Calculating Molar Solubility of Calcium Fluoride

Key Concepts

• Ksp (Solubility Product Constant): Represents the maximum product of the ion concentrations that can exist in a solution without precipitating.
• Molar Solubility: The concentration of a solute that dissolves to form a saturated solution.

Problem Statement

• Calculate the molar solubility of calcium fluoride (CaF2) with a Ksp value of 3.9 x 10^-11 at 25°C.

Steps for Calculation

1. Write the Dissolution Equation

• Calcium fluoride (CaF2) dissolves in water:

  [ \text{CaF}2 (s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^-(aq) ]

2. Set Up the ICE Table

• Initial Concentration (I):

  • [ \text{[Ca}^{2+}] = 0 ]
  • [ \text{[F}^-] = 0 ]

• Change in Concentration (C):

  • Represented as (-X) for CaF2
  • (+X) for [ \text{Ca}^{2+} ]
  • (+2X) for [ \text{F}^- ]

• Equilibrium Concentration (E):

  • [ \text{[Ca}^{2+}] = X ]
  • [ \text{[F}^-] = 2X ]

3. Write the Ksp Expression

• Ksp expression from the balanced equation:

  [ Ksp = [\text{Ca}^{2+}][\text{F}^-]^2 ]

• Plugging in equilibrium concentrations:

  [ 3.9 \times 10^{-11} = X (2X)^2 ]

4. Solve for X

• Simplify the expression:

  [ 3.9 \times 10^{-11} = X \times 4X^2 = 4X^3 ]

• Solve for X by dividing by 4 and taking the cube root:

  [ X = \sqrt[3]{\frac{3.9 \times 10^{-11}}{4}} ]

• Calculation gives:

  [ X = 2.1 \times 10^{-4} \text{ M} ]

5. Interpretation of X

• X represents the equilibrium concentration of ( \text{Ca}^{2+} ) ions.
• Also equals the molar solubility of CaF2.

6. Calculate Concentration of Fluoride Anions

• [ \text{[F}^-] = 2X = 2(2.1 \times 10^{-4}) = 4.2 \times 10^{-4} \text{ M} ]

Conclusion

• Molar Solubility of CaF2: 2.1 x 10^-4 M
• Equilibrium Concentration of Fluoride Ions: 4.2 x 10^-4 M
• Reminder: Molar solubility is determined at constant temperature (25°C).