6E.1 Textbook Question

[IsabelMurillo3K]

Calculate the pH of 0.15 M H2SO4(aq) at 25 °C.
Hey guys!
I was wondering if anyone could walk me through the steps of this problem?
Thank you!

[Hazelle Gunawan 3G]

Since H2SO4 is a strong diprotic acid, the first dissociation will be a direct 1:1 ratio so you'd end up with [H+] = 0.15M. After that, I don't have the problem pulled up but if they give you the Ka for the second dissociation, you do the ICE box like you'd do for any weak acid. Hope that helps!

[magalysantos_1F]

Because it is asking you for the pH rather than the pOH of the strong sulfuric acid, there are not many steps.
[H3O+] = 0.15 M
pH = -log(0.15)
= 0.8

[SavannahScriven_1F]

H2SO4 is pretty much the only strong diprotic acid. First you get the concentration of H3O+ from the concentration of H2SO4 itself because it completely dissociates into H3O+ and HSO4-. Now, you have to take into account the contribution of H3O+ from HSO4-. You would make an ICE table like any other weak acid, w/ HSO4-(initial) = 0.15, H3O+(initial)=0.15, and SO4^2-(initial)=0. HSO4-(final) = 0.15-x, H3O+(final)=0.15+x, and SO4^2-(final)=x. Solve for x using the Ka given in the textbook. Then use H3O+(final)=0.15+x to get the concentration, and calculate pH.