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Back titration of Vanadium with Iodine Starch complex.

Hey folks, I'm totally stuck on a lab report for uni. Have to get it done for after new year and I've just hit a mental block. I reduced a standard solution 2.2 vanadium solution (25ml) diluted to 50ml. Added 50ml 1M sulphuric acid and 1 g of potassium iodide. I then titrated the liberated Iodine with 0.1 M sodium thiosulphate using starch as an indicator. My titre was 13.1cm^3.

How do I work out the oxidation state of vanadium before titration?

Any help would be amazing! Fankooo!
Don't know if this helps but vanadium has 4 common oxidation states which can be distinguished by its colour:
yellow = +5
blue = +4
green = +3
violet = +2
Original post by Baq99
Hey folks, I'm totally stuck on a lab report for uni. Have to get it done for after new year and I've just hit a mental block. I reduced a standard solution 2.2 vanadium solution (25ml) diluted to 50ml. Added 50ml 1M sulphuric acid and 1 g of potassium iodide. I then titrated the liberated Iodine with 0.1 M sodium thiosulphate using starch as an indicator. My titre was 13.1cm^3.

How do I work out the oxidation state of vanadium before titration?

Any help would be amazing! Fankooo!


Not enough info in your question.

What does "standard solution 2.2 vanadium solution" mean?
Reply 3
Hi folks,

I think I have it. As the end point was blue that means the oxidation state was 4+.
I work out the moles of the 0.1 M sodium thiosulphate and the moles of vanadium. I then use the half equation of sodium thiosulphate:
2S2O3(2-) ---> S4O6(2-) + 2e-
to figure out I have 2 moles of electrons for every one mole of S4O6(2-). I then multiply the moles of sodium thiosulphate by 2 to work out the moles of electrons transferred and ratio the moles of electrons against the moles of vanadium. This results in 1:2 ratio meaning I take 2 from the end point 4 to figure out vanadium was in the 2+ state before titration...

I think. I thought I'd post my solution just in case anyone does anything similar in the future.
(edited 8 years ago)

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