Question

$5mL$ of $8N$ nitric acid, $4.8mL$ of $5N$ hydrochloric acid and a certain volume of $17M$ sulphuric acid are mixed together and made up to $2litres.30mL$ of this acid mixture exactly neutralise $42.9mL$ of sodium carbonate solution containing $onegram$ of $N{a}_{2}C{O}_3\cdot 10H_{2}O$ in $100mL$ of water. Calculate the amount (in $grams$) of the sulphate ions in solution.

Answer 1

Molecular mass of $N{a}_{2}C{O}_3\cdot 10H_{2}O=286$
Equivalent mass of $N{a}_{2}C{O}_3\cdot 10H_{2}O=\frac{286}{2}=143$.
$100mL$ solution of sodium carbonate contains $=1g$
$1000mL$ solution of sodium carbonate contains $=10g$
Normality of the solution $=\frac{10}{143}$
Applying the formula,
Normality of acid solution $\times$ its volume
$=$ Normality of sodium carbonate solution $\times$ its volume,
Normality of the acid solution $=\frac{10\times 42.9}{143\times 30}=0.1$
Let $VmL$ be the volume of $H_{2}S{O}_4$ taken.
$8\times 5+4.8\times 5+34\times V=0.1\times 2000$
$V=4mL$
Amount of $S{O}_4^{2-}=\frac{Normality\times Eq.mass\times Volume}{1000}$
$=\frac{34\times 48\times 4}{1000}=6.528g$.