Question

An aqueous solution containing $0.10\text{g}KI{O}_3$ (formula weight $=214$) was treated with excess of $KI$ solution. The solution was acidified with $HCl$. The liberated $I_2$ consumed $45.0\text{mL}$ of thiosulphate solution to decolourise the blue strach-iodine complex. Calculate the molarity of the sodium thiosulphate solution. (Report your answer upto two decimal only)

Answer 1

The reaction that has taken place to liberate $I_2$ is:
$K\stackrel{+5}{I}{O}_3+5K\stackrel{-1}{I}+6HCl\to 6KCl+3H_{2}O+3\stackrel{0}{I_2}$
i.e. $2\stackrel{+5}{I}+10e^{-}\to I_{2}2\stackrel{-1}{I}\to I_2+2e^{-}$

The liberated $I_2$ reacts with $N{a}_2{S}_2{O}_3$ as follows:
$2N{a}_2{\stackrel{+2}{S}}_2{O}_3+{\stackrel{0}{I}}_2⟶N{a}_2{\stackrel{+5/2}{S}}_4{O}_6+2Na\stackrel{-1}{I}$

The millimoles ratio of $I_2:N{a}_2{S}_2{O}_3$ is $1:2$
$\therefore \text{millimoles of}{I}_2\text{liberated}=\frac{\text{millimoles of}N{a}_2{S}_2{O}_3\text{liberated}}{2}=\frac{45}{2}\times M$
[where $M$ is molarity of $S_2{O}_3^{2-}$]
Also millimoles of $KI{O}_3=\frac{0.1}{214}\times 1000$
Now, millimoles ratio of $KI{O}_3:I_2$ is $1:3$
Thus,
$\frac{\frac{0.1}{214}\times 1000}{\frac{45}{2}M}=\frac{1}{3}$

$\Rightarrow M=\frac{0.1\times 1000\times 2\times 3}{214\times 45}=0.062\text{M}$