Naked Science Forum
Non Life Sciences => Chemistry => Topic started by: Chemistry4me on 21/02/2009 06:01:09
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When ammonium vanadate (NH4VO3) is dissolved in dilute sulfuric acid and the solution is treated with an excess of sulphur dioxide, a blue solution is formed...
What is the blue solution? I have searched high and low and still can't find a definitive answer... [>:(][>:(]
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When ammonium vanadate (NH4VO3) is dissolved in dilute sulfuric acid and the solution is treated with an excess of sulphur dioxide, a blue solution is formed...
What is the blue solution? I have searched high and low and still can't find a definitive answer... [>:(][>:(]
SO2 in acidic solution reduces VO3- (n.o. of vanadium = +5) to VO2+ (n.o. of vanadium = +4) which is blue.
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Thanks lightarrow [:)]
I was wondering if it was VO2+
It appears that it is.
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You know what the SO2 is oxidised to? SO32-? SO42-?
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You know what the SO2 is oxidised to? SO32-? SO42-?
The last you have said. It's the only possibility, think about it. Which is the n.o. of S in SO2 and in SO32-? It's the same. [:)]
So S is oxidized to +6, which means SO42- in water solution (or HSO4- in strong acidic solution).
By the way, the reaction from vanadate and SO2 is exploited industrially to make H2SO4; vanadium +4 is then re-oxidized by air to vanadium +5 and so it acts as catalyzer.
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Thank you [:)] again.
Haha, obviously I was NOT thinking!