Question

$1.03g$ mixture of sodium carbonate and calcium carbonate require $20mLNHCl$ for complete neutralisation. Calculate the percentage of sodium carbonate and calcium carbonate in the given mixture.

Answer 1

$\underset{106}{N{a}_{2}C{O}_3}+\underset{2\times 36.5}{2HCl}\to 2NaCl+H_{2}O+C{O}_2$
Eq. mass $\underset{1geq.}{53}\underset{1geq.}{36.5}$
$\underset{100}{CaC{O}_3}+\underset{2\times 36.5}{2HCl}\to CaC{l}_2+H_{2}O+C{O}_2$
Eq. mass $\underset{1geq.}{50}\underset{1geq.}{36.5}$
Let $xgCaC{O}_3$ be present in the mixture.
Mass of $N{a}_{2}C{O}_3$ in the mixture $=(1.03-x)g$
No. of $g$ equivalent of $CaC{O}_3=\frac{x}{50}$
No. of $g$ equivalents of $N{a}_{2}C{O}_3=\frac{(1.03-x)}{53}$
No. of $g$ equivalents in $20mLNHCl=\frac{Normality\times Vol.}{1000}$
$=\frac{1\times 20}{1000}=\frac{1}{50}$
At equivalence point,
No. of $g$ equivalent of $CaC{O}_3+$ No. of $g$ equivalent of $N{a}_{2}C{O}_3=$ No. of gram equivalent of $HCl$
$\frac{x}{50}+\frac{1.03-x}{53}=\frac{1}{50}$
or $x=0.50$
$CaC{O}_3=0.50g$, % $CaC{O}_3=\frac{0.50}{1.03}\times 100=48.54$
$N{a}_{2}C{O}_3=0.53g$, % $N{a}_{2}C{O}_3=\frac{0.53}{1.03}\times 100=51.46$.