How can we tell if an ion is amphiprotic [closed]

If I had the following ions: $$\ce{HCO3^-}$$ $$\ce{H2PO4^-}$$ $$\ce{CH3COO^-}$$

Which ones would be amphiprotic?

Looking at my $$K_A$$ chart, it looks like each ion is formed by the disassociation of weak acids, so It looks like all of them are amphiprotic because they could all technically act as a bronstead-lowry base.

The answer is the first two ions, but why?

Thanks for answering :)

closed as off-topic by MaxW, Mithoron, A.K., Tyberius, Todd MinehardtApr 19 at 22:24

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• Shouldn't the third choice be just $\ce{CH3COO-}$? – MaxW Apr 18 at 20:14
• yes, I fixed that :) – Horatio Nelson Apr 18 at 20:43
• How does each anion change if you add $\ce{H+}$ and if you add $\ce{OH-}$? – MaxW Apr 18 at 20:46

To be amphiprotic means that the chemical species can donate or accept H+ ions.

$$\ce{H2CO3<=>[H+] HCO3^- <=>[OH-] CO3^{2-}}$$ $$\ce{H3PO4 <=>[H+] H2PO4^- <=>[OH-] HPO4^{2-}}$$ $$\ce{CH3COOH <=>[H+] CH3COO^- <=>[OH-] \text{No Change}}$$

Thus since the acetate anion can't donate a proton, it is not amphiprotic.

Good question. If you show these three ions (as written) to a super intelligent being person who never took chem or physics, they will not be able to tell you which ion is amphoteric. In short, there is no simplr short-cut! This requires some knowledge of some formula writing conventions.

For simple inorganic ions, if H is written with a p-block element, that H is ionizable and it is an acid in water

HCl, HBr, H2SO4, H2S, H3PO4, H2CO3

From HCl, we can remove one H+

From H2SO4, we can remove two H+, one at a time

From H3PO4, we can remove three H+, one at a time.

So if you happen to see ions of these acids, SO4(2-), you can immediately guess from the minus two charge that this ion can accept two protons.

If you see, HPO4 (2-), you should be able to guess that HPO4(2-) can accept one or two proton(s), however it can also lose the last one.

For small organic ions, it is slightly tricky, the ionizable H is written in the end with "COO" group

CH3COOH, HCOOH