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6.1.12Additional lessons

Solving Acid-Base Dissociation Problems (HL) — IB Chemistry HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 126 of 172

📚 IB Chemistry HL Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Part 1 of 5The Approximation

Because dissociation is slight, a simple approximation gives the hydrogen ion concentration.

Set-up
  • Set-up For HA ⇌ H⁺ + A⁻, the dissociated amount is x, so [H⁺] = [A⁻] = x and [HA] = c − x.
Water ignored
  • Water ignored The H⁺ from the autoionisation of water is negligible compared with that from the acid.
Small x
  • Small x Since a weak acid dissociates only slightly, [HA] ≈ c, which simplifies Ka to x²/c.
Result
  • Result Rearranging gives [H⁺] = √(Ka × c), from which pH = −log₁₀[H⁺].

Part 2 of 5Weak Bases and Checks

Weak Bases and Checks
Weak base
  • Weak base In the same way, for a weak base [OH⁻] = √(Kb × c), then pOH is found and pH = 14.00 − pOH.
Check
  • Check The approximation is valid when the percentage dissociation is below about 5%.
Percentage dissociation
  • Percentage dissociation This equals (x / c) × 100, the fraction of acid molecules that have ionised.
Dilution
  • Dilution As a weak acid is diluted, the percentage dissociation increases, even though [H⁺] decreases.
TypeApproximate formula
Weak acid[H⁺] = √(Ka × c)
Weak base[OH⁻] = √(Kb × c)
Percentage dissociation(x / c) × 100

Part 3 of 5Procedure

1
Write the equilibrium and the expression for Ka or Kb, then define x as the amount that dissociates.
2
State the assumptions: the ions have equal concentration, and the change in the original concentration is negligible.
3
Solve for x, which is [H⁺] for an acid or [OH⁻] for a base.
4
Convert to the pH, using pOH and the relation pH + pOH = 14.00 for a base.
5
Check the assumption by calculating the percentage dissociation.

Part 4 of 5Worked Example

Question Calculate the pH and the percentage dissociation of 0.100 mol dm⁻³ ethanoic acid (Ka = 1.74 × 10⁻⁵ mol dm⁻³).

  1. Ka = x² / 0.100, so x² = 1.74 × 10⁻⁶ and x = 1.32 × 10⁻³ mol dm⁻³.
  2. pH = −log₁₀(1.32 × 10⁻³) = 2.88.
  3. Percentage dissociation = (1.32 × 10⁻³ / 0.100) × 100 = 1.3%, so the approximation is valid.

Answer pH = 2.88; 1.3% dissociated

Part 5 of 5Exam Checklist & Tips

6.1.12 – Key Points for the Exam

  • Percentage dissociation The proportion of the original acid molecules that have ionised, as a percentage.
  • Approximation Treating the equilibrium concentration of the weak acid as equal to its initial concentration.
  • Weak base A base that reacts only partly with water to form OH⁻.
  • x The concentration of H⁺ (or OH⁻) produced by dissociation.

Common Mistakes

  • Using the strong acid method [H⁺] = c for a weak acid.
  • Forgetting to take the square root of Ka × c.
  • Taking the pH directly from [OH⁻] for a weak base without converting via pOH.
  • Using the approximation when the dissociation is a large fraction of the acid.

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