Question

A 25-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?

Answer 1

Answer: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)

$25mL*\frac{0.0104mmol IO3}{1mL} *\frac{3 mmol I_{2} }{1mmol IO3}*\frac{2mmol S2O3}{1mmol I_{2} }$

There is 1.56 mol S2O3

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

[Na2S2O3]=  $\frac{1.56}{17.27mL}$

[Na2S2O3]=0.09M