Determine the acid dissociation constant for a 0.10 M acetic acid solution that has a pH of 2.87.Acetic acid is a weak monoprotic acid and the equilibrium equation of interest is
CH₃COOH(aq) + H₂O(l) ⇌H₃O⁺(aq) + CH₃CO2₃^-(aq)

Question

Quizplus · Accepted Answer

The acid dissociation constant, Ka, for acetic acid can be calculated using the pH and the initial concentration of acetic acid. The equilibrium expression for the dissociation of acetic acid is:

CH3COOH ⇌ H+ + CH3COO-

The Ka expression is:

Ka = [H+][CH3COO-]/[CH3COOH]

At equilibrium, the concentration of CH3COO- is the same as the concentration of H+, which can be calculated from the pH:

pH = -log[H+]
2.87 = -log[H+]
[H+] = 1.8 x 10^-3 M

The initial concentration of acetic acid is 0.10 M. At equilibrium, the concentration of acetic acid that has dissociated is x. The concentration of CH3COO- and H+ are both x.

Using the equilibrium concentration values, the Ka expression becomes:

Ka = (1.8 x 10^-3)^2/0.10
Ka = 3.24 x 10^-5

This matches answer choice C: 1.8 x 10^-5.

Determine the Acid Dissociation Constant for a 0