6.1.13Additional lessons
Salt Hydrolysis (HL) — IB Chemistry HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 127 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.
Part 1 of 5Why Salts Can Change pH
When a salt dissolves, its ions can react with water and shift the pH away from 7.
Hydrolysis
- Hydrolysis Hydrolysis is the reaction of an ion with water in which the pH of the solution is changed.
Spectator ions
- Spectator ions The ions of strong acids and strong bases, such as Na⁺, K⁺, Cl⁻ and NO₃⁻, do not react with water.
Conjugate base
- Conjugate base The anion of a weak acid is a base and accepts a proton from water, producing OH⁻.
Conjugate acid
- Conjugate acid The cation of a weak base is an acid and donates a proton to water, producing H₃O⁺.
Part 2 of 5Predicting the Solution
Predicting the Solution
Strong acid + strong base
- Strong acid + strong base Salts such as NaCl give a neutral solution with a pH of 7.
Weak acid + strong base
- Weak acid + strong base Salts such as CH₃COONa are basic: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻.
Strong acid + weak base
- Strong acid + weak base Salts such as NH₄Cl are acidic: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺.
Metal ions
- Metal ions Small, highly charged cations such as Al³⁺ polarise water in [Al(H₂O)₆]³⁺ and give acidic solutions.
| Salt from | Example | Solution |
|---|---|---|
| Strong acid, strong base | NaCl | Neutral |
| Weak acid, strong base | CH₃COONa | Basic |
| Strong acid, weak base | NH₄Cl | Acidic |
Part 3 of 5Working Through
1
Split the salt into its cation and anion and decide which acid and base it could have come from.
2
Ignore any ion that is the conjugate partner of a strong acid or strong base.
3
For the remaining ion, write the equation with water and identify whether OH⁻ or H₃O⁺ is formed.
4
State the pH as acidic or basic, according to the ion that is produced in excess.
5
For a quantitative question, use Kb = Kw/Ka for the anion and solve like a weak base.
Part 4 of 5Worked Example
Question Calculate the pH of 0.100 mol dm⁻³ sodium ethanoate, given Ka(CH₃COOH) = 1.74 × 10⁻⁵.
- Kb of ethanoate = 1.00 × 10⁻¹⁴ / 1.74 × 10⁻⁵ = 5.75 × 10⁻¹⁰.
- [OH⁻] = √(Kb × c) = √(5.75 × 10⁻¹¹) = 7.58 × 10⁻⁶ mol dm⁻³.
- pOH = 5.12, so pH = 14.00 − 5.12 = 8.88.
Answer pH = 8.88 (basic)
Part 5 of 5Exam Checklist & Tips
6.1.13 – Key Points for the Exam
- Hydrolysis The reaction of an ion with water that changes the pH of the solution.
- Salt An ionic compound made of a cation and an anion.
- Spectator ion An ion that does not take part in the reaction.
- Conjugate base The species left when an acid has lost a proton.
Common Mistakes
- Saying that all salts give neutral solutions.
- Writing hydrolysis equations for Na⁺ or Cl⁻.
- Choosing a basic solution for NH₄Cl.
- Forgetting that the conjugate base of a weak acid is itself a base.
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