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Topic: Oxalic Acid Dihydrate titration question...  (Read 10905 times)

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Offline UnconciousPilot

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Oxalic Acid Dihydrate titration question...
« on: November 18, 2006, 03:33:08 PM »
Simple question.. I just can't figure out what it goes to when titrated with NaOH,

I thought it was just something like H2C2O4•2H2O + NaOH --> NaHC2O4•2H2O + H2O however according to my experiment results, there should be a 2:1 ratio of the NaOH and the acid and I have no idea how to deal with the attached water molecules...

Any input is appreciated, thanks.

Offline Alberto_Kravina

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Re: Oxalic Acid Dihydrate titration question...
« Reply #1 on: November 18, 2006, 03:39:58 PM »
Forget about the attached water molecules in the crystal, as they are irrelevant during the reaction.
So just write H2C2O4*2H2O as H2C2O4 and that's it.

Offline UnconciousPilot

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Re: Oxalic Acid Dihydrate titration question...
« Reply #2 on: November 18, 2006, 03:56:07 PM »
Alright fair enough, I just figured out why I needed a 2:1 ratio too... 2 protons... Titration calls for a complete neutralization  ::) whoops.

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