What is the pH of a 0.300 M solution of sodium acetate?

Question

The acetate ion is the conjugate base of acetic acid. The K_a of acetic acid is 1.76×10^–5.

What is the pH of a 0.300 M solution of sodium acetate?

Answer 1

Acetic acid = CH₃COOH (acid)

Sodium acetate = CH₃COONa (basic salt)

Acetate ion = CH₃COO^- (conjugate base)

When the acetate ion (CH₃COO^-) is placed in H₂O, we have the following hydrolysis reaction:

CH₃COO^- + H₂O ==> CH₃COOH + OH^- (the Na⁺ ion is merely a spectator)

Since CH₃COO^- is acting as a base (and H₂O is the acid), we need to find the Kb for CH₃COO^-

KaKb = Kw = 1x10^-14

Kb = 1x10^-14 / Ka

kb = 1x10^-14 / 1.76x10^-5

kb = 5.68x10^-10

Next, we write the Kb expression for the above reaction:

Kb = [CH₃COOH][OH^-] / [CH₃COO^-]

5.68x10^-10 = (x)(x) / 0.300 - x and assuming x is small relative to 0.300, ignore it in the denominator

5.68x10^-10 = x² / 0.300

x² = 1.70x10^-10

x = [OH^-] = 1.31x10^-5 (note: this is insignificant relative to 0.300 M so above assumption was valid)

pOH = -log [OH^-] = -log 1.31x10^-5

pOH = 4.88

pH = 14 - pOH = 14 - 4.88

pH = 9.12