Calculating molar solubility from the solubility product constant (Ksp) involves understanding the stoichiometry of the dissolution reaction of a sparingly soluble ionic compound. Molar solubility, often denoted as 's', represents the concentration of the dissolved compound in a saturated solution, typically expressed in moles per liter (mol/L).
Understanding Ksp and Molar Solubility
The solubility product constant (Ksp) is an equilibrium constant that describes the extent to which an ionic solid dissolves in water. For a generic sparingly soluble ionic compound, MₓAᵧ, its dissolution in water can be represented by the following equilibrium:
MₓAᵧ(s) ⇌ xMʸ⁺(aq) + yAˣ⁻(aq)
The Ksp expression for this reaction is:
Ksp = [Mʸ⁺]ˣ[Aˣ⁻]ʸ
Molar solubility (s) is defined as the number of moles of the solid that dissolve to form a liter of saturated solution. When 's' moles of MₓAᵧ dissolve, they produce 'x' moles of Mʸ⁺ ions and 'y' moles of Aˣ⁻ ions per liter. Therefore, the equilibrium concentrations of the ions can be expressed in terms of 's':
- [Mʸ⁺] = x * s
- [Aˣ⁻] = y * s
By substituting these expressions into the Ksp equation, you can solve for 's'.
Steps to Calculate Molar Solubility from Ksp
Follow these systematic steps to determine molar solubility:
- Write the Balanced Dissolution Equation: Represent the dissolution of the ionic compound in water as an equilibrium reaction, showing the solid reactant and its dissociated ions in solution.
- Define Molar Solubility (s): Let 's' be the molar solubility of the compound.
- Express Ion Concentrations in Terms of 's': Based on the stoichiometry of the balanced equation, determine the equilibrium concentrations of each ion in terms of 's'.
- Substitute into the Ksp Expression: Plug the ion concentrations (expressed in terms of 's') into the Ksp expression for the compound.
- Solve for 's': Algebraically rearrange the equation to solve for 's'.
Examples of Ksp to Molar Solubility Calculations
The method varies depending on the stoichiometry of the ionic compound.
1. For 1:1 Stoichiometry (e.g., AgCl, BaSO₄)
For compounds like silver chloride (AgCl), where one cation and one anion are produced:
Reaction: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
Concentrations in terms of 's':
- [Ag⁺] = s
- [Cl⁻] = s
Ksp Expression: Ksp = [Ag⁺][Cl⁻] = (s)(s) = s²
Solving for 's': s = √Ksp
Example: If Ksp for AgCl is 1.8 × 10⁻¹⁰ at 25°C:
s = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol/L
2. For 1:2 or 2:1 Stoichiometry (e.g., PbI₂, CaF₂)
For compounds like lead(II) iodide (PbI₂) or calcium fluoride (CaF₂), where one cation produces two anions, or vice versa:
Reaction (PbI₂): PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq)
Concentrations in terms of 's':
- [Pb²⁺] = s
- [I⁻] = 2s
Ksp Expression: Ksp = [Pb²⁺][I⁻]² = (s)(2s)² = s(4s²) = 4s³
Solving for 's': s = ³√(Ksp/4)
Example: If Ksp for PbI₂ is 8.5 × 10⁻⁹ at 25°C:
s = ³√(8.5 × 10⁻⁹ / 4) = ³√(2.125 × 10⁻⁹) = 1.29 × 10⁻³ mol/L
3. For 1:3 or 3:1 Stoichiometry (e.g., Al(OH)₃)
For compounds like aluminum hydroxide (Al(OH)₃):
Reaction: Al(OH)₃(s) ⇌ Al³⁺(aq) + 3OH⁻(aq)
Concentrations in terms of 's':
- [Al³⁺] = s
- [OH⁻] = 3s
Ksp Expression: Ksp = [Al³⁺][OH⁻]³ = (s)(3s)³ = s(27s³) = 27s⁴
Solving for 's': s = ⁴√(Ksp/27)
Example: If Ksp for Al(OH)₃ is 3 × 10⁻³⁴ at 25°C:
s = ⁴√(3 × 10⁻³⁴ / 27) = ⁴√(1.11 × 10⁻³⁵) = 1.03 × 10⁻⁹ mol/L
Summary Table of Ksp-Molar Solubility Relationships
The relationship between Ksp and molar solubility depends directly on the stoichiometry of the dissolved compound:
| Compound Type | Example | Dissolution Reaction | Ksp Expression (in terms of 's') | Molar Solubility (s) Formula |
|---|---|---|---|---|
| 1:1 (AB) | AgCl | AB(s) ⇌ A⁺(aq) + B⁻(aq) | s² | √Ksp |
| 1:2 (AB₂) | PbI₂ | AB₂(s) ⇌ A²⁺(aq) + 2B⁻(aq) | 4s³ | ³√(Ksp/4) |
| 2:1 (A₂B) | Ag₂S | A₂B(s) ⇌ 2A⁺(aq) + B²⁻(aq) | 4s³ | ³√(Ksp/4) |
| 1:3 (AB₃) | Al(OH)₃ | AB₃(s) ⇌ A³⁺(aq) + 3B⁻(aq) | 27s⁴ | ⁴√(Ksp/27) |
| 3:1 (A₃B) | Ag₃PO₄ | A₃B(s) ⇌ 3A⁺(aq) + B³⁻(aq) | 27s⁴ | ⁴√(Ksp/27) |
| 2:3 (A₂B₃) | Ca₃(PO₄)₂ | A₂B₃(s) ⇌ 2A³⁺(aq) + 3B²⁻(aq) | 108s⁵ | ⁵√(Ksp/108) |
Factors Affecting Molar Solubility
While Ksp is a useful constant for ideal solutions, the actual molar solubility of a compound can be influenced by several factors in a real solution, making the calculation more complex:
- Common Ion Effect: The presence of a common ion (an ion already present in the solution that is also produced by the dissolution of the sparingly soluble salt) will decrease the molar solubility of the salt.
- pH of the Solution: If the anion or cation of the sparingly soluble salt is a conjugate acid or base, the pH of the solution can significantly affect its solubility. For example, hydroxides (like Mg(OH)₂) are more soluble in acidic solutions.
- Complex Ion Formation: The formation of soluble complex ions with a Lewis acid (like a metal ion) and a Lewis base (ligand) can increase the solubility of a sparingly soluble salt.
- Temperature: Ksp values are temperature-dependent. Most ionic solids become more soluble as temperature increases, leading to a higher Ksp.
Understanding these factors is crucial for predicting and calculating solubility in diverse chemical environments. For further details on solubility and Ksp, explore resources on chemical equilibrium and solubility product constants.
