6.1.11Additional lessons
Acid and Base Dissociation Constants (HL) — IB Chemistry HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 125 of 172
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.
Section 1 of 5Ka and Kb
Equilibrium constants for weak acids and bases measure their strength quantitatively.
Weak acid
- Weak acid For HA ⇌ H⁺ + A⁻, the acid dissociation constant is Ka = [H⁺][A⁻] / [HA].
Weak base
- Weak base For B + H₂O ⇌ BH⁺ + OH⁻, the base dissociation constant is Kb = [BH⁺][OH⁻] / [B].
Water omitted
- Water omitted Water is the solvent and its concentration is effectively constant, so it does not appear in either expression.
Strength
- Strength A larger Ka or Kb means a stronger acid or base, and the constants are given at a stated temperature.
Section 2 of 5pKa and pKb
pKa and pKb
Definitions
- Definitions pKa = −log₁₀ Ka and pKb = −log₁₀ Kb, so a smaller pK value means a stronger acid or base.
Reverse
- Reverse Ka = 10⁻ᵖᴷᵃ, which allows a conversion between the two forms.
Conjugate pair
- Conjugate pair For a conjugate acid–base pair, Ka × Kb = Kw, which is 1.00 × 10⁻¹⁴ at 298 K.
pK relation
- pK relation Taking logarithms gives pKa + pKb = 14.00 at 298 K.
| Strength of acid | Ka | pKa | Conjugate base |
|---|---|---|---|
| Stronger | Larger | Smaller | Weaker |
| Weaker | Smaller | Larger | Stronger |
Section 3 of 5Working with Constants
1
Write the equilibrium equation for the acid or base in water, and then the expression for Ka or Kb.
2
Use the data booklet value for Ka or pKa, converting between them with the logarithm.
3
To find the Kb of the conjugate base, divide Kw by Ka.
4
To compare two acids, compare their pKa values: the smaller value is the stronger acid.
5
Remember that the constants change with temperature, as for any equilibrium constant.
Section 4 of 5Worked Example
Question
- The pKa of ethanoic acid is 4.76.
- Calculate Ka and the Kb of the ethanoate ion at 298 K.
- Ka = 10⁻⁴·⁷⁶ = 1.74 × 10⁻⁵ mol dm⁻³.
- Kb = Kw / Ka = 1.00 × 10⁻¹⁴ / 1.74 × 10⁻⁵ = 5.75 × 10⁻¹⁰ mol dm⁻³.
- As a check, pKb = 14.00 − 4.76 = 9.24.
Answer
- Ka = 1.74 × 10⁻⁵;
- Kb = 5.75 × 10⁻¹⁰ mol dm⁻³
Section 5 of 5Exam Checklist & Tips
6.1.11 – Key Points for the Exam
- Ka The acid dissociation constant, [H⁺][A⁻]/[HA].
- Kb The base dissociation constant, [BH⁺][OH⁻]/[B].
- pKa The negative logarithm of Ka.
- Conjugate pair relationship Ka × Kb = Kw.
Common Mistakes
- Including [H₂O] in the Ka or Kb expression.
- Thinking that a higher pKa means a stronger acid.
- Multiplying Ka by Kw instead of dividing when finding Kb.
- Forgetting that Ka × Kb = Kw only holds for a conjugate pair.
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