Question

A sample of sodium carbonate contain sodium sulphate also. $1.5g$ of the sample is dissolved in water and volume raised to $250mL$. $25mL$ of this solution requires $20mL$ of $\frac{N}{10}{H}_{2}S{O}_4$ solution for neutralisation. Calculate percentage of sodium carbonate in the sample.

Answer 1

Only $N{a}_{2}C{O}_3$ will react with $H_{2}S{O}_4$.
Applying $\underset{\left(N{a}_{2}C{O}_3\right)}{N_1{V}_1}=\underset{\left(H_{2}S{O}_4\right)}{N_2{V}_2}$
$N_1\times 25=20\times \frac{1}{10}$
$N_1=\frac{20}{25\times 10}=0.08$
Eq. mass of $N{a}_{2}C{O}_3=\frac{Mol.mass}{2}=\frac{106}{2}=53$
Mass of $N{a}_{2}C{O}_3$ present in $250mL0.08N$ solution
$=\frac{N\times E\times V}{1000}=\frac{0.08\times 53\times 250}{1000}=1.06g$$
Percentage of $N{a}_{2}C{O}_3$ in the mixture $=\frac{1.06}{1.50}\times 100=70.67$.