Solubility Product and Common Ion Effect
Equilibrium: Solubility Product and Common Ion Effect
Solubility Product and Common Ion Effect
Solubility Product and Common Ion Effect
What you'll learn
- Define Ksp and write its expression for any sparingly soluble salt
- Relate molar solubility s to Ksp for different salt formulae
- Predict whether precipitation occurs using the ionic product Q
- Explain and calculate the common ion effect on solubility
- Apply fractional precipitation to separate two ions selectively
- Solve multi-step precipitation and solubility problems
Key concepts
Level 1 — Foundations
Solubility equilibrium: When a sparingly soluble salt dissolves to saturation: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
Ksp = [Ag⁺][Cl⁻]
The solid AgCl is not included (pure solid, activity = 1).
General Ksp expressions:
| Salt formula | Dissolution | Ksp |
|---|---|---|
| AB (1:1) | A⁺ + B⁻ | [A⁺][B⁻] |
| AB₂ (1:2) | A²⁺ + 2B⁻ | [A²⁺][B⁻]² |
| A₂B (2:1) | 2A⁺ + B²⁻ | [A⁺]²[B²⁻] |
| A₂B₃ (2:3) | 2A³⁺ + 3B²⁻ | [A³⁺]²[B²⁻]³ |
Level 2 — JEE depth
Molar solubility s from Ksp:
For AgCl: s = [Ag⁺] = [Cl⁻] Ksp = s × s = s² → s = √Ksp
For BaCl₂: BaCl₂ ⇌ Ba²⁺ + 2Cl⁻ [Ba²⁺] = s, [Cl⁻] = 2s Ksp = s × (2s)² = 4s³ → s = (Ksp/4)^(1/3)
For Ca₃(PO₄)₂: Ca₃(PO₄)₂ ⇌ 3Ca²⁺ + 2PO₄³⁻ [Ca²⁺] = 3s, [PO₄³⁻] = 2s Ksp = (3s)³(2s)² = 27s³ × 4s² = 108s⁵ → s = (Ksp/108)^(1/5)
Precipitation condition — Ionic Product Q: Q = [A^n+][B^m-] (using actual concentrations, not equilibrium)
| Condition | Meaning |
|---|---|
| Q < Ksp | Unsaturated; more salt can dissolve |
| Q = Ksp | Saturated; at equilibrium |
| Q > Ksp | Supersaturated; precipitate forms |
Common ion effect: Adding a common ion (one already in the Ksp expression) shifts equilibrium left → less salt dissolves → solubility decreases.
Example: Solubility of AgCl in 0.1 M NaCl: AgCl ⇌ Ag⁺ + Cl⁻ Let solubility = s; [Ag⁺] = s, [Cl⁻] = 0.1 + s ≈ 0.1 (since s << 0.1) Ksp = s × 0.1 → s = Ksp/0.1
Compare to pure water: s_pure = √Ksp >> Ksp/0.1 (much larger)
Fractional precipitation: Two ions in solution precipitate at different Q > Ksp thresholds as a precipitating agent is added. The one with smaller Ksp precipitates first (lower [precipitating agent] needed).
Example: Separating Cl⁻ and CrO₄²⁻ by adding Ag⁺: AgCl starts precipitating when [Ag⁺] = Ksp(AgCl)/[Cl⁻] Ag₂CrO₄ starts when [Ag⁺]² = Ksp(Ag₂CrO₄)/[CrO₄²⁻] Compare these [Ag⁺] thresholds — whichever is smaller triggers first.
Effect of pH on solubility: Salts of weak acids (e.g., CaCO₃) dissolve more in acidic solution because H⁺ reacts with CO₃²⁻, removing it from equilibrium → Q drops below Ksp → more dissolves.
JEE trap: s = √Ksp is ONLY for 1:1 salts like AgCl. For any other formula, re-derive from the dissolution equation.
JEE trap: Do not equate solubility (g/L or mol/L) directly with Ksp without converting through the stoichiometric relationship.
