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Topic: Karl Fischer question  (Read 3767 times)

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Offline LC

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Karl Fischer question
« on: July 12, 2007, 06:56:07 PM »
Hi, Can someone clear this up for me please.
When you do a water content by Karl Fischer on something that has H2O in its formula, is this H2O included in the Karl Fischer result? For eg. 2% H2O for the sample C22H24N2O8.HCl.1/2(H2O).1/2(C2H6O). Does this mean 2% other than the incorporated or 1.75% for the incorporated and an extra 0.25% moisture? Thanks

Offline Kellen

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Re: Karl Fischer question
« Reply #1 on: July 16, 2007, 01:01:15 PM »
I'm not very familiar with the Karl Fischer method, but my guess would be that the H2O would be counted. It may be incorporated into the crystalline structure of a solid, but it is still water - I would think the Fischer method would pull that water out of the crystals. Plus, there are crystal structures with and without incorporated water, but the one that's "anhydrous" won't have any water in the crystalline structure.

Offline profmsg

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Re: Karl Fischer question
« Reply #2 on: July 24, 2007, 12:00:28 PM »
all the water including the water in the crystalline state(in the molecular structure) and the trapped moisture are included. when the titration is carried out, all the crystalline or powder sample is first dissolved in the solvent in the vessel which ultimately results in loss of water from the moelcule due to the high concentration of solvent. Then only,titrant added to give to get 100% water. hence during kf, the sample must be properly dissolved before conforming the addition of sample. the proper dissolution of sample and solvent also gives smooth graph.

Offline LC

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Re: Karl Fischer question
« Reply #3 on: August 08, 2007, 10:32:25 PM »
Thats what I would have thought, thanks. ;)

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