U8.3 · Weak Acid and Base Equilibria (Ka, Kb)Practice · 弱酸弱碱的平衡（Ka、Kb）
Lessons covered:U8.3 Weak Acid and Base Equilibria (Ka, Kb)
- 1Numeric
What is the pH of 0.20 M HCN? Ka = 4.9 × 10−10.
ExplanationKa is tiny, so [H⁺] = x ≈ √(Ka·c) = √(4.9 × 10−10 × 0.20) = 9.9 × 10−6 M, only 5 × 10−3% of 0.20, so the shortcut is fine. pH = 5.00.
- 2Numeric
What is the percent ionization of acetic acid in a 0.010 M solution? Ka = 1.8 × 10−5.
%Explanationx² ÷ (0.010 − x) = 1.8 × 10−5. The shortcut x ≈ √(1.8 × 10−7) = 4.24 × 10−4 gives 4.24%; the quadratic gives x = 4.15 × 10−4, so percent ionization = 4.15 × 10−4 ÷ 0.010 × 100% = 4.15%. Both are under 5% and close to each other.
- 3Numeric
What is the pH of 0.30 M aqueous ammonia at 25 °C? Kb = 1.8 × 10−5.
ExplanationNH₃ + H₂O ⇌ NH₄⁺ + OH⁻, so [OH⁻] ≈ √(1.8 × 10−5 × 0.30) = 2.31 × 10−3 M, pOH = 2.64, and pH = 14.00 − 2.64 = 11.36. For a weak base you find OH⁻ first; don’t forget to convert to pH.
- 4Numeric
At 25 °C, a 0.10 M solution of a monoprotic weak acid HA has pH = 2.85. What is Ka for HA?
× 10Explanation[H⁺] = [A⁻] = 10−2.85 = 1.41 × 10−3 M, and [HA] = 0.10 − 1.41 × 10−3 ≈ 0.0986 M. Ka = (1.41 × 10−3)² ÷ 0.0986 = 2.0 × 10−5.
- 5Multiple choice
A 0.10 M weak acid solution is diluted with water to 0.010 M. Which statement is correct?
ExplanationDilution lowers Q = [H⁺][A⁻]/[HA] (both concentrations on top shrink), so Q < Ka and the ionization shifts right: the percent ionization goes up. Overall [H⁺] still falls, so the pH rises, but by less than 1 (about 0.5). Ka changes only with temperature.
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