Calculate the pH of 0.58 M solution of sodium benzoate?

Question

What is the pH of a 0.58 M solution of sodium benzoate, NaC₆H₅COO_(aq)? K_a = 6.3x10^-5 for benzoic acid (weak acid), HC₆H₅COO_(aq).

Answer 1

The pH will be determined by the benzoate anion dissolving in water, ie hydrolysis of C6H5COO-.

C6H5COO- + H2O ==> C6H5COOH + OH- (pH will be alkaline, i.e. >7)

Kb = [C6H5COOH][OH-] / [C6H5COO-]

Kb = 1x10^-14 / Ka = 1x10^-14 / 6.3x10^-5 = 1.59x10^-10

1.59x10^-10 = (x)(x) / 0.58 - x (ignoring x in the denominator to simplify the math) we have...

x2 = 1.59x10^-10 * 0.58 = 9.2x10^-11

x = 9.6x10^-6 (note this is small relative to 0.58 so our above assumption was valid)

[OH-] = 9.6x10^-6

pOH = -log 9.6x10^-6 = 5.0

pH = 14 - pOH

pH = 8.9 or 9 (depending on sig, figs,)