Question

The pKa of acetic acid is 4.76. If you make a 0.1M solution of acetic acid and adjust the pH to 4.56, what will the concentrations of acetic acid and acetate at equilibrium?

Answer 1

[CH3COOH] = 0.06131M

[CH3COO^-] = 0.03869M

Explanation

Henderson - Hasselbalch equation which give pH of a buffer solution is

pH = pKa + log([A^-] /[HA])

pKa = pKa of weak acid, 4.76

A^- = conjucate base , CH3COO^-

HA = weak acid , CH3COOH

adjusted pH is 4.56

4.56 = 4.76 + log([CH3COO^-]/[CH3COOH])

log([CH3COO^-]/[CH3COOH]) = -0.20

[CH3COO^-]/[CH3COOH] = 0.6310

[CH3COO^-] = 0.6310 [CH3COOH]

buffer concentration = 0.10M

so,

[CH3COO^-] + [CH3COOH] = 0.10M

0.6310 × [CH3COOH] + [CH3COOH] = 0.10M

1.6310× [CH3COOH] = 0.10M

[CH3COOH] = 0.06131M

[CH3COO^-] = 0.10M - 0.06131M = 0.03869M