Buffer questions

Question

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). (a) What is the pH of the benzoic acid solution prior to adding sodium benzoate? (b) How many grams of sodium benzoate should be added to prepare the buffer?

Answer 1

pH benzoic acid:

C₆H₅COOH ==> C₆H₅COO^- + H⁺

Ka = [C₆H₅COO^-][H⁺] / [C₆H₅COOH]

6.46x10^-5 = (x)(x) / 0.02 - x

x² = 1.29x10^-6 - 6.46x10^-5x

x² + 6.46x10^-5x - 1.29x10^-6 = 0 (use quadratic to solve for x)

x = 0.0011 M = [H⁺]

pH = -log 0.0011

pH = 2.96

Henderson Hasselbalch equation: pH = pKa + log [salt] / [acid]

pH = 4.00

pKa = -log Ka = -log 6.46x10^-5 = 4.19

[salt] / [acid] = ?

4.00 = 4.19 + log [salt] / [acid]

log [salt] / [acid] = -0.19

[salt] / [acid] = 0.646 = [C₆H₅COO^-] / [C₆H₅COOH]

[C₆H₅COOH] = 0.0200 M

Thus, we have [C₆H₅COO^-] / 0.0200 = 0.646

[C₆H₅COO^-] = 0.0129 M

molar mass C₆H₅COONa = 144 g / mol

0.0129 mol / L x 1.50 L x 144 g / mol = 2.79 g C₆H₅COONa needed

(be sure to check all of the math)