What is the molar solubility of barium fluoride (BaF2) in a 0.15 M NaF solution, given that the solubility product constant (Ksp) for BaF2 is 2.45 × 10^-5?

• A. 5.00 × 10^-3 M
• B. 2.45 × 10^-5 M
• C. 1.63 × 10^-3 M
• D. 3.92 × 10^-4 M

Answer: (C)

Verified step by step guidance

1. Understand the concept of molar solubility and the common ion effect. Molar solubility refers to the number of moles of a solute that can dissolve in a liter of solution to reach saturation. The common ion effect occurs when a salt's solubility is reduced due to the presence of a common ion in the solution.
2. Write the dissolution equation for barium fluoride (BaF2). The equation is: BaF2(s) ⇌ Ba²⁺(aq) + 2F⁻(aq). This shows that one mole of BaF2 produces one mole of Ba²⁺ ions and two moles of F⁻ ions in solution.
3. Express the solubility product constant (Ksp) for BaF2. The Ksp expression is: Ksp = [Ba²⁺][F⁻]². Given that Ksp = 2.45 × 10^-5, this expression will be used to find the molar solubility.
4. Consider the initial concentration of fluoride ions from the NaF solution. Since NaF is a strong electrolyte, it dissociates completely, providing an initial concentration of F⁻ ions as 0.15 M.
5. Set up the equilibrium expression incorporating the common ion effect. Let the molar solubility of BaF2 be 's'. Then, [Ba²⁺] = s and [F⁻] = 0.15 + 2s. Substitute these into the Ksp expression: Ksp = s(0.15 + 2s)². Solve this equation for 's' to find the molar solubility of BaF2.