AP Chemistry Unit 8 review: Acids and Bases
Unit 8 applies equilibrium to acids, bases, buffers and titrations. It is 11–15% of multiple choice. Expect to identify the species present before choosing a pH calculation, read titration curves and explain how molecular structure or added acid changes equilibrium.
Unit 8 on the May 2027 exam
| 2027 exam | |
|---|---|
| Weight | 11–15% of the multiple-choice section |
| Exam date | Thursday, May 6, 2027, afternoon session |
| Multiple choice | 60 questions in 90 minutes, worth 50% of the exam score |
| Free response | 7 questions in 105 minutes, worth 50%: three long questions (10 points each) and four short questions (4 points each). |
| Calculator | Both sections allow scientific or approved graphing calculators. Four-function calculators are allowed but not recommended. Bluebook includes both Desmos scientific and graphing calculators. |
| Class time | About 14–16 class periods in College Board's pacing guide |
| 2026 results | 15% of students scored a 5 on the whole exam, and about 76% scored 3 or higher |
Sources: College Board's course and exam description, exam page, 2027 exam dates and score distributions.
What Unit 8 covers
College Board's course description splits the unit into 11 topics. The second column is what you should be able to do for each one.
| Topic | What you should be able to do |
|---|---|
| 8.1 Introduction to Acids and Bases | Identify Brønsted–Lowry acids, bases and conjugate pairs. |
| 8.2 pH and pOH of Strong Acids and Bases | Calculate pH and pOH from complete dissociation of strong acids and bases. |
| 8.3 Weak Acid and Base Equilibria | Use Ka or Kb to calculate weak-acid or weak-base equilibrium concentrations. |
| 8.4 Acid-Base Reactions and Buffers | Account for neutralization before applying a buffer equilibrium calculation. |
| 8.5 Acid-Base Titrations | Identify titration regions, equivalence, half-equivalence and suitable indicators. |
| 8.6 Molecular Structure of Acids and Bases | Explain acid strength from bond polarity, bond strength and conjugate-base stability. |
| 8.7 pH and pKa | Relate pH, pKa and the relative amounts of an acid and its conjugate base. |
| 8.8 Properties of Buffers | Explain how a buffer reacts with added strong acid or base. |
| 8.9 Henderson-Hasselbalch Equation | Calculate buffer pH using the conjugate-base-to-acid ratio. |
| 8.10 Buffer Capacity | Compare buffers’ resistance to added acid or base from component amounts. |
| 8.11 pH and Solubility | Predict how acid–base reactions alter the solubility of salts. |
Choose the pH method from the species left after reaction
First carry out any nearly complete neutralization using moles. Then identify the equilibrium that controls the remaining solution.
| Solution or titration region | Method | Check |
|---|---|---|
| Strong acid or base in excess | Remaining H₃O⁺ or OH⁻ moles / total volume | Use pH + pOH = 14.00 only at 25 °C. |
| Weak acid alone | Ka = x²/(C − x) | Check x/C before using a small-x approximation. |
| Buffer: HA and A⁻ present | pH = pKa + log([A⁻]/[HA]) | Both buffer components must remain. |
| Half-equivalence, weak acid + strong base | pH = pKa | Half the original acid has become conjugate base. |
| Equivalence, weak acid + strong base | Kb = Kw/Ka for A⁻ hydrolysis | The solution is basic at 25 °C, not necessarily pH 7. |
| Beyond equivalence | Calculate excess strong base first | The excess OH⁻ usually controls pH. |
| Indicator choice | Transition range lies in the steep pH change | Endpoint is an observation; equivalence is a mole relationship. |
Common mistakes
Each of these loses points on questions you otherwise understand.
- Treating half-equivalence, equivalence and endpoint as the same point. Half-equivalence sets equal weak-acid and conjugate-base amounts; equivalence matches stoichiometry; endpoint is the indicator response. CED printed p. 143.
- Choosing an indicator only because its color is familiar. Its transition range must fit the sharp part of the titration curve. CED p. 143.
- Using Henderson–Hasselbalch before neutralizing added strong acid or base. Update buffer moles first. CED topics 8.4 and 8.9.
