Question

In a sample of sodium carbonate some sodium sulphate is also mixed. $1.25g$ of this sample is dissolved and the volume made up to $250mL$. $25mL$ of this solution neutralises $20mL$ of $\frac{N}{10}$ sulphuric acid. Calculate the percentage of sodium carbonate in the sample.

Answer 1

$25mL$ of sample solution neutralises
$=20mL\frac{N}{10}{H}_{2}S{O}_4$
$250mL$ of sample solution will neutralise
$=200mL\frac{N}{10}{H}_{2}S{O}_4$
$200mL\frac{N}{10}{H}_{2}S{O}_4\equiv 200mL\frac{N}{10}N{a}_{2}C{O}_3$ solution
Amount of $N{a}_{2}C{O}_{3}present=\frac{E\times N\times V}{1000}$
$=\frac{53\times 200}{10\times 1000}=1.06$
% of $N{a}_{2}C{O}_3$ in the sample $=\frac{1.06}{1.25}\times 100=84.8$.