Chemical Forums

Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: UnconciousPilot on November 18, 2006, 03:33:08 PM

Title: Oxalic Acid Dihydrate titration question...
Post by: UnconciousPilot 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.
Title: Re: Oxalic Acid Dihydrate titration question...
Post by: Alberto_Kravina 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.
Title: Re: Oxalic Acid Dihydrate titration question...
Post by: UnconciousPilot 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.