What is the expression for the solubility product constant Ksp of calcium fluoride, CaF2, which dissolves according to CaF2(s) <=> Ca^2+(aq) + 2 F^-(aq)?

Answer and explanation

Correct answer: B. Ksp = [Ca^2+] * [F^-]^2

  1. Step 1: Write the dissolution equilibrium: CaF2(s) <=> Ca^2+(aq) + 2 F^-(aq).
  2. Step 2: Pure solids have unit activity and are excluded from the equilibrium expression.
  3. Step 3: Each concentration is raised to its stoichiometric coefficient: Ksp = [Ca^2+]^1 * [F^-]^2.
Answer
Ksp = [Ca^2+] * [F^-]^2.
Why not the others: 'Ksp = [Ca^2+] [F^-]' omits the stoichiometric power 2 on fluoride; 'Ksp = [Ca^2+]^2 [F^-]' squares the wrong ion concentration; 'Ksp = [Ca^2+] * [2 F^-]' confuses stoichiometric coefficients in expressions with equilibrium solubility relationships.

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