Solubility Formula
Quantifying dissolution equilibrium and maximum concentration in solution
Solubility Product Constant (Ksp)
General Equilibrium
AₓBᵧ(s) ⇌ xA⁺(aq) + yB⁻(aq)
Ksp = [A⁺]ˣ[B⁻]ʸ
Molar Solubility (s)
Maximum moles that dissolve per liter
s = moles dissolved / L solution
Common Ksp Expressions
AB Type (1:1)
AgCl ⇌ Ag⁺ + Cl⁻
Ksp = [Ag⁺][Cl⁻] = s²
s = √Ksp
AB₂ Type (1:2)
CaF₂ ⇌ Ca²⁺ + 2F⁻
Ksp = [Ca²⁺][F⁻]² = s(2s)² = 4s³
s = ∛(Ksp/4)
A₂B Type (2:1)
Ag₂CrO₄ ⇌ 2Ag⁺ + CrO₄²⁻
Ksp = [Ag⁺]²[CrO₄²⁻] = (2s)²s = 4s³
s = ∛(Ksp/4)
A₃B Type (3:1)
Ca₃(PO₄)₂ ⇌ 3Ca²⁺ + 2PO₄³⁻
Ksp = [Ca²⁺]³[PO₄³⁻]² = (3s)³(2s)² = 108s⁵
s = ⁵√(Ksp/108)
Worked Examples
Example 1: Simple 1:1 Solubility
Problem: AgCl has Ksp = 1.8 × 10⁻¹⁰. Find molar solubility.
Solution:
AgCl ⇌ Ag⁺ + Cl⁻
Ksp = [Ag⁺][Cl⁻] = s × s = s²
1.8 × 10⁻¹⁰ = s²
s = √(1.8 × 10⁻¹⁰)
s = 1.3 × 10⁻⁵ M
(Very low solubility - AgCl is "insoluble")
Example 2: 1:2 Solubility (CaF₂)
Problem: CaF₂ has Ksp = 3.9 × 10⁻¹¹. Find molar solubility.
Solution:
CaF₂ ⇌ Ca²⁺ + 2F⁻
If s mol/L dissolves:
[Ca²⁺] = s, [F⁻] = 2s
Ksp = [Ca²⁺][F⁻]² = s(2s)² = 4s³
3.9 × 10⁻¹¹ = 4s³
s³ = 9.75 × 10⁻¹²
s = 2.1 × 10⁻⁴ M
Example 3: Common Ion Effect
Problem: Find solubility of AgCl in 0.10 M NaCl. Ksp(AgCl) = 1.8 × 10⁻¹⁰
Solution:
Initial [Cl⁻] = 0.10 M from NaCl
AgCl ⇌ Ag⁺ + Cl⁻
[Ag⁺] = s, [Cl⁻] = 0.10 + s ≈ 0.10 M (s very small)
Ksp = s(0.10)
1.8 × 10⁻¹⁰ = 0.10s
s = 1.8 × 10⁻⁹ M
140× less soluble than in pure water!
Precipitation Prediction
Reaction Quotient (Q)
Q < Ksp: Unsaturated - no precipitation
Q = Ksp: Saturated equilibrium
Q > Ksp: Supersaturated - precipitation occurs
Common Mistakes
Wrong Ksp Expression
For CaF₂: Ksp = s(2s)², NOT s³. Must account for stoichiometric coefficients!
Ignoring Common Ion
Solubility decreases in presence of common ion - use initial concentration
Ksp ≠ Solubility
Ksp is equilibrium constant. Solubility (s) is calculated FROM Ksp.
Temperature Dependence
Ksp varies with temperature - values given are typically at 25°C