In the most simple case, an iodine-thiosulphate titration can be used to determine the amount of iodine. However, the techique is actually a lot more powerful than that and can be used to determine the amount of a much broader range of oxidising agent.
The basic principle is that we use an oxidising agent to first convert an excess of I– into I2, we then use the iodine-thiosulphate titration to work out how much iodine is produced. We can then work backwards to work out how much of the oxidising agent was present.
Reaction 1: Any reaction that converts I– into I2….
Reaction 2: 2 S2O32- + I2 -> S4O62- + 2 I–
Here is a very straightforward example:
An excess of KI is added to a solution containing Cu2+ ions. The solution is then titrated with exactly 0.080 mol S2O32-. Calculate the moles of Cu2+ in the original solution.
Reaction 1: 2 Cu2+ + 4 I– –> 2 CuI + I2
Reaction 2: 2 S2O32- + I2 -> S4O62- + 2 I–
If the titration required 0.080 mol of S2O32- then there must have been 0.040 mol of I2 at the start of Reaction 2 and also at the end of Reaction 1 (since we’ve analysed the entire mixture, not just a sample). If 0.080 mol of I2 are formed in Reaction 1, then the must have been 0.040 mol of Cu2+ at the start!
You can probably think of quite a few ways that this question could now be made more complicated e.g., by giving pairs of concentration-volumes, asking for a mass of Cu to be determined, or perhaps an Mr of a soluble Cu2+ salt, etc…
| Example: A 0.82 g sample of CuCl2.xH2O is dissolved in distilled water and made into a 250 cm3 solution. A 100 cm3 sample is taken by pipette and an excess of KI is added to this sample. The resulting solution is then titrated with 19.2 cm3 of 0.1 mol/dm3 Na2S2O3. Determine the value of x to the nearest whole number. Reaction 1: 2 Cu2+ + 4 I– –> 2 CuI + I2 Reaction 2: 2 S2O32- + I2 -> S4O62- + 2 I– Step 1 We know that moles = concentration x volume: n S2O32- = 0.1 x (19.2 / 1000) = 1.92 x 10-3 mol Step 2 Reaction 2 tells us that S2O32- reacts with I2 in a 2:1 ratio, so we can work out how much I2 was made in the 10 cm3 sample. n I2 = 0.5 x 1.92 x 10-3 = 9.6 x 10-4 mol Step 3 Reaction 1 tells us that the ratio between I2 produced in the 10 cm3 sample and Cu2+ reacted is 1:2. n Cu2+ =2 x 1.92 x 10-3 = 1.92 x 10-3 mol Step 4 The total is 2.5x bigger than the sample (i.e., 250 / 100): n Cu2+ = 2.5 x 1.92 x 10-3 = 0.0048 mol Step 5 We have the mass (0.82 g) and moles (0.0048 mol) of the CuCl2.xH2O, so we can now work out its Mr : Mr CuCl2.xH2O = 0.82 / 0.0048 = 170.8 Step 6 As in other similar questions, we then subtract the Mr of the anhydrous salt (CuCl2 = 134.5) to find the Mr of xH2O. Mr xH2O = 170.8 – 134.5 = 36.3 The Mr of water is 18, so clearly x = 2 (to the nearest whole number). |
Note that Reaction 1 will change depending on which oxidising agent is being analysed. The reaction in the first step is often quite hard to predict as transition metals usually have variable oxidation states, and also many of the metal iodides are actually insoluble! You will generally be given Reaction 1 as part of the question.
In contrast, Reaction 2 will ALWAYS be the same, so it is perhaps worth remembering!