A25-ml aliquot of a 0.0104 m kio3 solution is titrated to the stoichiometric point with 17.27 ml of a sodium thiosulfate, na2s2o3, solution. what is the molar concentration of the na2s2o3 solution?

0.09 M step by step in explanation

Explanation:

First We need to considerated the equation of titrated of the KIO3

Then:

IO3 +  5I-  +  6H+  ---> 3I2  +  2H2O

I2  + 2 S2O3  --> 2I- + S4O62-

Then We have that one mmol of IO3 produced 3 mmol of I2 and one mmol of I2 reacted with 2 mmol of S2O3( thiosulfate ion)

There is 1.56 mol S2O3

then we divide the result by 17.27mL to obtain the concentration of the thiosulphate solution

[Na2S2O3]=  

[Na2S2O3]=0.09M

explanation:

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