The Hook: Why Do Caves Have Stalactites?
Stalactites and stalagmites in caves form because of solubility equilibrium! Water containing dissolved CaCO₃ drips from the ceiling. When CO₂ escapes, equilibrium shifts, and CaCO₃ precipitates—building these magnificent formations over thousands of years.
Other everyday examples:
- Hard water: CaCO₃ and Mg(OH)₂ precipitate in pipes (low solubility!)
- Kidney stones: Calcium oxalate crystals form when Ksp is exceeded
- Toothpaste with fluoride: CaF₂ precipitates on teeth, making them stronger
- Photography: AgBr (low Ksp) was used in film
The question: How do we predict when a solid will dissolve or precipitate?
JEE Reality: Solubility product is a high-yield topic (2-4 questions/year). It’s the perfect blend of equilibrium, ionic chemistry, and mathematical calculations. Master Ksp, and qualitative analysis becomes easy!
The Core Concept
What is Solubility Product (Ksp)?
For a sparingly soluble salt (one that barely dissolves):
$$A_xB_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq)$$At equilibrium (saturated solution):
$$\boxed{K_{sp} = [A^{y+}]^x [B^{x-}]^y}$$Key points:
- Solid is NOT included in the expression (activity = 1)
- Only applies to saturated solutions (solid in equilibrium with ions)
- Ksp is temperature-dependent (like all K values)
- Smaller Ksp → Lower solubility (harder to dissolve)
In simple terms: Ksp tells us how much of a solid can dissolve before it starts precipitating.
Relationship Between Ksp and Solubility (S)
Solubility (S): Maximum amount of solute that dissolves in a given amount of solvent at a particular temperature.
Units: mol/L, g/L, or g/100g water
Case 1: AB Type Salt
$$AB(s) \rightleftharpoons A^+(aq) + B^-(aq)$$If S mol/L dissolves:
- [A⁺] = S
- [B⁻] = S
Examples: AgCl, BaSO₄, PbSO₄
Case 2: AB₂ Type Salt
$$AB_2(s) \rightleftharpoons A^{2+}(aq) + 2B^-(aq)$$If S mol/L dissolves:
- [A²⁺] = S
- [B⁻] = 2S
Examples: CaF₂, PbCl₂, Mg(OH)₂
Case 3: A₂B Type Salt
$$A_2B(s) \rightleftharpoons 2A^+(aq) + B^{2-}(aq)$$If S mol/L dissolves:
- [A⁺] = 2S
- [B²⁻] = S
Examples: Ag₂S, Ag₂CrO₄
Case 4: A₂B₃ Type Salt
$$A_2B_3(s) \rightleftharpoons 2A^{3+}(aq) + 3B^{2-}(aq)$$If S mol/L dissolves:
- [A³⁺] = 2S
- [B²⁻] = 3S
Examples: Al₂(SO₄)₃, Fe₂(SO₄)₃
Memory Tricks & Patterns
Mnemonic for Ksp-S Relationships
| Salt Type | Formula | Ksp-S Relation | Example |
|---|---|---|---|
| AB | 1:1 | $K_{sp} = S^2$ | AgCl |
| AB₂ or A₂B | 1:2 or 2:1 | $K_{sp} = 4S^3$ | CaF₂, Ag₂S |
| AB₃ or A₃B | 1:3 or 3:1 | $K_{sp} = 27S^4$ | AlF₃ |
| A₂B₃ | 2:3 | $K_{sp} = 108S^5$ | Al₂(SO₄)₃ |
Pattern Recognition:
- Count total ions (x+y)
- Power of S = (x+y)
- Coefficient = Product of (coefficient × each ion)^(its power)
Example: AB₂ → Total ions = 1+2 = 3
- S³ term
- Coefficient: 1 × 2² = 4
- Ksp = 4S³
Mnemonic for Precipitation Prediction
“Quotient Questions Ksp: If Greater, Precipitate!”
