Chemistry Question?

When oxalic acid, H2C2O4, reacts with NaOH, is this the chemical formula?

H2C2O4 + 2NaOH --> 2H2O + Na2C2O4 ??

If it is so, why doesn't only one H+ ion transfer to the other side like other acid-base reactions like:

H2C2O4 + NaOH --> H2O + NaHC2O4; or is it because it's a double replacement reaction

Answer:
The equation you have written is the complete neutralization of oxalic acid:

H2C2O4 + 2NaOH --> 2H2O + Na2C2O4

You need two moles of OH- (from NaOH) to react with two moles of H+ (from H2C2O4). But this happens step wise. The first one is

H2C2O4 + OH- --> H2O + HC2O4-

and then the second one is

HC2O4- + OH- --> H2O + C2O4^2-

This is also true in other polyprotic acids like H3PO4. Since H3PO4 has 3Hs, neutralization is 3 steps. And it is also important to note that the degree of dissociation (or the dissociation constants) differs in every step (that's why there is k1, k2, etc).
here is the balanced equation:H2C2O4 + 2NaOH=2H2O +Na2C2O4
If you're clever, you CAN just remove one of the hydrogens. But clever is not a good recipe for a reliable titration. Titrations work best if they're idiot-proof. That's why you just take off both of the hydrogens.

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