Question

2.84 g of $KCl{O}_3$ are dissolved in cone. 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 100 mL of hypo. 12.3 mL of same hypo solution required 24.6 mL of 0.5 N iodine for complete neutralization. Calculate % purity of $KCl{O}_3$ sample.

Answer 1

$2.84g$ of $KCl{O}_3$

$\begin{array}{ccccccccccc}1& \to & {KclO}_3& +& 6Hcl& \to & Kcl& +& 3{cl}_2& +& 3H_{2}O\\ & & n_f=6& & & & & & n_f=2& & \\ 2& \to & 2KI& +& {cl}_2& \to & 2Kcl& +& I_2& & \\ & & & & n_f=2& & & & n_f=2& & \\ 3& \to & N{a}_2{S}_2{O}_3& +& I_2& \to & 2NaI& +& {Na}_2{S}_4{O}_6& & \\ & & n_f=1& & n_f=2& & & & & & \end{array}$

From $3^{rd}$ equation

equivalent of hypo = equivalent of $I_2$

$\Rightarrow N_{hypo}\times V_{hypo}=N_{I_2}\times V_{I_2}$

$\Rightarrow N_{hypo}\times 12.3=0.5\times 24.6$

$\Rightarrow N_{hypo}=1$

Now, equivalent of $Kcl{O}_3=$ equivalent of $c{l}_2=$ equivalent of $I_2=$ equivalent of hypo

$\therefore$ equivalent of $Kcl{O}_3=$ equivalent of hypo

$\Rightarrow N_{Kcl{O}_3}\times Kcl{O}_3=N_{N{a}_2{S}_2{O}_3}\times V_{N{a}_2{S}_2{O}_3}$

$\Rightarrow n_f\times M_{Kcl{O}_3}=1\times 100\times {10}^{-3}$

$\Rightarrow n_f\times moles=100\times {10}^{-3}$

$\Rightarrow$ No. of moles $=100/6\times {10}^{-3}$

$\therefore$ weight $=\frac{100}{6}\times {10}^{-3}\times 158.11=2.63gm$

$\therefore \%parity=\frac{2.63}{2.84}\times 100\simeq 92.7\%$