4 = 100 - 84.4 = 15.2$