Calculating Acetic Acid Ionization Degree in Buffer

The ionization of acetic acid ($ CH_3COOH $) is an equilibrium process:

$ CH_3COOH \rightleftharpoons H^+ + CH_3COO^- $

The acid dissociation constant ($ K_a $) expression is:

$ K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} $

We are given:

• Initial concentration of acetic acid: $ [CH_3COOH]_{initial} = 0.02 $ M
• Concentration of acetate ions from sodium acetate (common ion): $ [CH_3COO^-]_{initial} = 0.01 $ M
• $ K_a = 1.8 \times 10^{-5} $

Let $ \alpha $ be the degree of ionization. The concentration of $ H^+ $ produced from the ionization of acetic acid is $ x = [CH_3COOH]_{initial} \times \alpha $. At equilibrium:

• $ [CH_3COOH]_{eq} = [CH_3COOH]_{initial} - x = 0.02 - x $
• $ [CH_3COO^-]_{eq} = [CH_3COO^-]_{initial} + x = 0.01 + x $
• $ [H^+]_{eq} = x $

Substitute these equilibrium concentrations into the $ K_a $ expression:

$ 1.8 \times 10^{-5} = \frac{x(0.01 + x)}{0.02 - x} $

Since acetic acid is a weak acid and the presence of the common ion ($ CH_3COO^- $) suppresses further ionization, $ x $ is expected to be very small compared to $ 0.01 $ and $ 0.02 $. We can make the following approximations:

• $ 0.01 + x \approx 0.01 $
• $ 0.02 - x \approx 0.02 $

The equation simplifies to:

$ 1.8 \times 10^{-5} \approx \frac{x(0.01)}{0.02} $

Solving for $ x $ (which represents $ [H^+]_{eq} $):

$ x \approx (1.8 \times 10^{-5}) \times \frac{0.02}{0.01} $

$ x \approx (1.8 \times 10^{-5}) \times 2 $

$ x \approx 3.6 \times 10^{-5} \text{ M} $

The degree of ionization ($ \alpha $) is calculated using the definition:

$ \alpha = \frac{x}{[CH_3COOH]_{initial}} $

$ \alpha = \frac{3.6 \times 10^{-5}}{0.02} $

$ \alpha = \frac{3.6 \times 10^{-5}}{2 \times 10^{-2}} $

$ \alpha = 1.8 \times 10^{-3} $

The degree of ionization of 0.02 M acetic acid containing 0.01 M sodium acetate is $ 1.8 \times 10^{-3} $.