Diprotic Acid ? - Two Ionizations ?

Question

Oxalic acid (C₂H₂O₄) is a diprotic acid. Write the two ionization reactions for C₂H₂O₄ to C₂O₄²⁻.

Can you please explain how this works and create the reactions?

Answer 1

When you have a diprotic acid, that means there are TWO ionizable hydrogen atoms. Each one of these H atoms has it's own Ka, i.e. one H⁺ is donated more easily than the other and so it "comes off" or dissociates first. A simple example would be the generic diprotic acid, H₂A. It would ionize as follows:

H₂A ==> H⁺ + HA^-

HA^- ==> H⁺ + A^2-

So, for C₂H₂O₄, we can look at it as H₂C₂O₄, to make it easier to visualize the ionization of the H atoms.

H₂C₂O₄ ==> H⁺ + HC₂O₄^-

HC₂O₄^- ==> H⁺ + C₂O₄^2-

Hope this makes sense and helps you understand the process. If not, please leave a comment.

Answer 2

A diprotic acid has two acidic, or "removable", hydrogens. Each hydrogen has a dissociation constant (K_a - the smaller the Ka value, the harder it is to remove the hydrogen) associated with it, with one ionizable hydrogen being more acidic than the other.

Generally, it is significantly easier for a diprotic acid to lose the first acidic hydrogen (for oxalic acid, the Ka₁ is 5.6 x 10^-2) - the loss of the second acidic hydrogen is a much less favorable process (again for oxalic acid, Ka₂ is 5.1 x 10^-5). The reason given for this is that when the first hydrogen is removed, you are pulling H away from the rest of the molecule, pulling something neutral away from something else neutral, generating two charged species (H+ and HC₂O₄^-). But when you go to remove the second acidic hydrogen, you are trying to pull something increasingly positive (H⁺) away from something negative (HC₂O₄^-). According to electrostatics principles, positively charged and negatively charged particles are attracted to each other and require more energy to separate than two neutral or neutral-ish particles.

You already have an answer that does a nice job of generating the reactions you requested, so I won't replicate that work here. I hope this answer gave you some more depth into this topic.