Calculating the pH of a Polyprotic Solution

[Jar-Yee 3L]

So in HW 12.79, we're asked to find the pH of a .15 M H₂SO₄ solution.

Since sulfuric acid is a strong acid, I assumed maximum ionization/deprotonation/dissociation and just calculated pH with -log[H₃O⁺], and did -log(.15)=.82.

However, upon looking at the solutions guide, they started a new ICE table with .15 M HSO₄^- and H₃O⁺, and calculated pH based on K_a2.
In the end, their answer was .80, which was significantly close to my answer (which I didn't do a lot of work for).

I understand that their answer is more precise, but is it necessary for us to do the further calculation? Sulfuric acid is already incredibly strong, and the second calculation barely made a difference to the pH.

[Johnny Yu]

They calculate a 2nd Ka2 value because H2SO4 can donate 2 H+. For polyprotic acids, the Ka2 values will usually be way smaller than the Ka1 values (which is why your answer was similar to the solution manual's). The reason why the Ka2 values are smaller is because the H+ is bonded to a negative ion (in this case HSO4-) and the ion will want to keep the H+ more. This results in a way smaller Ka2 value and when calculated with the Ka1 value it'll be closer to the Ka1 value. Hope this helped!