AP Chemistry / Topic
Acids and Bases
Unit 8 covers pH and pOH, strong versus weak acids and bases, Ka and Kb equilibria, buffers and Henderson-Hasselbalch, titration curves, and salt hydrolysis.
What the exam asks
- Compute pH for strong acids/bases directly and for weak ones via ICE tables with Ka or Kb
- Identify conjugate acid-base pairs and compare strengths with Ka values
- Explain how buffers resist pH change and compute buffer pH
- Read titration curves: equivalence point, half-equivalence point, and indicator choice
- Predict whether a salt solution is acidic, basic, or neutral
Key formulas and rules
pH = −log[H3O+]; pOH = −log[OH−]; pH + pOH = 14 at 25 °C
Kw = [H3O+][OH−] = 1.0×10⁻¹⁴ at 25 °C; Ka·Kb = Kw
[H3O+] ≈ √(Ka·C) when x is small
pH = pKa + log([A−]/[HA])
pH = pKa (buffer region midpoint)
strong-strong: 7; weak acid + strong base: > 7; weak base + strong acid: < 7
Question bank breakdown
46
3
11 easy · 19 medium · 16 hard
Skills covered
Every question in the bank comes with a per-choice explanation, so you learn why each wrong answer is wrong — not just the key.
Sample question
From the free tier of the ScoreMint AP Chemistry bank — try it, then check the answer.
A 0.10 M solution of a weak acid HA has a pH of 3.00 at 25°C. What is the acid dissociation constant, Ka, for HA?
- A. 1.0 × 10⁻³
- B. 1.0 × 10⁻⁵
- C. 1.0 × 10⁻⁶
- D. 1.0 × 10⁻²
Show answer and explanation
Answer: B. Correct. pH = 3.00 means [H⁺] = 10⁻³·⁰⁰ = 1.0 × 10⁻³ M. The dissociation: HA ⇌ H⁺ + A⁻. At equilibrium: [H⁺] = [A⁻] = 1.0 × 10⁻³ M. [HA] = 0.10 - 0.001 = 0.099 M ≈ 0.10 M (valid since x << initial concentration). Ka = [H⁺][A⁻] / [HA] = (1.0 × 10⁻³)(1.0 × 10⁻³) / 0.10 = 1.0 × 10⁻⁶ / 0.10 = 1.0 × 10⁻⁵. ✓ Note: Using the exact [HA] = 0.099 gives Ka = 1.01 × 10⁻⁵, essentially the same.
Study tip
The fastest buffer check: a buffer needs comparable amounts of a weak conjugate pair. After a titration adds strong base, recount moles — if some weak acid and some conjugate base remain, use Henderson-Hasselbalch.