A30.0-ml volume of 0.50 m ch3cooh (ka=1.8×10−5) was titrated with 0.50 m naoh. calculate the ph after addition of 30.0 ml of naoh at 25 ∘c.

According to this reaction:
CH3COOH + OH-↔ CH3COO- + H2O
we have to get the moles of acetic acid:
moles of acetic acid = volume of acetic acid per L * molarity
                                  = 0.03 L * 0.5 M = 0.015 m
and the moles of NaOH = volume of NaOH * molarity
                                         = 0.03 L * 0.5M = 0.015 m
and when the total volume = volume of acetic acid + volume of NaOH
                                              = 0.03L + 0.03L = 0.06 L
So we can get the molarity of CH3COO- = moles of CH3COO/ total volume
                                                                     = 0.015 mol / 0.06 L =0.25 M
and for the reaction :
                       CH3COO- + H2O ↔ CH3COOH + OH-
initial                0.25                                 0               0
equilibrium (0.25 - X)                               X                X

and when Kw = 1 X 10^-14 and Ka = 1.8 X 10^-5 and Kb = Kw/Ka so:
Kw/Ka = [CH2COOH][OH]/[CH3COO]
(1x10^-14/1.8x10^-5) = X^2 / (0.25-X)
5.6x10^-10 = X^2/(0.25-X)
1.12 - (5.6x10^-10 X) = X^2
∴X = 1.2x10^-5 ∴[OH-] = 1.2x10^-5 M
when POH = - ㏒[OH-]
         POH = -㏒(1.2x10^-5) = 4.9

and when POH + PH = 14 
∴PH = 14- POH = 14 - 4.9 = 9.1

Ithink about 50 % of an original isotope's number of atoms remain