An aqueous solution containing 1.07 g KIO₃ was treated with an excess of KI solution. The solution was acidified with HCl. The liberated I₂ consumed 50 ml of thiosulphate solution to decolourize the b

Question

An aqueous solution containing 1.07 g KIO₃ was treated with an excess of KI solution. The solution was acidified with HCl. The liberated I₂ consumed 50 ml of thiosulphate solution to decolourize the blue starch-iodine complex. The molarity of the sodium thiosulphate solution is x M, then the value of 1000x is (K = 39, I = 127)

TestHub · Accepted Answer

KIO₃ = 1.07/214 = 5 mmol. IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O gives 15 mmol I₂. With I₂ + 2S₂O₃²⁻, thiosulphate = 30 mmol in 50 ml, i.e. x = 0.6 M. 1000x = 0600.

Chemistry - Redox Reactions Question with Solution | TestHub

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