Question

Calculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid (C6H5COOH, Ka = 6.5 × 10-5) to produce a pH of 4.00. The ionization (dissociation) equation is shown below.

C6H5COOH(aq) + H2O (aq) <--> C6H5COO– (aq) + H3O+ (aq)

Answer 1

concentration of sodium benzoate =?

concentration of benzoic acid = 0.20 M

pH = 4.00

Acid dissociation constant Ka = 6.5*10^-5

pKa = - log Ka = - log (6.5*10^-5) = 4.187

From Henderson-Hasselbalch equation equation

pH = pKa + log ([sodium benzoate] /[benzoic acid])

4 = 4.187 + log ([sodium benzoate] /[benzoic acid])

4 - 4.187 = log ([sodium benzoate] /[benzoic acid])

-0.187 = log ([sodium benzoate] /[benzoic acid])

0.65 = [sodium benzoate] /[0.20]

[sodium benzoate] = 0.20 * 0.65 = 0.130 M