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

Step-by-step 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*(0.0104mmol IO3)/(1mL) *(3 mmol I_(2) )/(1mmol IO3)*(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]=
`(1.56)/(17.27mL)`

[Na2S2O3]=0.09M