Question

An aqueous solution containing $0.10$ g $KI{O}_3$ (formula weight $=214.0$) 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. $0.0623$
• B. $0.126$
• C. $1.123$
• D. None of the above

Answer 1

The correct option is A $0.0623$

The reaction is as follows:

$KI{O}_3+5KI\to K_{2}O+3I_2$
Moles of $KI{O}_3=\frac{0.1}{214}$
Moles of iodine liberated $=3\times \frac{0.1}{214}$

$2N{a}_2{S}_2{O}_3+I_2\to 2NaI+N{a}_2{S}_4{O}_6$

Moles of $N{a}_2{S}_2{O}_3$ required $=3\times \frac{0.1}{214}\times 2$

Molarity$=\frac{\text{Number of moles}}{{\text{Volume}}_{\text{mL}}}\times 1000$ $=3\times \frac{0.1}{214}\times 2\times \frac{1}{45}\times 1000$ $=0.0623$