- Q (ionic product) vs Ksp
- If Q > Ksp → Precipitation occurs
- If Q = Ksp → Saturated (equilibrium)
- If Q < Ksp → Unsaturated (more can dissolve)
Common Ion Effect on Solubility
Common Ion Effect: The solubility of a sparingly soluble salt decreases when a common ion is added.
Why Does This Happen?
Le Chatelier’s Principle in action!
$$AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$$Add NaCl (provides Cl⁻):
- [Cl⁻] increases
- Equilibrium shifts left (toward solid)
- [Ag⁺] decreases
- Solubility decreases
Quantitative Treatment
For AgCl in pure water:
$$K_{sp} = S^2$$ $$S = \sqrt{K_{sp}}$$For AgCl in 0.1 M NaCl:
Let solubility = s (much smaller than S)
- [Ag⁺] = s
- [Cl⁻] = 0.1 + s ≈ 0.1 (since s « 0.1)
Since Ksp is constant, s « S
Solubility in presence of common ion is much lower!
Ionic Product (Q) vs Ksp
Ionic Product (Q): Same expression as Ksp but for non-equilibrium conditions.
$$Q = [A^{y+}]^x [B^{x-}]^y$$(at any instant)
Decision Tree: Will Precipitation Occur?
| Comparison | Meaning | Result |
|---|---|---|
| Q < Ksp | Solution is unsaturated | No precipitation, more can dissolve |
| Q = Ksp | Solution is saturated | At equilibrium, no net change |
| Q > Ksp | Solution is supersaturated | Precipitation occurs until Q = Ksp |
Precipitation questions always use Q vs Ksp comparison!
Steps:
- Calculate Q from given ion concentrations
- Compare Q with Ksp
- If Q > Ksp → Precipitation occurs
- If Q < Ksp → No precipitation
Example: Will AgCl precipitate if [Ag⁺] = 10⁻⁴ M and [Cl⁻] = 10⁻⁶ M?
Given: Ksp(AgCl) = 1.8×10⁻¹⁰
Q = [Ag⁺][Cl⁻] = (10⁻⁴)(10⁻⁶) = 10⁻¹⁰
Q = 10⁻¹⁰ > Ksp = 1.8×10⁻¹⁰?
Actually, 10⁻¹⁰ < 1.8×10⁻¹⁰
No precipitation! (Q < Ksp)
Selective Precipitation
Concept: Using differences in Ksp to separate ions by sequential precipitation.
How It Works
If a solution contains multiple cations (like Ag⁺ and Pb²⁺), add an anion (like Cl⁻) slowly:
- The salt with lower Ksp precipitates first
- Continue adding until that ion is essentially removed
- Then the second salt starts precipitating
Example: Separating Ag⁺ and Pb²⁺
Solution contains: 0.01 M Ag⁺ and 0.01 M Pb²⁺
Given:
- Ksp(AgCl) = 1.8×10⁻¹⁰
- Ksp(PbCl₂) = 1.7×10⁻⁵
Add Cl⁻ slowly:
For AgCl to precipitate:
$$[Cl^-] = \frac{K_{sp}}{[Ag^+]} = \frac{1.8 \times 10^{-10}}{0.01} = 1.8 \times 10^{-8} \text{ M}$$For PbCl₂ to precipitate:
$$[Cl^-]^2 = \frac{K_{sp}}{[Pb^{2+}]} = \frac{1.7 \times 10^{-5}}{0.01} = 1.7 \times 10^{-3}$$ $$[Cl^-] = \sqrt{1.7 \times 10^{-3}} = 4.1 \times 10^{-2} \text{ M}$$Result:
- AgCl precipitates first (needs only 1.8×10⁻⁸ M Cl⁻)
- PbCl₂ precipitates much later (needs 4.1×10⁻² M Cl⁻)
- Excellent separation possible!
