The solubility product (Ksp) of calcium fluoride (CaF2) at 25 °C is 3.2 × 10⁻¹¹ mol³ L⁻³.
Understanding the Solubility Product (Ksp) of CaF2
The solubility product constant (Ksp) is a measure of the solubility of an ionic compound in water. For slightly soluble salts like CaF2, it represents the equilibrium between the undissolved solid and its dissolved ions in a saturated solution. A smaller Ksp value indicates lower solubility.
Calcium fluoride dissociates in water according to the following equilibrium:
CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq)
The expression for the solubility product constant for CaF₂ is:
Ksp = [Ca²⁺][F⁻]²
This equation signifies that the Ksp is the product of the molar concentrations of its constituent ions, each raised to the power of its stoichiometric coefficient in the balanced dissociation equation.
Key Aspects of CaF2 Solubility
- Temperature Dependence: Like most equilibrium constants, the Ksp of CaF2 is temperature-dependent. The value of 3.2 × 10⁻¹¹ is specific to 25 °C. Higher temperatures generally increase the solubility of most ionic solids, thus increasing their Ksp.
- Low Solubility: The very small Ksp value (3.2 × 10⁻¹¹) indicates that CaF2 is sparingly soluble in water. This low solubility is crucial in various applications, including its use in toothpastes and water fluoridation.
- Ion Stoichiometry: The Ksp expression for CaF2 includes [F⁻] raised to the power of 2 because two fluoride ions are produced for every one calcium ion when CaF2 dissolves.
Calculating Molar Solubility from Ksp
The solubility product constant allows us to calculate the molar solubility (s) of CaF2, which is the number of moles of CaF2 that dissolve per liter of solution.
Let 's' be the molar solubility of CaF2.
At equilibrium:
[Ca²⁺] = s
[F⁻] = 2s
Substituting these into the Ksp expression:
Ksp = (s)(2s)²
Ksp = s(4s²)
Ksp = 4s³
Given Ksp = 3.2 × 10⁻¹¹ at 25 °C:
3.2 × 10⁻¹¹ = 4s³
s³ = (3.2 × 10⁻¹¹) / 4
s³ = 0.8 × 10⁻¹¹
s³ = 8 × 10⁻¹²
s = ³√(8 × 10⁻¹²)
s = 2 × 10⁻⁴ mol/L
Therefore, the molar solubility of CaF2 at 25 °C is approximately 2 × 10⁻⁴ mol/L.
Factors Influencing CaF2 Solubility
Several factors can affect the observed solubility of CaF2:
- Common Ion Effect: The presence of a common ion (either Ca²⁺ or F⁻) in the solution will decrease the solubility of CaF2, shifting the equilibrium towards the solid phase according to Le Chatelier's principle.
- pH: The solubility of CaF2 is affected by pH because fluoride ions (F⁻) are the conjugate base of a weak acid (HF). In acidic solutions (lower pH), F⁻ ions can react with H⁺ ions to form HF. This removes F⁻ from the solution, shifting the equilibrium to dissolve more CaF2, thus increasing its solubility.
- Complex Ion Formation: If calcium ions can form stable complex ions with other species present in the solution, this can also increase the solubility of CaF2 by reducing the free Ca²⁺ concentration.
Summary of CaF2 Solubility Data
| Property | Value | Conditions |
|---|---|---|
| Solubility Product (Ksp) | 3.2 × 10⁻¹¹ mol³ L⁻³ | 25 °C |
| Molar Solubility (s) | 2 × 10⁻⁴ mol/L | 25 °C |
For more detailed information on solubility products and chemical equilibrium, you can refer to resources like Chemistry LibreTexts on Solubility Product.
