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Grade upto college level Physical Chemistry

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 HCI. 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.

Profile image of Amit Saxena
12 Years agoGrade upto college level
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1 Answer

Profile image of Navjyot Kalra
12 Years ago
Write the reactions taking place, balance them and equate moles of I2 and Na22O3.
KIO3 + 5KI → 3K2O + 3I2
i.e., 2.I5+ + 10e- → I02
2I- → I02 + 2e-
Now liberated I2 reacts with Na2S2O3
I2 + 2e- → 2I-
2S2O32- →S3O62- + 2e-
∴millimole ratio of I2 : S2O3 1 : 2
Thus, m mole of I2 liberated
= m mole of Na2S4O6 used * 1/2 = 45 * M * 1/2
[M is molarity of thiosulphate]
Also m mole of KIO3 = 0.1/214 * 1000
Now m mole ratio of KIO3 : I2 = 1 : 3
Thus, (0.1/214) *1000/(45M/2) = 1/3
M = 0.1 *100 *3 *2/214 *45 = 0.062