1.00 M sodium acetate + 1.00 M acetic acid is a BUFFER solution.
You are correct in the initial pH because when [salt] = [acid], the pH = pKa which for acetic acid is 4.74.
Now, to find final pH, you need to find final concentrations of acetate (Ac-) and acetic acid (HAc) and plug them back into the Henderson Hasselbalch equation of pH = pKa + log [acetate]/[acetic acid]
Initial moles Ac- = 1 L x 1 mole/L = 1 mole
Initial moles HAc = 1 L x 1 mole/L = 1 mole
Reaction: NaOH + HAc ==> NaAc + H2O SO NOTE THE [HAc] DECREASES AND [Ac-] INCREASES
moles OH- added = 0.041 L x 0.540 mol/L = 0.0221 moles
Final moles Ac- = 1 + 0.0221 = 1.0221 moles Ac-
Final moles HAc = 1 - 0.0221 = 0.9779 moles HAc
Final volume = 1 L + 0.041 L = 1.041 L
Final [Ac-] = 1.0221 moels/1.041 L = 0.9818 M
Final [HAc] = 0.9779 moles/1.041 L = 0.9394 M
Final pH = 474 + log (0.9818/0.9394)
pH = 4.74 + 0.0192
pH = 4.76
If you've learned about ICE Tables, you can do it easier that way.