Solubility Equilibria and Solubility Product (Ksp)
Understanding solubility equilibria and the solubility product constant is essential for predicting whether precipitates will form and controlling precipitation reactions. Let’s explore these important concepts.
What is Solubility?
Definition: Solubility is the maximum amount of a solute that can dissolve in a given amount of solvent at a specific temperature to produce a saturated solution.
Solubility Equilibrium
When a slightly soluble salt dissolves in water, an equilibrium is established:
AB(s) ⇌ A⁺(aq) + B⁻(aq)
For example:
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) PbCl₂(s) ⇌ Pb²⁺(aq) + 2Cl⁻(aq)
Solubility Product Constant (Ksp)
Definition: The solubility product (Ksp) is the product of the molar concentrations of the ions in a saturated solution of a slightly soluble salt, each concentration raised to the power of its stoichiometric coefficient in the dissociation equation.
For a Salt of Type AB (1:1 ratio)
AB(s) ⇌ A⁺(aq) + B⁻(aq) Ksp = [A⁺][B⁻]
Example: AgCl
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) Ksp = [Ag⁺][Cl⁻]
For a Salt of Type AB₂ (1:2 ratio)
AB₂(s) ⇌ A²⁺(aq) + 2B⁻(aq) Ksp = [A²⁺][B⁻]²
Example: PbCl₂
PbCl₂(s) ⇌ Pb²⁺(aq) + 2Cl⁻(aq) Ksp = [Pb²⁺][Cl⁻]²
For a Salt of Type A₂B₃
A₂B₃(s) ⇌ 2A³⁺(aq) + 3B²⁻(aq) Ksp = [A³⁺]²[B²⁻]³
Example: Ca₃(PO₄)₂
Ca₃(PO₄)₂(s) ⇌ 3Ca²⁺(aq) + 2PO₄³⁻(aq) Ksp = [Ca²⁺]³[PO₄³⁻]²
Relationship Between Solubility and Ksp
For AB type salt:
If solubility (s) = x mol/dm³ Then [A⁺] = x and [B⁻] = x Ksp = x · x = x² Therefore: x = √Ksp
For AB₂ type salt:
If solubility (s) = x mol/dm³ Then [A²⁺] = x and [B⁻] = 2x Ksp = x(2x)² = 4x³ Therefore: x = ∛(Ksp/4)
Worked Examples
Example 1: Calculating Solubility from Ksp (AB Type)
Problem: The Ksp of AgCl is 1.8 × 10⁻¹⁰. Calculate the solubility of AgCl in pure water.
Solution:
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) If solubility = x mol/dm³ [Ag⁺] = x [Cl⁻] = x Ksp = [Ag⁺][Cl⁻] = x · x = x² 1.8 × 10⁻¹⁰ = x² x = √(1.8 × 10⁻¹⁰) = 1.3 × 10⁻⁵ mol/dm³ Solubility of AgCl = 1.3 × 10⁻⁵ mol/dm³
Example 2: Calculating Solubility from Ksp (AB₂ Type)
Problem: The Ksp of PbCl₂ is 1.6 × 10⁻⁵. Calculate the solubility of PbCl₂ in pure water.
Solution:
PbCl₂(s) ⇌ Pb²⁺(aq) + 2Cl⁻(aq) If solubility = x mol/dm³ [Pb²⁺] = x [Cl⁻] = 2x Ksp = [Pb²⁺][Cl⁻]² 1.6 × 10⁻⁵ = x(2x)² 1.6 × 10⁻⁵ = 4x³ x³ = 4.0 × 10⁻⁶ x = 1.6 × 10⁻² mol/dm³ Solubility of PbCl₂ = 0.016 mol/dm³
Example 3: Calculating Ksp from Solubility
Problem: The solubility of Ca₃(PO₄)₂ is 7.1 × 10⁻⁹ mol/dm³. Calculate the Ksp.
Solution:
Ca₃(PO₄)₂(s) ⇌ 3Ca²⁺(aq) + 2PO₄³⁻(aq) If solubility = 7.1 × 10⁻⁹ mol/dm³ [Ca²⁺] = 3 × 7.1 × 10⁻⁹ = 2.13 × 10⁻⁸ mol/dm³ [PO₄³⁻] = 2 × 7.1 × 10⁻⁹ = 1.42 × 10⁻⁸ mol/dm³ Ksp = [Ca²⁺]³[PO₄³⁻]² Ksp = (2.13 × 10⁻⁸)³ × (1.42 × 10⁻⁸)² Ksp = 5.5 × 10⁻⁴⁹
The Common Ion Effect
Definition: The common ion effect is the decrease in solubility of a salt when a common ion is added to the solution.
Worked Example: Common Ion Effect
Problem: Calculate the solubility of AgCl in a 0.1M NaCl solution. (Ksp of AgCl = 1.8 × 10⁻¹⁰)
Solution:
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) Initial [Cl⁻] from NaCl = 0.1 mol/dm³ If solubility of AgCl = x mol/dm³ [Ag⁺] = x [Cl⁻] = 0.1 + x ≈ 0.1 (since x is very small) Ksp = [Ag⁺][Cl⁻] 1.8 × 10⁻¹⁰ = x × 0.1 x = 1.8 × 10⁻⁹ mol/dm³ Solubility in 0.1M NaCl = 1.8 × 10⁻⁹ mol/dm³ Compare with solubility in pure water: 1.3 × 10⁻⁵ mol/dm³ The solubility decreased significantly due to common ion effect.
Predicting Precipitation
To determine if a precipitate will form, calculate the **ion product (Q)** and compare with Ksp:
- Q < Ksp: Solution is unsaturated, no precipitate forms
- Q = Ksp: Solution is saturated, equilibrium established
- Q > Ksp: Solution is supersaturated, precipitate forms
Worked Example: Will a Precipitate Form?
Problem: If 0.01M Pb(NO₃)₂ and 0.02M NaCl are mixed, will PbCl₂ precipitate form? (Ksp of PbCl₂ = 1.6 × 10⁻⁵)
Solution:
When mixed, the concentrations become:
[Pb²⁺] = 0.01/2 = 0.005 M [Cl⁻] = 0.02/2 = 0.01 M
Calculate Q:
Q = [Pb²⁺][Cl⁻]² Q = 0.005 × (0.01)² Q = 0.005 × 0.0001 Q = 5 × 10⁻⁷
Compare with Ksp:
Q (5 × 10⁻⁷) > Ksp (1.6 × 10⁻⁵) Since Q > Ksp, PbCl₂ precipitate will form
Summary Table: Ksp Expressions
| Compound | Dissociation | Ksp Expression | Solubility Relationship |
|---|---|---|---|
| AB (AgCl) | A⁺ + B⁻ | [A⁺][B⁻] | s = √Ksp |
| AB₂ (PbCl₂) | A²⁺ + 2B⁻ | [A²⁺][B⁻]² | s = ∛(Ksp/4) |
| A₂B₃ (Ca₃(PO₄)₂) | 2A³⁺ + 3B²⁻ | [A³⁺]²[B²⁻]³ | s = ∛(Ksp/108) |
Important Notes About Ksp
- Ksp is temperature dependent
- Ksp applies only to slightly soluble salts
- The common ion effect decreases solubility
- Ksp is a constant only when all ions come from the salt
- Always write the correct Ksp expression based on dissociation
