5. Consider the dissociation of acetic acid in water at equilibrium

Question

you start with .05 moles of acetic acid in 500 mL of water. At equilibrium, the pH of the solution is 2.873. What is the equilibrium constant of this reaction?

Answer 1

Acetic acid dissociates in water as follows: CH3COOH + H2O ====== > CH3COO^- + H3O⁺.

The equilibrium expression is K = [ CH3COO^- ] [ H3O⁺] / [CH3COOH]

As acetic acid is a weak acid K can be simplified to K = [ H3O⁺]² / [CH3COOH]

So K = [ H3O⁺]² / 0.1 M (0.05 / 500ml x 1000 mL = 0.1 M)

As the pH = 2.873, the [ H3O⁺] is the antilog of -2.873 or 1.34 x 10^-3 M.

Put this last value into the simplified equilibrium expression and solve for K.

The answer I got is 1.80 x 10^-5.

Answer 2

CH3COOH <==> CH3COO- + H+

The equilibrium expression can be written as: Keq = [CH3COO-][H+] / [CH3COOH]

From the pH, we can obtain the [H+] which is also the [CH3COO-].

pH = 2.873

[H+] = 1x10^-2.873 = 1.34x10-3 M

[CH3COOH] = 0.05 moles / 0.5 L = 0.1 M

Solving for Keq, we have..,

Keq = (1.34x10-3)(1.34x10-3) / 0.1

Keq = 1.8x10^-5