# DOUBT in titrations!

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 ml of thiosulphate solution to decolorize the blue starch iodine complex. 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 Rudraksh Singh
1 year, 5 months ago

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@Prakhar Bindal , @Aniket Sanghi , @Shubham Dhull , @Harsh Shrivastava , @A E please guys help me out here ,,,

- 1 year, 5 months ago

- 1 year, 5 months ago

Thanks for tagging me. Let me try, then i will see ur solution, Then if it is different than urs, i will post it here.

- 1 year, 5 months ago

http://exxamm.com/QuestionBank/practice/1119367210 THIS is the link bro ,, but sorry it is not the full length paper

- 1 year, 5 months ago

have u tried it ??

- 1 year, 5 months ago

Hey this is iit 1998 problem ... Thanks for responding bro !!

- 1 year, 5 months ago