Question

$480g$ of $KCl{O}_3$ are dissoved in conc. $HCl$ and the solution was boiled. Chlorine gas evolved in the reaction was then passed through a solution of $KI$ and liberated iodine was titrated with $100mL$ of hypo. $12.3mL$ of same hypo solution required $24.6mL$ of $0.5N$ iodine for complete neutralization. Calculate % purity of $KCl{O}_3$ sample.

Answer 1

Chemical reactions for the given equation will be:

$2KCl{O}_3+12HCl\to 2KCl+6H_{2}O+6C{l}_2$

$C{l}_2+2KI\to 2KCl+I_2$

$I_2+\underset{\left(hypo\right)}{2N{a}_2{S}_2{O}_3}\to N{a}_2{S}_4{O}_6+2NaI$

For $12.3$ml hypo solution,

$N_1{V}_1=N_2{V}_2$ ($N_1{V}_1\to$ normality & volume of hypo solution, $N_2{V}_2\to$ normality & volume of Iodine)

$N_1\times 12.3=0.5\times 24.6$

$N_1=\frac{0.5\times 24.6}{12.3}$

$N_1=0.25$N

Equivalents of $N{a}_2{S}_2{O}_3=\underset{volumeremained}{87.7\times {10}^{-3}}\times \underset{normality}{0.25}=0.02192$ (eq)

$2$ equivalents $\left(N{a}_2{S}_2{O}_3\right)\to 1$ eq. of $I_2$

$\therefore 0.02192$ eq. of $N{a}_2{S}_2{O}_3\to \frac{1}{2}\times 0.02192=0.01096$

from the reaction, equivalents of $C{l}_2=$ equivalents of $I_2$

$=0.01096$

Also, $3\times$(eq of $C{l}_2$)$=1$ eq. of $KCl{O}_3$

$\therefore 0.01096$ eq. $C{l}_2=\frac{0.01096}{3}=3.65\times {10}^{-3}$ eq. of $KCl{O}_3$

moles of $KCl{O}_3=3.65\times {10}^{-3}\times 5=0.01825$ mol

Mass of $KCl{O}_3=\underset{\left(molarmass\right)}{122.55}\times \underset{(no.ofmoles)}{0.01825}$

$=2.23$g of $KCl{O}_3$

$\%$ $KCl{O}_3=\frac{weight\times 100}{givenwt.}=\frac{2.23}{480}\times 100=0.464\%$.