Effect of pH on Solubility
For salts containing basic anions (like OH⁻, S²⁻, CO₃²⁻, PO₄³⁻), pH affects solubility.
Example: Mg(OH)₂
$$Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$$ $$K_{sp} = [Mg^{2+}][OH^-]^2$$Effect of adding acid (lowering pH):
- Acid neutralizes OH⁻: H⁺ + OH⁻ → H₂O
- [OH⁻] decreases
- Equilibrium shifts right to replace OH⁻
- More Mg(OH)₂ dissolves
- Solubility increases
Effect of adding base (increasing pH):
- [OH⁻] increases (common ion effect)
- Equilibrium shifts left
- Solubility decreases
Metal Sulfides: Qualitative Analysis
Metal sulfides (like CuS, ZnS, FeS) have different Ksp values and dissolve at different pH:
$$MS \rightleftharpoons M^{2+} + S^{2-}$$In acidic solution:
$$S^{2-} + H^+ \rightarrow HS^-$$ $$HS^- + H^+ \rightarrow H_2S$$- Acid removes S²⁻
- Shifts equilibrium right
- More sulfide dissolves
JEE Application: Group analysis in qualitative analysis uses this!
- Group II (CuS, PbS): Precipitate even in acidic medium (very low Ksp)
- Group IV (ZnS, MnS): Precipitate only in basic medium (higher Ksp)
Common Mistakes to Avoid
Common Error: Using $S = \sqrt{K_{sp}}$ for all salts
Correct: Relationship depends on salt type!
| Salt Type | Correct Formula |
|---|---|
| AB | $S = \sqrt{K_{sp}}$ |
| AB₂ or A₂B | $S = \sqrt[3]{K_{sp}/4}$ |
| A₂B₃ | $S = \sqrt[5]{K_{sp}/108}$ |
JEE Trap: Mixing up AB₂ and A₂B (both give 4S³ but different examples!)
Wrong: Calculating solubility of AgCl in NaCl solution using $S = \sqrt{K_{sp}}$
Correct: Account for Cl⁻ from NaCl!
For AgCl in 0.1 M NaCl:
- [Cl⁻] = 0.1 + s ≈ 0.1 (common ion dominates)
- $K_{sp} = s \times 0.1$
- $s = K_{sp}/0.1$
Always check if a common ion is present!
Common Error: “Q > Ksp means more will dissolve”
Correct: Q > Ksp means precipitation occurs (solution is supersaturated)
Memory: “Q questions Ksp. If Q is greater, solid gets heavier (precipitates)!”
Wrong: For PbCl₂, writing Q = [Pb²⁺][Cl⁻]
Correct: Q = [Pb²⁺][Cl⁻]²
Rule: Always raise concentrations to their stoichiometric coefficients!
Practice Problems
Level 1: Foundation (NCERT)
The solubility of AgCl in water is 1.3×10⁻⁵ M. Calculate Ksp.
Solution:
AgCl is AB type: $AgCl \rightleftharpoons Ag^+ + Cl^-$
If S = 1.3×10⁻⁵ M:
- [Ag⁺] = 1.3×10⁻⁵ M
- [Cl⁻] = 1.3×10⁻⁵ M
Ksp of BaSO₄ is 1.0×10⁻¹⁰. Calculate its solubility in mol/L.
Solution:
BaSO₄ is AB type: $BaSO_4 \rightleftharpoons Ba^{2+} + SO_4^{2-}$
$$K_{sp} = S^2$$ $$S = \sqrt{K_{sp}} = \sqrt{1.0 \times 10^{-10}}$$ $$S = 1.0 \times 10^{-5} \text{ mol/L}$$Will AgCl precipitate when 100 mL of 10⁻⁴ M AgNO₃ is mixed with 100 mL of 10⁻⁴ M NaCl?
