Question

An aqueous solution of acetic acid is found to have the following concentrations at 25°C: [CH₃COOH] = 1.65 x 10-2 M; [H30+] = 5.44 x 10^-4 M; and [CH₃COO-] = 5.44 x 10+ M. Calculate the equilibrium at 25°C. The reaction is:

CH₃COOH(aq) + H₂O(I) ⇆ H₃O(aq) + CH₃COO-(aq)

Strategy:

Using the balanced chemical equation, write the equilibrium constant expression, Kc, then substitute the given equilibrium concentrations to it.

Answer 1

CH₃COOH + H₂O = H₃O⁺ + CH₃COO^-

$K_c = \cfrac {[H_3O^+][CH_3COO^-]}{[CH_3COOH]}$

$K_c = \frac {[5,44*10^{-4}][5,44*10^{-4}]}{[1,65*10^{-2}]}= 1,8*10^{-5}$

ANSWER: 1,8 * 10^-5