Titrations - Weak acid with strong base

A 35.0 mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. Calculate the pH after 50.0 mL of NaOH have been added.

In the original question, I answered what the pH would be with no NaOH added, 34.5 mL, and 35.0 mL (equivalence point). I'm just not sure how to set everything up after the equivalence point has been reached.

For example, I used CH3COOH + H2O <---> CH3COO- + H3O+ because this is the reaction without any base added and used molarity and Ka to ultimately find the pH. And then for 34.5 mL, I did CH3COOH + OH- --> CH3OO- + H2O. I used the moles of CH3COOH and OH- and used the henderson-hasselbalch equation to find pH.

I'm just not sure what to do after equilibrium. Thanks!

Since the volume added is greater than that needed for the equivalence point then that means the solution is now basic (Adding a base) and has passed its buffer point (the buffer phase exists before the equivalence point) so you don't use the henderson equation to find the pH after 50ml was added.

You find the number of moles of NaOH that were added:
0.15*50/1000 = 0.0075 moles
Then the number of moles of the acid that reacted with the added base:
0.15*35/1000 = 0.00525 moles
The number of moles left unreacted: (remember all the acid has reacted so only OH- ions are left)
Unreacted moles = .0075 - .00525 = 0.00225 moles of OH- exist in the solution

Therefore pOH = -log [OH-] = -log[.00225] = 2.6478...
pH= 14 - pOH = 14 - 2.6478.. = 11.35