Again Doubt in titrations!

An aqueous solution containing 0.15 g KIO3 ( formula weight =220.0) was less treated with an excess of KI solution. The solution was acidified with HCl . The liberated I2 consumed 50 ml of thiosulphate solution to decolorize the blue starch iodine complex. Please help me write my paper how calculate the molarity of Na2S2O3 ??

My solution is Redox change IO3- ---> I2 2IO3- + 12H+ + 10e- = I2 + 6H2O n factor of KIO3 = 5 Eq wt of KIO3 = 214/5 Gm eq of KIO3 = 0.10/(214/5) = Gm eq of I2 Let the molarity of Na2S2O3 = x N factor of Na2S2O3 = 1 (0.10 x 5)/(214) = (45/1000) x (x) x 1 or x = 0.052M But the solution is 0.062M What is wrong in my solution.

Note by Nick Yavorski
1 year, 3 months ago

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