how to calculate pka and pkb (of acetate ion) from ph? -acetic acid/NaOH titration?

Consider the titration (neglect the dissociation of CH₃COOH and that of CH₃COO⁻):
NaOH(aq) + CH₃COOH(aq) → CH₃COONa(aq) + H₂O(l)

No. of milli-moles of NaOH reacted = (0.1154 mmol/ml) × (24.3 ml) = 2.80 mmol
No. of milli-moles of CH₃COOH reacted = (0.112 mmol/ml) × (25 ml) = 2.80 mmol
No. of milli-moles of CH₃COO⁻ formed = 2.80 mmol
Volume of final solution = (24.3 + 25) ml = 49.3 ml
Molarity of CH₃COO⁻ in the final solution = (2.80 mmol) / (49.3 ml) = 0.0568 M


Consider the dissocation (hydrolysis) of CH₃COO⁻ :
CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq) ...... Kb(CH₃COO⁻)

Initial concentrations :
[CH₃COO⁻]ₒ = 0.0568 M
[CH₃COOH]ₒ = [OH⁻]ₒ = 0 M

At equilibrium :
pOH = 14.00 - pH = 14.00 - 9.33 = 4.67
[OH⁻] = 10⁻⁴·⁶⁷
Hence, [CH₃COO⁻] = (0.0568 - 10⁻⁴·⁶⁷) M
[CH₃COOH] = [OH⁻] = 10⁻⁴·⁶⁷ M

Kb(CH₃COO⁻)
= [CH₃COOH] [OH⁻] / [CH₃COO⁻]
= (10⁻⁴·⁶⁷)² / (0.0568 - 10⁻⁴·⁶⁷)
= 8.05 × 10⁻⁹ (M)


Ka(CH₃COOH)
= Kw / Kb(CH₃COO⁻)
= (1.00 × 10⁻¹⁴) / (8.05 × 10⁻⁹)
= 1.24 × 10⁻⁶ (M)