Given: Ksp(AgCl) = 1.8×10⁻¹⁰
Solution:
After mixing (dilution): Total volume = 200 mL
$$[Ag^+] = \frac{10^{-4} \times 100}{200} = 5 \times 10^{-5} \text{ M}$$ $$[Cl^-] = \frac{10^{-4} \times 100}{200} = 5 \times 10^{-5} \text{ M}$$Calculate Q:
$$Q = [Ag^+][Cl^-] = (5 \times 10^{-5})(5 \times 10^{-5}) = 25 \times 10^{-10} = 2.5 \times 10^{-9}$$Compare: Q = 2.5×10⁻⁹ > Ksp = 1.8×10⁻¹⁰
Answer: YES, precipitation will occur! (Q > Ksp)
Level 2: JEE Main
Calculate the solubility of CaF₂ in water. Given: Ksp = 3.2×10⁻¹¹
Solution:
CaF₂ is AB₂ type: $CaF_2 \rightleftharpoons Ca^{2+} + 2F^-$
If S mol/L dissolves:
- [Ca²⁺] = S
- [F⁻] = 2S
Calculate the solubility of AgCl in:
- Pure water
- 0.01 M AgNO₃ solution
Given: Ksp(AgCl) = 1.8×10⁻¹⁰
Solution:
1. In pure water:
$$S_1 = \sqrt{K_{sp}} = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5} \text{ M}$$2. In 0.01 M AgNO₃:
AgNO₃ provides Ag⁺ = 0.01 M (common ion)
Let solubility = s (much smaller than 0.01)
- [Ag⁺] = 0.01 + s ≈ 0.01 M
- [Cl⁻] = s
Comparison:
- In pure water: 1.34×10⁻⁵ M
- In 0.01 M AgNO₃: 1.8×10⁻⁸ M
Solubility decreased by factor of ~740! (Common ion effect)
Interactive Demo: Visualize Equilibrium Constant and Solubility
Explore how Ksp affects solubility and observe the common ion effect in action.
A solution contains 0.01 M each of Cl⁻ and CrO₄²⁻. Ag⁺ is added gradually. Which will precipitate first?
Given:
- Ksp(AgCl) = 1.8×10⁻¹⁰
- Ksp(Ag₂CrO₄) = 1.9×10⁻¹²
Solution:
For AgCl to precipitate:
$$K_{sp} = [Ag^+][Cl^-]$$ $$[Ag^+] = \frac{K_{sp}}{[Cl^-]} = \frac{1.8 \times 10^{-10}}{0.01} = 1.8 \times 10^{-8} \text{ M}$$For Ag₂CrO₄ to precipitate:
$$K_{sp} = [Ag^+]^2[CrO_4^{2-}]$$ $$[Ag^+]^2 = \frac{K_{sp}}{[CrO_4^{2-}]} = \frac{1.9 \times 10^{-12}}{0.01} = 1.9 \times 10^{-10}$$ $$[Ag^+] = \sqrt{1.9 \times 10^{-10}} = 1.38 \times 10^{-5} \text{ M}$$Comparison:
- AgCl needs: 1.8×10⁻⁸ M Ag⁺
- Ag₂CrO₄ needs: 1.38×10⁻⁵ M Ag⁺
Answer: AgCl precipitates first! (needs less Ag⁺)
JEE Insight: This is the basis of chromate test for Cl⁻ detection!
Level 3: JEE Advanced
The Ksp of Ag₂CrO₄ is 1.9×10⁻¹². What is the solubility of Ag₂CrO₄ in:
- Pure water
- 0.1 M AgNO₃
Solution:
1. In pure water:
Ag₂CrO₄ is A₂B type: $Ag_2CrO_4 \rightleftharpoons 2Ag^+ + CrO_4^{2-}$
If S mol/L dissolves:
- [Ag⁺] = 2S
- [CrO₄²⁻] = S
2. In 0.1 M AgNO₃:
AgNO₃ provides Ag⁺ = 0.1 M (common ion)
Let solubility = s
- [Ag⁺] = 0.1 + 2s ≈ 0.1 M (since s « 0.1)
- [CrO₄²⁻] = s
Comparison:
- In pure water: 7.8×10⁻⁵ M
- In 0.1 M AgNO₃: 1.9×10⁻¹⁰ M
Solubility decreased by factor of ~400,000! (Strong common ion effect for A₂B type)
Calculate the solubility of Mg(OH)₂ at pH = 12.
