A buffer containing 1.2169 M of acid, HA, and 0.1431 M of its conjugate base, A-, has a pH of 3.96. What is the pH after 0.0017 mol NaOH is added to 0.5000 L of this solution?

Question

I can't figure this out for the life of me, i'm sure its just some simple mistake i'm making but can someone please help me?

HA ==> H^+ + A^-

The Henderson-Hasselbalch equation is
pH = pKa + log[(base)/(acid)]
Plug in the value for pH and (base) and (acid) and calculate the pKa for the acid since they didn't give that to you in the problem.
Then we add the 0.0017 mol KOH.
That will neutralize 0.0017 mol of the acid. So (acid) will decrease by 0.0017 mol and the conjugate base, (A^-) will increase by 0.0017 mol. Plug in those new values for the acid and base along with the value you found for pKa and obtain the pH of the new solution.

I think I should point out that this problem really has no solution unless we make some assumptions. The problem states the molarity of the acid and its base BUT not how much of the acid and its conjugate base. Therefore, we really don't know the number of mols of the acid OR its base. We can make an assumption and take any amount of base and acid we wish and work the problem but the value of Ka will change depending upon how much of the acid/base mixtures we take. And of course that will change the pH of the final mixture after the KOH is added.