Sodium gluconate and cerium(IV) sulphate

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Offline Stephen

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Sodium gluconate and cerium(IV) sulphate
« on: 16/04/2008 13:58:53 »
Determination of the concentration of sodium gluconate may be undertaken by boiling the acidified solution with 0.1M cerium(IV) sulphate solution, backtitrating with ammonium iron(II) sulphate solution.

A number of oxidation products are possible.  Factors are quoted for turning the titration results into concentrations of sodium gluconate but they seem to be empirical.

Can anyone quote the stoichiometric equation (and references) for the reaction?

Stephen

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Offline Bored chemist

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Sodium gluconate and cerium(IV) sulphate
« Reply #1 on: 16/04/2008 18:27:37 »
As you say, there are lots of possible products and I guess quite a lot of them get made. In principle you could carefully determine exaclty what gets formed and then calculate the cluconate concentration from that. However that would be more difficult than finding another way to measure the gluconate in the first place.
I doubt there is one equation that would really cover this rection.
If I were doing this I'd get a known concentration of cluconate somehow and use that to prepare a set of samples, analyse them and calculate my own empirical factor. (actually, I'd get my colleague who does ion chromatography to measure the stuff, but I don't supose you have that option).
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Offline Stephen

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Sodium gluconate and cerium(IV) sulphate
« Reply #2 on: 17/04/2008 16:57:28 »
I have checked the analytical method using commercially available sodium gluconate and 0.1M cerium(IV) sulphate.  The latter is very stable and was standardized here using iron(II) ethylenediammonium sulphate and ferroin indicator.  Three close results come out to a mole ratio of 1:12 (actually 1 to 12.12, but the sodium gluconate was declared as 98% to 102% possibly because of excess gluconic acid).  Presumably the C=O is not involved in the oxidation leaving six C-O-H groups to be oxidized by the twelve Ce(IV) ions.  But what is formed?

Stephen