Given: Ksp[Mg(OH)₂] = 1.8×10⁻¹¹
Solution:
At pH = 12:
$$pOH = 14 - 12 = 2$$ $$[OH^-] = 10^{-2} = 0.01 \text{ M}$$$Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^-$
Let solubility = s
- [Mg²⁺] = s
- [OH⁻] = 0.01 + 2s ≈ 0.01 M (common ion from pH)
Compare to pure water:
$$S_{water} = \sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{1.8 \times 10^{-11}}{4}} = 1.65 \times 10^{-4} \text{ M}$$In basic solution (pH 12), solubility is ~900 times lower!
JEE Application: This explains why adding NaOH precipitates Mg(OH)₂!
A saturated solution of H₂S contains 0.1 M H₂S. Calculate [S²⁻] at pH = 1.
Given:
- Ka₁(H₂S) = 1×10⁻⁷
- Ka₂(H₂S) = 1.3×10⁻¹⁴
Solution:
First ionization:
$$H_2S \rightleftharpoons H^+ + HS^-$$ $$K_{a1} = \frac{[H^+][HS^-]}{[H_2S]}$$At pH = 1: [H⁺] = 10⁻¹ M
$$1 \times 10^{-7} = \frac{10^{-1} \times [HS^-]}{0.1}$$ $$[HS^-] = \frac{1 \times 10^{-7} \times 0.1}{10^{-1}} = 1 \times 10^{-7} \text{ M}$$Second ionization:
$$HS^- \rightleftharpoons H^+ + S^{2-}$$ $$K_{a2} = \frac{[H^+][S^{2-}]}{[HS^-]}$$ $$1.3 \times 10^{-14} = \frac{10^{-1} \times [S^{2-}]}{1 \times 10^{-7}}$$ $$[S^{2-}] = \frac{1.3 \times 10^{-14} \times 1 \times 10^{-7}}{10^{-1}} = 1.3 \times 10^{-20} \text{ M}$$Key Insight: At low pH (acidic), [S²⁻] is extremely small! This is why Group II sulfides (CuS, PbS with very low Ksp) precipitate even in acidic medium, but Group IV sulfides (ZnS, MnS with higher Ksp) don’t.
Qualitative analysis connection!
Quick Revision Box
| Salt Type | Ksp Expression | Ksp-S Relation | Example |
|---|---|---|---|
| AB | $[A^+][B^-]$ | $S^2$ | AgCl, BaSO₄ |
| AB₂ | $[A^{2+}][B^-]^2$ | $4S^3$ | CaF₂, PbCl₂ |
| A₂B | $[A^+]^2[B^{2-}]$ | $4S^3$ | Ag₂S, Ag₂CrO₄ |
| AB₃ | $[A^{3+}][B^-]^3$ | $27S^4$ | AlF₃ |
| A₂B₃ | $[A^{3+}]^2[B^{2-}]^3$ | $108S^5$ | Al₂(SO₄)₃ |
Precipitation Rules:
- Q < Ksp → Unsaturated (no precipitation)
- Q = Ksp → Saturated (equilibrium)
- Q > Ksp → Supersaturated (precipitation occurs)
Common Ion Effect:
- Adding common ion → Solubility decreases
- Higher pH → Metal hydroxide solubility decreases
- Lower pH → Basic salt solubility increases
When to Use This
Step 1: Identify the question type
Given Ksp, find solubility?
