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Topic: Titrating using ammonia cerium (IV) sulphate  (Read 6863 times)

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Offline zilalti

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Titrating using ammonia cerium (IV) sulphate
« on: November 06, 2005, 01:52:01 PM »
I am trying to work out the purity of a sample of 4 - hydroxy benzene and i thought the easiest way would be to titrate with ammonia cerium (IV) sulphate. I have little experience with titarion so I have been trying to teach myself the basics. I know once i have found out the number of moles that is required to neutralize the sample i can go from there.
  My question is basically theroetically if I have a 100% pure sample, is the number of moles of the ammonium cerium sulphate (or Ce4+ ions) required to netralise the sample, the same number of moles in the sample?  Or in other words if I were to write neutralisation equation would it be

1((NH4)2Ce(SO4)3) + 1(C6H6OHNH2) ==> Product

I know I havn't worded it very well but help would be much appreciated.

Offline FeLiXe

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Re:Titrating using ammonia cerium (IV) sulphate
« Reply #1 on: November 06, 2005, 02:10:33 PM »
you are probably talking about 4-Amino-phenol (according to your sum formula) and you will oxidize it to 4-Nitro-phenol

The nitrogen changes from -III to +III = 6
The cerium changes from +IV to +II = -2

That means you need three times as much cerium as nitrogen

(that's my guess anyway. if your titration works it probably works like this)
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Offline zilalti

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Re:Titrating using ammonia cerium (IV) sulphate
« Reply #2 on: November 06, 2005, 02:32:45 PM »
Yea sorry I did mean 4-Amino phenol, my mistake.
So when im trying to work out the number of moles in my sample of 4-amino phenol it will be a third of the number of moles required to neutralise the sample, correct?
« Last Edit: November 06, 2005, 02:36:53 PM by zilalti »

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