The solubility of calcium fluoride, CaF2, in pure water is 2.15 × 10^-4 M. What is the value of Ksp?
The Correct Answer and Explanation is:
To calculate the Ksp (solubility product constant) of calcium fluoride (CaF₂) from its solubility, we follow these steps:
Step 1: Write the dissociation equation
Calcium fluoride dissociates in water as follows:
CaF₂ (s) ⇌ Ca²⁺ (aq) + 2 F⁻ (aq)
Step 2: Define solubility
Let the solubility of CaF₂ be s = 2.15 × 10⁻⁴ M.
From the balanced equation:
- [Ca²⁺] = s = 2.15 × 10⁻⁴ M
- [F⁻] = 2s = 2 × 2.15 × 10⁻⁴ = 4.30 × 10⁻⁴ M
Step 3: Write the Ksp expression
Ksp = [Ca²⁺][F⁻]²
Substitute the concentrations:
Ksp = (2.15 × 10⁻⁴) × (4.30 × 10⁻⁴)²
First calculate (4.30 × 10⁻⁴)²:
(4.30 × 10⁻⁴)² = 18.49 × 10⁻⁸ = 1.849 × 10⁻⁷
Then multiply:
Ksp = (2.15 × 10⁻⁴) × (1.849 × 10⁻⁷)
= 3.975 × 10⁻¹¹
Final Answer:
Ksp = 3.98 × 10⁻¹¹
Explanation
The solubility product constant (Ksp) represents the equilibrium constant for a solid dissolving into its ions in a saturated solution. In this case, calcium fluoride (CaF₂) dissolves in water to form calcium ions (Ca²⁺) and fluoride ions (F⁻). The Ksp expression is based on the concentrations of these ions raised to the power of their stoichiometric coefficients.
When 1 mole of CaF₂ dissolves, it yields 1 mole of Ca²⁺ and 2 moles of F⁻. If the solubility (s) of CaF₂ is 2.15 × 10⁻⁴ M, this means the concentration of Ca²⁺ is also 2.15 × 10⁻⁴ M, while the concentration of F⁻ is twice that, or 4.30 × 10⁻⁴ M.
The Ksp is calculated using the equation:
Ksp = [Ca²⁺][F⁻]²
Substituting the concentrations derived from solubility gives:
Ksp = (2.15 × 10⁻⁴)(4.30 × 10⁻⁴)²
First, we square the fluoride ion concentration and then multiply by the calcium ion concentration. This gives a final Ksp value of approximately 3.98 × 10⁻¹¹.
This calculation is important in chemistry for predicting whether a precipitate will form in a solution. A low Ksp value like this indicates that CaF₂ is not very soluble in water, which aligns with what is observed experimentally.
