Question

Calculate the pH at 25°C of a 0.54M solution of sodium benzoate NaC6H5CO2. Note that benzoic acid HC6H5CO2 is a weak acid with a pKa of 4.20.Round your answer to 1 decimal place.

Answer 1

Given C6H5COONa is the salt of a strong base (NaOH) and weak acid (C6H5COOH), and thus the salt in aqueous solution will have a pH.

The hydrolysis of the cyanide ion is as follows,

C6H5COO- + H2O ==> C6H5COOH + OH-

pKa value of C6H5COOH is given, that is, pKa = 4.20, to find Ka

Ka = 10-pKa

Ka = 10-4.20

Ka = 6.3096 x 10-5

Using the formula Kw = KaKb

Kb = Kw/kb

Kb = 1 x 10-14/ 6.3096 x 10-5

Kb = 0.15849 x 10-9

Kb = 0.15849 x 10-9= [C6H5COOH][OH-]/[ C6H5COO-]

0.15849 x 10-9= (x)(x)/0.54M                  where, [C6H5COOH] = x, [OH-] = x, C6H5COO- = 0.54 M

x2 = 2.935 x 10-10

x =[OH-]= 1.7132 x 10-5

pOH = -log[OH-]

pOH = 4.7662

14 = pH + pOH

pH = 14 – pOH

pH = 14 – 4.7662

pH = 9.2338

pH = 9