Chemistry 112

Polyprotic acids:

How do we treat acids with more than one ionizable hydrogen atom?

In most cases, polyprotic acids can be treated in exactly the same fashion as monoprotic acids.

Treat each ionized hydrogen ion one step at a time.

Successive acid ionization constants are number sequentially: Ka1, Ka2, Ka3, ...



Examples


What is the pH of a 0.0050 M solution of sulfuric acid at 25 oC?

Strategy: The first proton of sulfuric acid is completely ionized – it's a strong acid.

Bisulfate ion is a weak acid, so we will need to use the usual weak acid procedures, remembering that the first step of the sulfuric acid ionization generates both hydronium ion and bisulfate ion.

The range of possible pH values is –log(2×0.0050) = –log(0.010) = 2.00, (assuming both hydrogen ions ionized completely) to 7.00 (pure water, assuming sulfuric acid were nonacidic).



What is the pH of a 0.037 M solution of phosphoric acid at 25 oC?

Strategy: H3PO4 has three ionizable hydrogen atoms, so this will be a three-step problem.

This is a weak acid, so treat each ionization reaction as an independent problem, using the results of previous steps, as appropriate.

The range of possible pH values is 0.96 (all three H+ donated, = –log(3×0.037) = –log(0.11)) to 7.00 (pure water, no H+ donated).

Since this is a weak acid, the dihydrogen phosphate and hydrogen phosphate ions should have significantly smaller Ka values, so more realistically, the lower extreme for the pH is probably 1.43 (= –log(0.037)).