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Chemistry Revision Guides

C2: Equilibria, Kp, pH and Buffer Solutions

Year 1 / ASYear 2 / A-Level All Boards (AQA, Edexcel, OCR, WJEC, CCEA) AQA

Gas equilibria Kp, acid-base theories, pH calculations, Ka, Kw, buffer mechanisms, titration curves.

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📋 Key Definitions and Core Concepts

Buffer Solution: A solution that maintains approximately constant pH when small quantities of acid or base are added.
Acid Dissociation Constant (Ka): Ka = [H⁺][A⁻]/[HA]; pKa = -log₁₀ Ka.

🔍 Key Principles & Specification Requirements

  • Kp expressions use partial pressures: pA = (mole fraction A) × P_total. Temperature is the only variable that alters Kp.
  • Weak acid pH: [H⁺] ≈ √(Ka · [HA]) => pH = -log₁₀[H⁺].
  • Buffer pH: [H⁺] = Ka · ([HA]/[A⁻]) => pH = pKa + log₁₀([A⁻]/[HA]).

💡 Worked Example Question

Exam-Style Question

Question:

Calculate the pH of a buffer with 0.20 M ethanoic acid (Ka = 1.74 × 10⁻⁵) and 0.15 M sodium ethanoate.

Model Solution & Mark Scheme:

[H⁺] = Ka × ([HA]/[A⁻]) = (1.74 × 10⁻⁵) × (0.20 / 0.15) = 2.32 × 10⁻⁵ M.
pH = -log₁₀(2.32 × 10⁻⁵) = 4.63.

❓ Practice Questions & Mark Schemes

Q1: Explain buffer action when small HCl is added to CH₃COOH/CH₃COO⁻.

Show Model Answer

Answer: Added H⁺ reacts with ethanoate ions: CH₃COO⁻ + H⁺ → CH₃COOH. Equilibrium shifts left, removing excess H⁺ and keeping pH stable.

Q2: Calculate the pH of 0.050 M Ba(OH)₂ at 298 K (Kw = 1.0 × 10⁻¹⁴).

Show Model Answer

Answer: [OH⁻] = 2 × 0.050 = 0.10 M. [H⁺] = 1.0 × 10⁻¹⁴ / 0.10 = 1.0 × 10⁻¹³ M => pH = 13.00.

🎬 Video Resources

📄 Past Papers & Exam Resources

🔗 Further Reading & Resources