Question

Titration of the I2 produced from .1045g of primary standard KIO3 required 30.72ml of sodium thiosulfate

Calculate the concentration of the Na2S2O3

Answer 1

Hi!

The first step for this problem is to determine the stoichiometric ratio in which the KIO3 and Na2S2O3 react to form the I2 product:

It turns out that this reaction is a little complicated. Here is a detailed description of the titration reactions. But, in brief:

For every 1 mole of IO3, you get 3 moles of I3

For every 1 mole of I3 you require 2 moles of S2O3

Thus, for every 1 mole of KIO3 that are present you require 6 moles of Na2S2O3

So, lets start by converting 0.1045g of KIO3 to moles:

0.1 045g × 214,-= 4.88 × 10-4 moles

Next, we multiply this by 6 to determine the required number of moles of Na2S2O3:

4.88 × 10-4-6 = 0.0029m○les

Finally, we divide 0.0029 moles of Na2S2O3 by the volume (in Liters) to get the final concentration:

0.0029moles 0.09Molar 0.03072L

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