# Identifying a polyprotic acid using titration

I have the following question from a high school chemistry lab:

If an unknown acid has more than one equivalence point, is there a reason to expect a specific $K_\mathrm{a}$ ($K_{\mathrm{a1}}$, $K_{\mathrm{a2}}$ or possibly $K_{\mathrm{a3}}$) to be a better value to identify the acid?

I understand that if an acid has multiple equivalence points, it must have more than one hydrogen disassociating. As we titrate with a strong base, we will get an equivalence point for each hydrogen. Similarly, we will get a $K_\mathrm{a}$ value for each hydrogen, but why would one of these values be better than another for identifying the acid? Would we expect one to be more accurately measured than the other(s)?

• No. For instance if Ka1 and Ka2 are close then the two dissociations are muddled and you don't get a good value for either. So if Ka3 is much different then it would be the better one to determine. If Ka2 and Ka3 are close, but Ka1 is much different than Ka2, then Ka1 would be the best. So it depends... – MaxW Apr 15 '16 at 17:28
• @MaxW What if there are only two equivalence points? Is there still an advantage of picking one over the other? – Alex S Apr 15 '16 at 17:44
• Yes. You pick whichever one works better. For $\ce{H2SO4}$ for example the first ionization is so favored that you'd never be able to see it in aqueous solution. – MaxW Apr 15 '16 at 19:00