The analysis of potassium iodate(V)

In this experiment we shall measure the purity of the sample of potassium iodate(V) that we made in the last experiment. A 100 cm³ standard solution of the impure, solid potassium iodate(V) is first made up and then 10 cm³ samples of this are reacted with an excess of 0.1 M potassium iodide solution and 1 M sulfuric acid solution in order to release iodine in the ratio shown in the following equation:

KIO₃(aq) + 5KI(aq) + 3H₂SO₄(aq) ⇒ 3I₂(aq) + 3H₂O(l) + 3K₂SO₄(aq)

The amount of iodine released in this reaction is then measured using a sodium thiosulfate titration. The titration result allows us to calculate the actual mass of potassium iodate(V) in the original sample and this can then be expressed as a percentage of the original impure mass. Watch the following video and then try to calculate the percentage purity of the sample. You can check your answer by clicking here. Normally at least two titrations would be done and the results averaged, but I have only done one to save time on the video.

Video - the analysis of potassium iodate(V)