Topic 11 · page 18 of 21
Core Practical 9a: iodine and propanone by titration, method
Mix propanone, acid and iodine, take samples at timed intervals, stop the reaction in each with sodium hydrogencarbonate, and titrate the iodine left with thiosulfate. The titre is proportional to the iodine concentration.
Read first: Iodination of propanone: data and predictions, Choosing a technique to follow a reaction
Aim and chemistry 11.1211.6(i)
The aim is to follow the rate of the acid-catalysed reaction between propanone and iodine by a titrimetric method, and to find the order with respect to iodine. Page 11 gave the reaction.
Iodine is the species followed. Samples of the mixture are taken at known times. Each sample is quenched, which stops the reaction, and the iodine left is titrated with sodium thiosulfate:
One mole of iodine reacts with two moles of thiosulfate. The titre of thiosulfate is therefore proportional to the amount of iodine in the sample, and so to [I₂] at the time of sampling.
The method 11.12
PracticalFollowing the reaction
- Measure into a conical flask 25.0 cm³ of propanone solution (1.00 mol dm⁻³), 25.0 cm³ of sulfuric acid (1.00 mol dm⁻³) and 25.0 cm³ of water.
- Add 25.0 cm³ of iodine solution (0.0200 mol dm⁻³), swirl, and start the stopwatch. The total volume is 100.0 cm³.
- In a series of conical flasks, put about 25 cm³ of sodium hydrogencarbonate solution (0.50 mol dm⁻³), an excess.
- At about 4 minutes, pipette 10.0 cm³ of the reaction mixture into one of the flasks, swirling at once. Record the time at which the sample is added.
- Titrate the iodine left with sodium thiosulfate solution (0.0100 mol dm⁻³). When the solution is pale yellow, add a few drops of starch and continue until the blue-black colour just disappears. Record the titre.
- Repeat steps 4 and 5 every 4 minutes, for about 24 minutes.
- Repeat the whole experiment with a different concentration of propanone, and then with a different concentration of acid, keeping the total volume at 100.0 cm³ by adjusting the volume of water.
Why each step is there 11.12
| Step | Reason |
|---|---|
| Iodine added last, stopwatch started at once | the reaction begins when all three reagents are together |
| Sodium hydrogencarbonate in excess | it neutralises the H⁺ catalyst, so the reaction stops; the titre shows [I₂] at the time of sampling |
| Time recorded at the quench | the reaction stops then, not when the titration is done |
| Starch added near the end point | starch gives an intense blue-black colour with iodine, but if it is added early it holds iodine and makes the end point sluggish |
| Constant total volume | concentrations stay proportional to the volumes you measured out |
| Constant temperature (a water bath or a quiet room) | k changes with temperature |
| Propanone and acid in large excess | their concentrations hardly change, so only iodine changes (see below) |
Checking the quantities 11.6(i)
Worked exampleConcentrations in the mixture
- Dilution: each reagent is in 25.0 cm³ of a total of 100.0 cm³, so each is diluted by a factor of 4.
- [CH₃COCH₃] = 1.00 ÷ 4 = 0.250 mol dm⁻³
- Each H₂SO₄ gives two H⁺ (taken as fully dissociated), so [H⁺] = 2 × 1.00 ÷ 4 = 0.500 mol dm⁻³
- [I₂] = 0.0200 ÷ 4 = 0.00500 mol dm⁻³
Worked exampleThe first titre
Calculate the thiosulfate titre for a 10.0 cm³ sample at the start, when [I₂] = 0.00500 mol dm⁻³.
- n(I₂) = 0.00500 × 0.0100 = 5.00 × 10⁻⁵ mol
- n(S₂O₃²⁻) = 2 × 5.00 × 10⁻⁵ = 1.00 × 10⁻⁴ mol
- V = n ÷ c = 1.00 × 10⁻⁴ ÷ 0.0100 = 0.0100 dm³ = 10.0 cm³
If all of the iodine reacted, 0.00500 mol dm⁻³ of propanone would be used, which is 0.00500 ÷ 0.250 = 2.0% of the propanone. Acid is not used up at all, because it is a catalyst. So the concentrations of propanone and H⁺ are effectively constant during a run, and the only concentration that changes is [I₂]. That is why the graph of titre against time shows the order with respect to iodine directly (page 19).
Check yourself
- Calculate the concentrations of propanone, H⁺ and iodine in the reaction mixture in the method above.
- Explain why each sample is run into sodium hydrogencarbonate solution.
- Why must the time be recorded at the moment the sample is quenched?
- A 10.0 cm³ sample contains iodine at a concentration of 0.00250 mol dm⁻³. Calculate the volume of 0.0100 mol dm⁻³ sodium thiosulfate it needs.
- Why is starch added only near the end point?
- What percentage of the propanone is used up if all the iodine reacts? Why does this matter for the experiment?
Answers cover these until you have tried the questions
- Each solution is diluted from 25.0 cm³ to 100.0 cm³, a factor of 4. [CH₃COCH₃] = 1.00 ÷ 4 = 0.250 mol dm⁻³. [H⁺] = 2 × 1.00 ÷ 4 = 0.500 mol dm⁻³. [I₂] = 0.0200 ÷ 4 = 0.00500 mol dm⁻³.
- Sodium hydrogencarbonate neutralises the H⁺ catalyst, so the reaction stops. The titre then gives the iodine concentration at the time the sample was taken. It is in excess so that all the acid is neutralised.
- The reaction continues until the sample is quenched. The titre measures the iodine at that moment, so that is the time that belongs to the titre. The titration itself takes place later, on a mixture in which nothing more is happening.
- n(I₂) = 0.00250 × 0.0100 = 2.50 × 10⁻⁵ mol. n(S₂O₃²⁻) = 2 × 2.50 × 10⁻⁵ = 5.00 × 10⁻⁵ mol. V = 5.00 × 10⁻⁵ ÷ 0.0100 = 5.00 × 10⁻³ dm³ = 5.00 cm³.
- Starch forms a strong blue-black complex with iodine. If it is added at the start, when there is a lot of iodine, some of the iodine stays bound to the starch and the end point is slow and unclear. Added when the solution is pale yellow, it shows the last of the iodine sharply.
- 0.00500 ÷ 0.250 × 100 = 2.0%. So the propanone concentration is effectively constant during a run (and the acid is not used up), which means only the iodine concentration changes and the graph shows the order with respect to iodine.