Question

$0.50g$ of a mixture of $K_{2}C{O}_3$ and $L{i}_{2}C{O}_3$ requires $30mL$ of a $0.25NHCl$ solution for neutralisation. What is the percentage composition of the mixture?

Answer 1

$\underset{\underset{Eq.mass69}{138}}{K_{2}C{O}_3}+\underset{\underset{36.5}{2\times 36.5}}{2HCl}\to 2KCl+H_{2}O+C{O}_2$
$\underset{\underset{Eq.mass37}{74}}{L{i}_{2}C{O}_3}+\underset{\underset{36.5}{2\times 36.5}}{2HCl}\to 2LiCl+H_{2}O+C{O}_2$
Let $xg$ of $K_{2}C{O}_3$ be present in the mixture.
Mass of $L{i}_{2}C{O}_3=(0.50-x)$
No. of $g$ equivalents of $K_{2}C{O}_3=\frac{x}{69}$
No. of $g$ equivalents of $L{i}_{2}C{O}_3=\frac{(0.50-x)}{37}$
No. of $g$ equivalents in $30mL$ of $0.25NHCl$
$=\frac{Normality\times Volume}{1000}=\frac{0.25\times 30}{1000}=\frac{3}{400}$
At equivalence point,
$\frac{x}{69}+\frac{(0.50-x)}{37}=\frac{3}{400}$
So, $x=0.48$
$K_{2}C{O}_3=0.48g$; or $96$%
$L{i}_{2}C{O}_3=0.02g$; or $4$%.