Acids and Bases — AP Chemistry Unit 8 practice.
This unit covers pH calculations, strong vs weak acids, buffers and titrations — essential concepts for AP Chemistry. Use our interactive study games to test your understanding, or review questions in traditional format below.
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This unit covers pH calculations, strong vs weak acids, buffers and titrations — essential concepts for AP Chemistry. Use our interactive study games to test your understanding, or review questions in traditional format below.
Key Concepts Breakdown
1 pH Calculations
pH is defined as -log[H⁺], and pOH as -log[OH⁻], with pH + pOH = 14 at 25°C. Students must be able to convert between [H⁺], [OH⁻], pH, and pOH in both directions. Understanding that a one-unit change in pH represents a tenfold change in [H⁺] is frequently tested.
Key Points
- pH = -log[H⁺]; pOH = -log[OH⁻]; pH + pOH = 14 (at 25°C)
- For strong acids/bases, [H⁺] or [OH⁻] equals the molar concentration directly
- Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C
- Acidic solution: pH < 7; basic: pH > 7; neutral: pH = 7 (at 25°C)
Calculate the pH of a 0.050 M HCl solution.
HCl is a strong acid and dissociates completely, so [H⁺] = 0.050 M. Applying pH = -log(0.050) gives pH = -log(5.0 × 10⁻²) = 2 - log(5.0) ≈ 1.30. No equilibrium calculation is needed because 100% dissociation is assumed.
2 Strong vs Weak Acids
Strong acids dissociate completely in water; weak acids establish an equilibrium described by Ka. Students must be able to set up and solve ICE tables for weak acid equilibria and calculate pH from Ka and initial concentration. The relationship pKa = -log(Ka) and the relative strength implied by Ka magnitude are both tested.
Key Points
- Strong acids (HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄, HClO₃): assume 100% dissociation
- Weak acid equilibrium: HA ⇌ H⁺ + A⁻; Ka = [H⁺][A⁻]/[HA]
- Use ICE table: if x << initial concentration, simplify; check that x < 5% of initial
- Larger Ka (smaller pKa) = stronger acid; comparing Ka values determines relative strength
Find the pH of 0.10 M acetic acid (Ka = 1.8 × 10⁻⁵).
Set up ICE: [H⁺] = [CH₃COO⁻] = x and [CH₃COOH] ≈ 0.10 − x ≈ 0.10 (assuming x is small). Solving Ka = x²/0.10 gives x = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ M. pH = -log(1.34 × 10⁻³) ≈ 2.87; verify the 5% approximation: 1.34 × 10⁻³/0.10 = 1.3%, so the approximation is valid.
3 Buffers
A buffer resists pH change and consists of a weak acid and its conjugate base (or weak base and conjugate acid) in comparable concentrations. The Henderson-Hasselbalch equation, pH = pKa + log([A⁻]/[HA]), is the primary calculation tool on the exam. Students must also be able to determine how adding strong acid or base shifts a buffer and calculate the new pH.
Key Points
- Buffer pH = pKa + log([conjugate base]/[weak acid]); optimal buffering when [A⁻] = [HA], so pH = pKa
- Adding strong acid converts A⁻ → HA; adding strong base converts HA → A⁻
- Buffer capacity is greatest when concentrations of both components are high and roughly equal
- Buffer breaks down when enough strong acid/base is added to consume one component entirely
A buffer contains 0.20 mol CH₃COOH and 0.30 mol CH₃COONa in 1.0 L. What is the pH? (pKa = 4.74)
Apply Henderson-Hasselbalch: pH = 4.74 + log(0.30/0.20) = 4.74 + log(1.5) = 4.74 + 0.18 = 4.92. Because the ratio of base to acid is greater than 1, the pH is above the pKa, which is the expected qualitative check. No ICE table is required when both components are present in significant amounts.
4 Titrations
Titrations involve the systematic neutralization of an acid or base, and students must be able to calculate pH at four key points: before the titrant is added, before the equivalence point, at the equivalence point, and after the equivalence point. Strong acid–strong base titrations yield pH = 7 at equivalence; weak acid–strong base titrations yield a basic equivalence point due to the conjugate base hydrolyzing. The half-equivalence point of a weak acid titration occurs where pH = pKa.
Key Points
- Equivalence point: moles of titrant added = moles of analyte; for strong/strong, pH = 7
- For weak acid titrated with strong base: at equivalence, solution contains conjugate base A⁻, pH > 7
- Half-equivalence point: exactly half the weak acid is neutralized, [HA] = [A⁻], so pH = pKa
- Indicator choice: endpoint color change should occur near the equivalence point pH
25.00 mL of 0.100 M acetic acid is titrated with 0.100 M NaOH. What is the pH at the half-equivalence point and at the equivalence point? (Ka = 1.8 × 10⁻⁵, pKa = 4.74)
At the half-equivalence point, 12.50 mL NaOH has been added; half the acetic acid is converted to acetate, so [CH₃COOH] = [CH₃COO⁻] and pH = pKa = 4.74 directly. At the equivalence point, 25.00 mL NaOH has been added, all acid is converted to 0.050 M CH₃COO⁻ in ~50.00 mL solution; using Kb = Kw/Ka = 5.6 × 10⁻¹⁰ in an ICE table gives [OH⁻] ≈ 5.3 × 10⁻⁶ M, pOH ≈ 5.28, and pH ≈ 8.72, confirming the basic equivalence point.
Questions, answered.
What is Acids and Bases?
Acids and Bases is Unit 8 of AP Chemistry, covering pH calculations, strong vs weak acids, buffers and titrations.
How to study for AP Chemistry Unit 8?
Start with the Quick Summary above, review the Key Concepts, then test yourself with our interactive study games. Aim for 80%+ accuracy before moving on.
How many questions are in this unit?
This unit has 30+ review questions across 5 different game modes.