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iliketorifk iliketorifk
wrote...
11 years ago
I don't understand how the 2H2O behaves when writing a chemical equation of oxalic acid dihydate dissolved in water and then titrated with NaOH to the end point. I'm having difficulty writing a balanced equation to eventually solve for molarity.
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wrote...
11 years ago
The dihydrate part of the reactant does not take part in the reaction. Throughout my training I was never taught to include it in any balanced equation. I do not think that students would be taught to do so today:

the balanced equation is:

(COOH)2 + 2NaOH → (COONa)2 + 2H2O

If you insist on writing oxalic acid as H2C2O4:

H2C2O4 + 2NaOH → Na2C2O4 + 2H2O

And if you insist on including the dihydrate:

H2C2O4.2H2O + 2NaOH → Na2C2O4 + 4H2O

The problem with this last equation is that the reaction is carried out in aqueous medium, and the water of crystalisation or hydration of the oxalic acid is no longer associated with the oxalic acid when it is in solution. secondly: The product sodium oxalate occurs as a trihydrate, so if you have to consider the oxalic acid as a dihydrate, then in the equation you should consider that the sodium oxalate is a trihydrate, and you would get:
H2C2O4.2H2O + 2NaOH → Na2C2O4.3H2O + H2O - which I think is actually quite ridiculous.
wrote...
11 years ago
H2C2O4 + 2 NaOH → Na2C2O4 + 2 H2O
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