I just have a question on how do you do the polyprotic bases.

Question: show how oxalate ion can be a polyprotic base:

My answer ->

  1. $$\ce{OOCCOO^{2-} + HOH -> HOOCCOO^{-} + OH ^{-}}$$

  2. $$\ce{HOOCCOO^{-} + HOH -> HOOCCOOH + OH-}$$

My question is that do we stop doing the reaction until the oxalate ion has no charge or do we keep going on to form $\ce{HOOCCOOHH+}$

Another question I have is why would Hypochlorite ion be a monoprotic base, rather than a polyprotic base?

Shall appreciate your answer/thoughts on this.


If you have a molecule or ion with multiple sites that can be protonated (as you do here), and the sites can interact with each other (as they can here), the affinity for protons drops rapidly as more sites are protonated.

The proton affinity of the ion in your second reaction is a couple of orders of magnitude less than in your first reaction because of this effect.

Piling a third proton on there would be far less favorable than that---oxalic acid is a weak acid, and you're forcing it to act as a base by making it accept a proton1. You should stop with the two reactions you have.

It's really the same thing with hypochlorite ion; putting one proton on is fine (you get HClO, which is a weak acid). Putting a second proton on is tough, because you'd be making an acid act as a base.

1. This kind of thing does happen, though such protonated acids will quickly lose their extra proton to the surrounding water molecules. If you're using glacial acetic acid as a solvent, your acid ion is $\rm CH_3COOH_2^+$ instead of $\rm H_3O^+$, and $\rm CH_3COO^-$ instead of $\rm OH^-$.

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  • $\begingroup$ oh ok, so for any polyprotic acid, you should stop at two hydrogen right? $\endgroup$ – Ethan Hunt Jan 23 '15 at 20:31
  • $\begingroup$ You should stop when you've added enough protons to its ions get a neutral molecule. For example, with $\rm PO_4^{-3}$ you'd add 3 protons one at a time to get neutral $\rm H_3PO_4$. $\endgroup$ – Fred Senese Jan 23 '15 at 20:48
  • $\begingroup$ okay that makes sense. Thank you so much Fred for clearing my doubts $\endgroup$ – Ethan Hunt Jan 23 '15 at 21:51

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