You are asked to prepare a pH=4.00 buffer starting from 1.50 L of 0.0200 M solution of benzoic acid (C6H5COOH) and an excess of sodium benzoate (C6H5COONa). You may want to reference (Pages 721 - 728) Section 17.2 while completing this problem. Part A What is the pH of the benzoic acid solution prior to adding sodium benzoate?

pH = 2.96

Explanation:

C6H5COOH ↔ C6H5COO - + H3O+

∴ Ka = 6.5 E-5 = [H3O+][C6H5COO-] / [C6H5COOH]

C6H5COONa → C6H5COO - + Na+

prior to adding C6H5COONa:

⇒ C C6H5COONa = 0 M = [Na+]

∴ mass balance:

⇒ C C6H5COOH = [C6H5COOH] + [C6H5COO-] = 0.0200 M

⇒ [C6H5COOH] = 0.0200 - [C6H5COO-]

charge balance:

⇒ [H3O+] = [C6H5COO-]

⇒ [C6H5COOH] = 0.0200 - [H3O+]

⇒ Ka = [H3O+]² / (0.0200 - [H3O+]) = 6.5 E-5

⇒ [H3O+]² + 6.5 E-5[H3O+] - 1.3 E-6 = 0

⇒ [H3O+] = 1.108 E-3 M

∴ pH = - Log [H3O+]

⇒ pH = 2.9554 = 2.96