- Equating strong with concentrated. Strength concerns ionization equilibrium; concentration is amount per volume. CED topics 8.1–8.3.
6 practice questions with answers
Written by Ashore for this page, in the style of the multiple-choice section. Try each one before you open the answer.
Question 1 · 8.2
At 25 °C, a solution contains 0.00250 M HCl. What is its pH?
- A2.60
- B11.40
- C3.60
- D2.40
Show the answer
A. HCl dissociates essentially completely, so [H₃O⁺] = 0.00250 M. pH = −log(0.00250) = 2.60. The pOH would be 11.40.
Question 2 · 8.3
A 0.100 M weak acid HA has Ka = 1.0 × 10⁻⁵. Using a justified small-x approximation, what is [H₃O⁺]?
- A1.0 × 10⁻⁶ M
- B1.0 × 10⁻⁴ M
- C1.0 × 10⁻³ M
- D1.0 × 10⁻² M
Show the answer
C. Ka ≈ x²/C gives x ≈ √(1.0 × 10⁻⁵ × 0.100) = 1.0 × 10⁻³ M. That is about 1% of the initial acid concentration, so the approximation is reasonable.
Question 3 · 8.9
A buffer has pKa = 4.80, [A⁻] = 0.200 M and [HA] = 0.100 M. What is its pH?
- A4.50
- B5.10
- C4.80
- D9.60
Show the answer
B. pH = 4.80 + log(0.200/0.100) = 4.80 + 0.301 = 5.10. More conjugate base than acid makes the pH higher than pKa.
Question 4 · 8.4
A buffer contains 0.0300 mol HA and 0.0200 mol A⁻. After adding 0.00500 mol HCl, what is the A⁻/HA mole ratio?
- A0.0250/0.0250
- B0.0200/0.0350
- C0.0250/0.0300
- D0.0150/0.0350
Show the answer
D. H⁺ reacts with A⁻ to form HA. A⁻ falls to 0.0150 mol while HA rises to 0.0350 mol. Since both share the same final volume, their mole ratio equals their concentration ratio.
Question 5 · 8.5
A 20.0 mL sample of 0.150 M monoprotic weak acid is titrated with 0.100 M NaOH. At what added volume is half-equivalence reached?
- A15.0 mL
- B20.0 mL
- C30.0 mL
- D7.50 mL
Show the answer
A. The acid amount is 0.0200(0.150) = 0.00300 mol. Equivalence needs 30.0 mL of base, so half-equivalence occurs at 15.0 mL. At that point pH = pKa under the buffer approximation.
Question 6 · 8.2
At 25 °C, 10.0 mL of 0.100 M HCl is mixed with 15.0 mL of 0.100 M NaOH. What is the final pH?
- A1.70
- B7.00
- C12.30
- D13.00
Show the answer
C. Initial amounts are 0.00100 mol H⁺ and 0.00150 mol OH⁻. Excess OH⁻ is 0.00050 mol in 0.0250 L, or 0.020 M. pOH = 1.70 and pH = 14.00 − 1.70 = 12.30.
How to study Unit 8
Divide a weak-acid titration curve into initial acid, buffer, equivalence and excess-base regions. For each region, list the dominant species and write the equation that determines pH. This prevents applying the buffer equation at equivalence.
Practice one buffer-addition calculation entirely in moles until neutralization is finished. Then calculate the new ratio. Compare two buffers with the same ratio but different total concentrations to separate pH from capacity.
Ashore's Chemistry course has a short lesson for every topic in this unit, each ending with a few questions on what it taught.
- AP Chemistry: all 9 units, exam format and scores
- AP Chemistry Unit 1: Atomic Structure and Properties
- AP Chemistry Unit 2: Compound Structure and Properties
- AP Chemistry Unit 3: Properties of Substances and Mixtures
- AP Chemistry Unit 4: Chemical Reactions
- AP Chemistry Unit 5: Kinetics
- AP Chemistry Unit 6: Thermochemistry
- AP Chemistry Unit 7: Equilibrium
- AP Chemistry Unit 9: Thermodynamics and Electrochemistry
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