Ok, moving on from my last question, you have 100mL of 0.10M acetic acid with ka = 1.8x10(-5) and add 50mL (0.10M) NaOH.

Answer 1

100 mL HAc x 0.1M = 10 mmoles.
50 mL NaOH x 0.1M = 5 mmoles.

.........HAc + OH^- => Ac^- + H2O
begin....10....0........0.......0
add............5.0.................
change..-5.0...-5.0.....+5.0....+5.0
equil....5.0.....0.......5.0
Then I would use the Henderson-Hasselbalch equation to solve for pH.
Since (base) = (acid), then log base/acid = log 1 = 0 so pH = pKa or about 4.74