Redox titration with iodine-thiosulfate to determine the amount of iodine

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Thiosulfate (S2O32-) is a strong reducing agent and so can be used in titrations to determine the amount of oxidising agent present:

Reduction: 2 S2O32- -> S4O62- + 2 e–

Unfortunately both S2O32- and S4O62- are colourless and there isn’t any suitable indicator for either. This means that, for a useful titration method, the other reaction involved in the tiration must contain something that is coloured, or has an indicator, so that we can see the end-point.

A classic example for A Level Chemistry is iodine (I2):

Reduction: 2 S2O32- -> S4O62- + 2 e–

Oxidation: I2 + 2 e– -> 2 I–

Overall: 2 S2O32- + I2 -> S4O62- + 2 I–

Imagine you have thiosulfate in the burette and an iodine solution in the conical flask. Initially, the iodine solution would be brown due to the presence of aqueous I2. As more thiosulphate is added, the iodine reacts and the brown colour fades to a very pale yellow (when only a little iodine remains) and then completely colourless (at the end point).

To help us see this end-point better, some starch is added when the solution is a very pale yellow and this causes the mixture to become deeply blue-black due to the formation of the iodine-starch complex. When the end-point is finally reached, the mixture becomes colourless. 

 

Question: 15 small crystals of solid I2 are dissolved into 250 cm3 of distilled water. A 25 cm3 sample is titrated with 24.9 cm3 of 0.01 mol/dm3 Na2S2O3 . Determine the average mass of each I2 crystal in mg.

Reaction: 2 S2O32- + I2 -> S4O62- + 2 I–

 

Step 1

We know that moles = concentration x volume:

n S2O32- = 0.01 x (24.9 / 1000) =  2.49 x 10-4 mol

Step 2

The overall balanced symbol equation tells us that S2O32- reacts with I2 in a 2:1 ratio.

n I2 = 0.5 x 2.49 x 10-4 = 1.25 x 10-4 mol

Step 3

The total is 10x bigger than the sample (i.e., 250 / 25):

n I2 = 10 x 1.25 x 10-4 = 1.25 x 10-3 mol

Step 4

We can now work out the total mass of dissolved I2 by multiplying by its Mr (253.8)

m I2 = 253.8 x 1.25 x 10-3 = 0.32 g

Step 5

We can now work out the average mass of each I2 crystal by dividing by the number of crystals (15).

m I2 = 0.32 / 15 = 0.021 g = 21 mg

 

(It turns out that iodine-thiosulfate titrations are actually FAR more useful than this question suggests. In fact, we can use an iodine-thiosulfate titration to determine the amount of a much wider range of oxidising agents – i.e., not just iodine!)