Respuesta :

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