Question

An aqueous solution containing 0.10 g KIO₃ (formula weight = 214.0) was treated with an excess of KI solution. The solution was acidified with HCl. The liberated I₂ consumed 45.0 mL of thiosulphate solution to decolourise the blue starch-iodine complex. Calculate the molarity of the sodium thiosulphate solution.

Answer 1

On reaction between KIO₃ and KI, I₂ is liberated

KIO₃ + 5KI  → 2K₂O + 3I₂

This liberated I₂ reacts with Na₂S₂O₃ 

2NaS₂O₃ + I₂ →  1NaI + Na₂S₄O₆

One mole KIO₃ liberates three moles of I₂

Mole of KIO₃ = 0.10 g/mol. wt.(214) = 0.000467

Liberated moles of I₂ = 3 x 0.000467

One mole of I₂ requires two moles of Na₂S₂O₃ for complete decolourisation of I₂