Question

The Fe³⁺ concentration is determined iodometrically by adding an excess of KI (20.0 ml of a 0.08 M) solution to 15.00 ml of an unknown Fe3+ solution.

2Fe³⁺ + 3I^- <> 2Fe²⁺ + I₃^-

The triiodide is titrated to a starch endpoint with a S2O32- solution. It required 8.20 ml of a 0.10 M thiosulfate solution. What is the concentration of the Fe3+ solution?

I₃^- + 2S₂O₃^2- <> I^- + S₄O₆^2-

Please answer only if you are sure about the answer. Thank you.

Answer 1

We need to solve this question by back calculation method.

In second step, the triiodide is titrated to a starch endpoint with a S2O32- solution.

Volume of thiosulfate solution used = 8.20 ml = 0.00820 L

Concentration of thiosulfate solution = 0.10 M

Moles of thiosulfate solution used = volume * concentration = 0.00820*0.10 = 0.00082 mol

In the reaction equation of triiodide with thiosulfate, one mole of triiodide reacts with 2 moles of thiosulfate.

So, for reaction of 0.00082 mol of thiosulfate solution, triiodide required = 0.00082/2 = 0.00041 mol

Now we come back to first step of the reaction, where Fe3+ is reacted with excess of KI.

As triiodide reacted in 2nd step is 0.00041 mol, this means that 0.00041 mol of I3- would have formed in 1st step.

In the reaction equation of Fe3+ with KI, 2 moles of Fe3+ gives 1 mole of I3-.

So, Fe3+ required in 1st step = 0.00041*2 = 0.00082 mol

Volume of Fe3+ used = 15.00 ml = 0.015 L

Concentration of Fe3+ used = moles/volume = 0.00082/0.015 = 0.055 M

Concentration of Fe3+ solution = 0.055 M