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Iodimetry using iodate-thiosulfate

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Presentation on theme: "Iodimetry using iodate-thiosulfate"— Presentation transcript:

1 Iodimetry using iodate-thiosulfate

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 H2O 10I– + 2IO3– H+ → 6I H2O 5I– + IO3– + 6H+ → 3I H2O Simplifying We 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 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

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 used V(S2O32–) = mL c(S2O32–) = mol L–1 =  10–3 L 0.100 mol L–1 29.73  10–3 L n(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 sample 5I– + IO3– + 6H+ → 3I H2O I S2O32– → 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

12 n(S2O32–) =  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 = = mol L–1 The concentration of the potassium iodate solution is mol L–1

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