Textbook problem 6E.1

[Beverly Lauring 1L]

Hi, the problem asks to calculate the pH of 0.15 M H2SO4 (aq) at 25 degrees C.
I understand that this is a polyprotic acid, but my thought was that since H2SO4 is a strong acid, you would also have 0.15 M of H3O+ at equilibirum. Then, I thought you would use this value to find the pH of H2SO4 by doing pH=-log(0.15).

However, the answer key goes one step further and does the ICE table and calculations for HSO4- +H20--> H3O+ + SO4 2-. After calculations, they get a final concentration of 0.16 M H3O+, so the pH=-log(0.16). could someone please explain why this extra step is necessary? because i thought strong acids will just completely dissociate.

[Dhruv Suresh 2J]

In its first dissociation, H2SO4 dissociates fully into HSO4- and H3O+. However, since it's a polyprotic acid, the relatively weaker HSO4- can still further dissociate into SO4 2- and H3O+ (although the concentration of H3O+ from the second dissociation would be much smaller because HSO4- is a weaker acid than H2SO4). This results in 0.16 M H3O+ from the two dissociations.