Calculate the pH of a 0.10 M solution of sodium benzoate

Question

Calculate the pH of a 0.10 M solution of sodium benzoate (NaC₆H₅COO) given that the K_a of benzoic acid (C₆H₅COOH) is 6.50 x 10^-5.

Answer 1

Benzoate is tge conjugate base of benzoic acid. In water, benzoate hydrolyses as follows...

C₆H₅COO^- + H₂O ==> C₆H₅COOH + OH^-

Kb = Kw/Ka = 1x10^-14 / 6.50x10^-5 = 1.54x10^-10

Kb = 1.54x10^-10 = [C₆H₅COOH][OH^-]/[C₆H₅COO^-]

1.54x10^-10 = (x)(x)/0.1-x and assuming x is small, we have....

1.54x10^-10 = x²/0.1

x² = 1.54x10^-11

x = 3.92x10^-6 = [OH-]

pOH = 5.41

pH = 14 - 5.41

pH = 8.59