Worked example
Ksp(AgCl) = 1.8×10⁻¹⁰ — find molar solubility in pure water
AgCl ⇌ Ag⁺ + Cl⁻
If s = molar solubility:
[Ag⁺] = s, [Cl⁻] = s
Ksp = [Ag⁺][Cl⁻] = s²
s² = 1.8×10⁻¹⁰
s = √(1.8×10⁻¹⁰)
s = 1.342×10⁻⁵ mol/L
Answer: s ≈ 1.34×10⁻⁵ mol/L
Solubility of AgCl in 0.1 M NaCl solution
AgCl ⇌ Ag⁺ + Cl⁻
[Cl⁻] from NaCl = 0.1 M (common ion)
Let additional Ag⁺ from AgCl dissolution = s
[Ag⁺] = s, [Cl⁻] = 0.1 + s ≈ 0.1 (since s << 0.1)
Ksp = [Ag⁺][Cl⁻]
1.8×10⁻¹⁰ = s × 0.1
s = 1.8×10⁻¹⁰ / 0.1 = 1.8×10⁻⁹ mol/L
Comparison:
- In pure water: s = 1.34×10⁻⁵ mol/L
- In 0.1 M NaCl: s = 1.80×10⁻⁹ mol/L
Solubility reduced by a factor of ~7500 due to common ion effect.
Answer: s ≈ 1.8×10⁻⁹ mol/L (about 10,000× less than pure water)
Common mistakes
| Mistake | Why it happens | Fix |
|---|---|---|
| s = √Ksp for all salts | Formula memorised without derivation | Always write dissolution equation and set up [ion] = f(s) before solving |
| Not accounting for stoichiometry in [Cl⁻] = 2s for BaCl₂ | Assuming 1:1 for every salt | Read the formula: BaCl₂ gives 2 Cl⁻ per formula unit |
| Comparing Ksp values directly to rank solubility | Works only for same formula type | AB > AB > AB: same type OK; AB vs AB₂: must compute s from Ksp first |
| Thinking common ion always makes s = 0 | Common ion greatly reduces s but doesn't eliminate dissolution | s = Ksp / [common ion]; always a small but non-zero solubility |
Quick check
- Q1: Write Ksp for PbCl₂ and find s if Ksp = 1.7×10⁻⁵.
- Q2: Ksp(BaSO₄) = 1.1×10⁻¹⁰. What is the solubility in g/L? (M = 233 g/mol)
- Q3: Will a precipitate form if 100 mL of 1×10⁻³ M BaCl₂ is mixed with 100 mL of 1×10⁻³ M Na₂SO₄? (Ksp(BaSO₄) = 1.1×10⁻¹⁰)
- Q4: Find solubility of Mg(OH)₂ (Ksp = 5.6×10⁻¹²) in a solution buffered at pH = 9.
- Stretch: Q5: A solution contains 0.1 M each of Cl⁻ and CrO₄²⁻. Ag⁺ is added dropwise. At what [Ag⁺] does each precipitate begin to form? Which precipitates first? (Ksp(AgCl) = 1.8×10⁻¹⁰, Ksp(Ag₂CrO₄) = 1.2×10⁻¹²)
NCERT Chapter 7 link: Chapter 7 (Class 11) — Sections 7.15 and 7.16 cover Ksp, molar solubility, common ion effect, and precipitation. Solved examples 7.28–7.32 illustrate exactly the types of problems that appear in JEE.
Exam connections: JEE Mains tests Ksp to s conversions (all formulae types), Q vs Ksp precipitation decisions, and common ion effect. JEE Advanced tests fractional precipitation ordering and pH-dependent solubility changes.
Study strategy: Make a formula card: for each salt type (AB, AB₂, A₂B, A₂B₃), write the dissolution and the s-Ksp formula. Practise deriving, not memorising. Then do precipitation prediction problems by always computing Q first.
Interactive Exploration Suggestions (Drishti Live Worlds)
- Use the platform-native live simulation or PhET-style tool for this topic.
- Mirror / body / home activity: dissolve salt until saturated, then add a common-ion salt and observe crystallisation — photograph and explain in terms of Q > Ksp.
- Voice or text reflection with AI Mentor: explain the concept to a younger student or family member.
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Key Takeaways (TL;DR)
- What you'll learn
- Key concepts
- Worked example
- Common mistakes
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