Calculating H+/OH- Concentrations after Second Deprotonation

[Sang_Han_1E]

In pretty much all of the questions that we've done in class or in the textbook we assume that the second Ka or Kb value is insignificant enough to where we can ignore second deprotonation. My question is how would we solve questions like the ones we've seen so far in the textbook but where the second Ka or Kb value is not insignificant and we actually do need to consider the second deprotonation? Would the first deprotonation act as the starting concentrations for the second deprotonation and we just continue to solve the question as normal or would both deprotonations need to be considered in the final answer and, if so, what would the steps be in solving that form of question? Sorry for the weird wording of my question!

[Miah Chao 2I]

Hi! We would use the first deprotonation to find the concentrations of H3O+ and HA-, and then use that for the starting concentrations for the second deprotonation. Then, we would solve as normal. Here's an example with H2SO4 that I did in my discussion:

H2SO4 starting concentration: 0.15M
H2SO4 + H2O -> H3O+ + HSO4-
Since H2SO4 is a strong acid, we can assume that all of the H2SO4 will dissociate to yield [H3O+]=0.15M and [HSO4-]=0.15M.

Now, for the second deprotonation, we use the values we found in the first deprotonation as our "initial" values:
HSO4- + H2O -> H3O+ + SO42-

At equilibrium, [HSO4-]=0.15-x, [H3O+]=0.15+x, [SO42-]=x. Given that Ka2=1.2x10-2, now we have 1.2x10-2=(0.15+x)(x)/(0.15-x). From there, we solve for x as usual to find the equilibrium concentrations.

Hope this helps!