Iodimetry using iodate-thiosulfate

Potassium iodate is a good primary standard: it is available in pure form, it does not react with the air, pick up water from the atmosphere or dry out on storage, and it reacts quickly and to completion.

Pipette out 20.0 mL aliquots of potassium iodate solution into each of three conical flasks.

Acidify each flask by adding about 20 mL of dilute sulfuric acid.

Add about 1 g of potassium iodide Add about 1 g of potassium iodide. This is most conveniently done by adding 10 mL of a 10% KI solution. The iodate oxidises the iodide to iodine, while itself being reduced to iodine.

The thiosulfate will reduce the iodine to iodide: I2 + 2e– → 2I– 2I– → I2 + 2e– 2IO3– + 12H+ + 10e– → I2 + 6H2O 10I– + 2IO3– + 12H+ → 6I2 + 6H2O 5I– + IO3– + 6H+ → 3I2 + 3H2O Simplifying We titrate the iodine liberated (‘freed’ – ie formed) against sodium thiosulfate solution. The thiosulfate will reduce the iodine to iodide: I2 + 2e– → 2I– 2S2O32– → S4O62- + 2e– I2 + 2S2O32– → 2I– + S4O62-

Colourless thiosulfate is added to the dark brown iodine solution. The colour slowly disappears.

When the solution is a pale yellow colour, add a few drops of starch solution. The iodine will turn blue-black. Keep on adding thiosulfate until the blue colour disappears and the solution is colourless.

20.0 mL of iodate. Add 20 mL acid And 10 mL of 10% KI solution. I2 forms. Titrate in thiosulfate to reduce I2 to I–. End-point colourless Colour fades Add starch

When the method described was used, an average of 29. 73 mL of 0 When the method described was used, an average of 29.73 mL of 0.100 mol L–1 thiosulfate was required to react with the iodine liberated by 10.0 mL of a potassium iodate solution. What is the concentration of the iodate solution? 1 Find the amount of thiosulfate used V(S2O32–) = 29.73 mL c(S2O32–) = 0.100 mol L–1 = 29.73  10–3 L 0.100 mol L–1 29.73  10–3 L n(S2O32–) = cV =  = 2.973  10–3 mol

2 Use the equations for the reactions occurring to determine the amount of iodate present in the 10.0 mL sample 5I– + IO3– + 6H+ → 3I2 + 3H2O I2 + 2S2O32– → 2I– + S4O62- 1 IO3– IO3– 3 I2 I2 1 I2 I2 2 S2O32– S2O32– 3 Write mole ratios for each equations separately, and multiply them together. Unknown on top n(IO3–) n(I2)  =  n(I2) n(S2O32–) n(IO3–) 1 Simplify Known on the bottom = n(S2O32–) 6 n(IO3–) Rearrange = n(S2O32–) 6

n(S2O32–) = 2.973  10–3 mol 2.973  10–3 mol n(S2O32–) n(IO3–) = 6 = 6 = 4.955  10–4 mol 4.955  10–4 mol V(IO3–) = 10.0 mL = 10.0  10–3 L 10.0  10–3 L n c(IO3–) = V = = 0.0496 mol L–1 The concentration of the potassium iodate solution is 0.0496 mol L–1