Question

An aqueous solution containing 0.10 g $KI{O}_3$ (formula wt. = 214.0) was treated with an excess of KI solution. The solution was acidified with HCl. The liberated $I_2$ consumed 45.0 ml of thiosulphate solution to decolourise the blue starch - iodine complex. Calculate the molarity of the sodium thiosulphate solution.

• A. 0.006 M
• B. 0.020 M
• C. 0.062 M
• D. 0.163 M

Answer 1

The correct option is C 0.062 M

The reaction is as follows
$KI{O}_3+5KI\to K_{2}O+3I_2$
From the reaction we know 1 mole of $KI{O}_3$ gives 3 mole of $I_2$
$\text{Number of moles of}KI{O}_3=\frac{\text{Mass}}{\text{Molecular mass}}$
$n_{KI{O}_3}=\frac{0.10}{214}$
$n_{KI{O}_3}=4.67\times {10}^{-4}mol$
$\therefore \text{Number of moles of}{I}_2\text{liberated}=3\times 4.67\times {10}^{-4}$
$n_{I_2}=1.40\times {10}^{-3}mol$
$2N{a}_2{S}_2{O}_3+I_2\to 2NaI+N{a}_2{S}_4{O}_6$
$\text{Moles of}N{a}_2{S}_2{O}_3\text{required}=2\times 1.40\times {10}^{-3}mol$
$\text{Molarity of}N{a}_2{S}_2{O}_3=\frac{\text{Number of Moles}}{\text{Volume (L)}}$
$\text{Molarity of}N{a}_2{S}_2{O}_3=\frac{2.80\times {10}^{-3}}{0.045}=0.0623M$