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Question

An aqueous solution containing 0.10 g KIO3 (formula wt. = 214.0) was treated with an excess of KI solution. The solution was acidified with HCl. The liberated I2 consumed 55 mL of thiosulphate solution to decolorize the blue starch-iodine complex. Calculate molarity of the sodium thiosulphate solution.

A
0.05 M
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B
0.25 M
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C
0.01 M
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D
0.05 M
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Solution

The correct option is D 0.05 M
KIO3+5KI3K2O+3I2
Moles of KIO3=0.10214.0=0.00047
Moles of I2 liberated =0.00047×3=0.00141
2Na2S2O3+I22NaI+Na2S4O6
Moles of Na2S2O3=2×0.00141=0.00282
Molarity =0.00282×100055=0.0512 M

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