Question

A mixture containing $KCl$ and $NaCl$ was dissolved and total halide was determined by titration with silver nitrate. A sample weighing $0.3250g$ required $51mL$ of $0.1N$ solution. Calculate the percentage of each salt in the sample.

Answer 1

In the mixture of $KClandNaCl$ :

74.5 g of $KCl$ requires 170 g of $AgN{O}_3$ and

58.5 g of $NaCl$ requires 170 g of $AgN{O}_3$

And in $51mLof0.1N$ solution :

weight of $AgN{O}_3$ = $\frac{0.1\times 170\times 51}{1000}=0.867g$

In solution weighing $0.3250g$ let us consider $x$ gm $KCl$ and the remaining $0.325-x$ gm is $NaCl$.

$\therefore mole{s}_{NaCl}+mole{s}_{KCl}=mole{s}_{AgN{O}_3}$

$\therefore \frac{x}{74.5}+\frac{0.325-x}{58.5}=\frac{0.867}{170}$

$\therefore x=0.1240gm=weightofKClinsample$

$\therefore 0.201gm=weightofNaClinsample$

$percentageofKCl=\frac{0.124\times 100}{0.325}=38.19\%$

$percentageofNaCl=\frac{0.201\times 100}{0.325}=61.81\%$