2 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.
3 Pipette out 20.0 mL aliquots of potassium iodate solution into each of three conical flasks.
4 Acidify each flask by adding about 20 mL of dilute sulfuric acid.
5 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.
6 The thiosulfate will reduce the iodine to iodide: I2 + 2e– → 2I– 2I– → I e–2IO3– H e– → I H2O10I– + 2IO3– H+ → 6I H2O5I– + IO3– + 6H+ → 3I H2OSimplifyingWe titrate the iodine liberated (‘freed’ – ie formed) against sodium thiosulfate solution.The thiosulfate will reduce the iodine to iodide:I e– → 2I–2S2O32– → S4O e–I S2O32– → 2I– + S4O62-
7 Colourless thiosulfate is added to the dark brown iodine solution. The colour slowly disappears.
8 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.
9 20.0 mL of iodate.Add 20 mL acidAnd 10 mL of 10% KI solution. I2 forms.Titrate in thiosulfate to reduce I2 to I–.End-point colourlessColour fadesAdd starch
10 When the method described was used, an average of 29. 73 mL of 0 When the method described was used, an average of mL of 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 usedV(S2O32–) = mL c(S2O32–) = mol L–1= 10–3 L0.100 mol L–129.73 10–3 Ln(S2O32–)=cV==2.973 10–3 mol
11 2 Use the equations for the reactions occurring to determine the amount of iodate present in the mL sample5I– + IO3– + 6H+ → 3I H2OI S2O32– → 2I– + S4O62-1IO3–IO3–3I2I21I2I22S2O32–S2O32–3 Write mole ratios for each equations separately, and multiply them together.Unknown on topn(IO3–)n(I2)=n(I2)n(S2O32–)n(IO3–)1SimplifyKnown on the bottom=n(S2O32–)6n(IO3–)Rearrange=n(S2O32–)6