iodimetry using iodate- thiosulfate. potassium iodate is a good primary standard: it is available in...
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![Page 1: 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](https://reader035.vdocuments.us/reader035/viewer/2022062313/56649c875503460f9493f5a4/html5/thumbnails/1.jpg)
Iodimetry using iodate-thiosulfate
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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.
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Pipette out 20.0 mL aliquots of potassium iodate solution into each of three conical flasks.
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Acidify each flask by adding about 20 mL of dilute sulfuric acid.
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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.
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2I– → I2 + 2e–
2IO3– + 12H+ + 10e– → I2 + 6H2O
10I– + 2IO3– + 12H+ → 6I2 + 6H2O
We titrate the iodine liberated (‘freed’ – ie formed) against sodium thiosulfate solution.
The thiosulfate will reduce the iodine to iodide:
I2 + 2e– → 2I–
2S2O32– → S4O6
2- + 2e–
I2 + 2S2O32– → 2I– + S4O6
2-
Simplifying5I– + IO3– + 6H+ → 3I2 +
3H2O
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Colourless thiosulfate is added to the dark brown iodine solution.
The colour slowly disappears.
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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.
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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–.
Colour fades
Add starch
End-point colourless
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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(S2O3
2–) = 0.100 mol L–1
= 29.73 10–3 Ln(S2O3
2–) cV=
=
=
0.100 mol L–
1
29.73 10–3 L
2.973 10–3 mol
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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– + S4O6
2-
n(IO3–)
n(I2)n(I2)
n(S2O32–)
n(S2O32–)
n(IO3–)
=
=
=
1
1
16
6
3
2
IO3– I2
I2 S2O32–
3 Write mole ratios for each equations separately, and multiply them together.
Unknown on top
Known on the bottom
IO3– I2
I2 S2O32–
Simplify
n(S2O32–)
n(IO3–) Rearrange
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n(S2O32–) = 2.973 10–3 mol
n(IO3–)
n(S2O32–)
2.973 10–3 mol
6
4.955 10–4 mol
=
=
=
6
V(IO3–) = 10.0 mL
= 10.0 10–3 L c(IO3
–) =
=
=
nV
10.0 10–3 L
4.955 10–4 mol
0.0496 mol L–1
The concentration of the potassium iodate solution is 0.0496 mol L–1