2Potassium 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.
3Pipette out 20.0 mL aliquots of potassium iodate solution into each of three conical flasks.
4Acidify each flask by adding about 20 mL of dilute sulfuric acid.
5Add 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.
6The 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-
7Colourless thiosulfate is added to the dark brown iodine solution. The colour slowly disappears.
8When 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.
920.0 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
10When 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
112 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