dissociate the salt: i assume it is NaCH3COO and NOT NaCH3COOH
NaCH3COO ⇒ Na⁺ + CH₃COO⁻
Note that [NaCH3COO] = [CH₃COO⁻] = 0.05 M
Na⁺ is a spectator so we ignore it
CH₃COO⁻ under goes hydrolysis and it will act as a base in solution
CH₃COO⁻ + H₂O ⇄ OH⁻ + CH₃COOH
Note that Ka converts to Kb:
Kb = Kw / Ka = 1.0 x 10^(-14) / (1.8 x 10^-5) = 5.56 x 10^-10
Set up a table
CH₃COO⁻ + H₂O ⇄ OH⁻ + CH₃COOH
ST 0.05 0 0
+Δ -x +x +x
-------------------------------------------------------------------------------------
EQ: 0.05 - x x x
Kb = [OH⁻][CH₃COOH] / [ CH₃COO⁻]
5.56 x 10^-10 = x² / (0.05 - x)
x = 5.27 x 10^-6 = [OH⁻]
pOH = -log[OH⁻]
pH = 14 - pOH = 14 + log[OH⁻]
pH = 14 + log(5.27 x 10^-6 ) = 8.721
the pH is 8.7