- Identify salt type (AB, AB₂, etc.)
- Check for common ion
- Use appropriate Ksp-S formula
Given solubility, find Ksp?
- Write equilibrium expression
- Find ion concentrations from S
- Calculate Ksp
Will precipitation occur?
- Calculate Q from given concentrations
- Compare Q with Ksp
- Q > Ksp → Yes, Q < Ksp → No
Which precipitates first?
- Calculate minimum ion concentration needed for each salt
- Lower concentration needed → Precipitates first
Effect of pH?
- If anion is basic (OH⁻, S²⁻, CO₃²⁻) → pH affects solubility
- Higher pH → Lower solubility (for metal hydroxides)
Qualitative Analysis Connection (JEE Advanced)
Group-wise Precipitation
| Group | Reagent | Precipitates | Ksp Range |
|---|---|---|---|
| I | Dil HCl | AgCl, PbCl₂, Hg₂Cl₂ | 10⁻¹⁰ to 10⁻¹⁸ |
| II | H₂S (acidic) | CuS, PbS, CdS, Bi₂S₃ | 10⁻³⁶ to 10⁻⁵² |
| III | NH₄OH + NH₄Cl | Al(OH)₃, Fe(OH)₃, Cr(OH)₃ | Medium Ksp |
| IV | H₂S (basic) | ZnS, MnS, NiS, CoS | 10⁻²² to 10⁻²⁶ |
| V | (NH₄)₂CO₃ | BaCO₃, CaCO₃, SrCO₃ | 10⁻⁹ to 10⁻¹⁰ |
Why different groups?
- Low Ksp (Group I, II): Precipitate easily, even in acidic medium
- Medium Ksp (Group III, IV): Need specific pH/conditions
- Higher Ksp (Group V): Precipitate only with strong reagents
Connection to Other Topics
Solubility product connects to:
- Ionic Equilibrium - Ksp is an equilibrium constant
- Le Chatelier’s Principle - Common ion effect explained
- pH and Buffers - pH affects solubility of basic salts
- Chemical Equilibrium - Uses Q vs K concept
- Acids and Bases - Anion basicity affects solubility
- Coordination Chemistry - Complex formation increases solubility
- Electrochemistry - AgCl reference electrode
Teacher’s Summary
- Ksp is the equilibrium constant for sparingly soluble salts - Smaller Ksp = Lower solubility
- Ksp-S relationship depends on salt type - AB: S², AB₂: 4S³, A₂B₃: 108S⁵
- Q > Ksp means precipitation - Compare ionic product with Ksp
- Common ion decreases solubility - Le Chatelier’s principle in action
- pH affects solubility of basic salts - Higher pH → Lower solubility of M(OH)ₙ
“Ksp is nature’s way of saying ’that’s all I can dissolve!’ Go beyond it, and the excess precipitates out.”
JEE Strategy:
- Master Ksp-S formulas for different salt types (AB, AB₂, A₂B, A₂B₃)
- Always check for common ions before calculating solubility
- Q vs Ksp comparison is the key to precipitation questions
- Link to qualitative analysis—know which groups precipitate when
- pH effect on solubility is favorite in JEE Advanced (H₂S, metal hydroxides)
- Selective precipitation problems test your numerical skills—practice them!
What’s Next?
Congratulations! You’ve completed the Equilibrium chapter. Here’s where to go next:
Within Chemistry:
- Chemical Kinetics - How fast reactions reach equilibrium
- Thermodynamics - Energy changes and spontaneity (ΔG° = -RT ln K)
- Electrochemistry - Nernst equation uses equilibrium concepts
Cross-subject connections:
- Logarithms - Essential for pH, pKa, Ksp calculations
- Differential Equations - Rate laws in kinetics
For comprehensive revision:
- Review all 6 equilibrium topics together
- Practice mixed problems combining multiple concepts
- Focus on numerical problem-solving speed