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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.

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Presentation on theme: "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."— 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. 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 2I – → I 2 + 2e – 2IO 3 – + 12H e – → I 2 + 6H 2 O 10I – + 2IO 3 – + 12H + → 6I 2 + 6H 2 O We titrate the iodine liberated (‘freed’ – ie formed) against sodium thiosulfate solution. The thiosulfate will reduce the iodine to iodide: I 2 + 2e – → 2I – 2S 2 O 3 2– → S 4 O e – I 2 + 2S 2 O 3 2– → 2I – + S 4 O 6 2- Simplifying 5I – + IO 3 – + 6H + → 3I 2 + 3H 2 O

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. I 2 forms. Titrate in thiosulfate to reduce I 2 to I –. Colour fades Add starch End-point colourless

10 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(S 2 O 3 2– ) = mL c(S 2 O 3 2– ) = mol L –1 =  10 –3 L n(S 2 O 3 2– ) cV = = = mol L –  10 –3 L  10 –3 mol 

11 2 Use the equations for the reactions occurring to determine the amount of iodate present in the 10.0 mL sample 5I – + IO 3 – + 6H + → 3I 2 + 3H 2 O I 2 + 2S 2 O 3 2– → 2I – + S 4 O 6 2- n(IO 3 – ) n(I2)n(I2) n(I2)n(I2) n(S 2 O 3 2– ) n(IO 3 – ) = = =   IO 3 – I2I2 I2I2 S 2 O 3 2– 3 Write mole ratios for each equations separately, and multiply them together. Unknown on top Known on the bottom IO 3 – I2I2 I2I2 S 2 O 3 2– Simplify n(S 2 O 3 2– ) n(IO 3 – ) Rearrange

12 n(S 2 O 3 2– ) =  10 –3 mol n(IO 3 – ) n(S 2 O 3 2– )  10 –3 mol  10 –4 mol = = = 6 V(IO 3 – ) = 10.0 mL = 10.0  10 –3 L c(IO 3 – ) = = = n V 10.0  10 –3 L  10 –4 mol mol L –1 The concentration of the potassium iodate solution is mol L –1


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