pH of a Weak Acid Solution

Given

• [CH3COOH] = 0.10 M

• Ka(acetic acid) = 1.8 x 10^-5 at 25 C

• Water autoionization is negligible beside the acid's own H+.

Find

The pH of the solution.

Principle

Weak acids ionize only slightly, so an ICE table with [H+] = [A-] = x and [HA] = C - x gives Ka = x^2/(C - x). When Ka is small, x is small enough that C - x ≈ C, simplifying the equation to x = sqrt(Ka x C). The approximation must be validated with the 5% rule: x/C must be below 0.05.

Formula

Solution

1. Let x = [H+] = [CH3COO-] at equilibrium. From the ICE table, [CH3COOH] = 0.10 - x.

2. Write Ka = x^2/(0.10 - x) = 1.8 x 10^-5. Because Ka is tiny, assume 0.10 - x ≈ 0.10, giving x^2 ≈ 1.8 x 10^-5 x 0.10 = 1.8 x 10^-6.

3. Solve: x = sqrt(1.8 x 10^-6) = 1.34 x 10^-3 M = [H+].

4. Validate the approximation: x/C = (1.34 x 10^-3)/0.10 = 0.0134 = 1.34%, which is under 5%, so the approximation is justified.

5. Compute pH = -log(1.34 x 10^-3) = -(log 1.34 + log 10^-3) = -(0.127 - 3) = 2.87.

Unit check

Ka has units of M (concentration ratio), so sqrt(Ka x C) has units sqrt(M x M) = M, giving [H+] in mol/L. The pH is dimensionless, as required.

What this means physically

Only about 1.3% of the acetic acid molecules ionized, which is why the pH is 2.87 rather than 1.00 (what a 0.10 M strong acid would give). Weak acids sit mostly as intact HA molecules, and the tiny ionized fraction is governed by Ka.