In a weak acid-strong base titration, a 25.0 ml sample of .100 M acetic acid (HC2H3O2 Ka=1.8x10^-5) is titrated with .200 M NaOH.?

Initial moles of HC₂H₃O₂ = (0.100 mol/L) × (25.0/1000 L) = 0.00250 mol
Moles of NaOH added = (0.200 mol/L) × (20.0/1000 L) = 0.00400 mol

Equation for the reaction:
HC₂H₃O₂ + NaOH → NaC₂H₃O₂ + H₂O
Mole ratio  HC₂H₃O₂ : NaOH = 1 : 1

Since (Moles of NaOH added) > (Initial moles of HC₂H₃O₂),
NaOH is in excess.

Excess moles of NaOH = (0.00400 - 0.00250) mol = 0.00150 mol
Volume of the final solution = (25.0 + 20.0) ml = 45.0 ml = 0.0450 L
In the final solution, [OH⁻] = (0.00150 mol) / (0.0450 L) = 0.0333 M

pOH = -log[OH] = -log(0.0333) = 1.48
pH = pKw - pOH = 14.00 - 1.48 = 12.52