What is the pH of a 0.02 M solution of a weak acid HA with a Ka = 1.8 x 10⁻⁵?
Correct: C
For a weak acid, [H⁺] ≈ √(Ka * [HA]). So, [H⁺] ≈ √(1.8 x 10⁻⁵ * 0.02) = √(3.6 x 10⁻⁷) ≈ 6 x 10⁻⁴ M. pH = -log[H⁺] = -log(6 x 10⁻⁴) ≈ 3.22. The closest choice is 3.37. However, it's important to verify the 5% rule: (6 x 10⁻⁴ / 0.02) * 100% = 3%. Since it's less than 5%, the approximation is valid.