The concentration of fluoride ions in a saturated solution of barium fluoride (BaF2) is ________ M. The solubility product constant (Ksp) of BaF2 is 1.7 x 10^-6. What is the concentration of fluoride ions?

• A. 1.0 x 10^-2 M
• B. 3.0 x 10^-2 M
• C. 2.0 x 10^-2 M
• D. 4.0 x 10^-2 M

Answer: (C)

Verified step by step guidance

1. Understand the dissolution process of barium fluoride (BaF2) in water. BaF2 dissociates into barium ions (Ba^2+) and fluoride ions (F^-). The balanced chemical equation is: BaF2(s) ⇌ Ba^2+(aq) + 2F^-(aq).
2. Define the solubility product constant (Ksp) expression for BaF2. The Ksp is given by the equation: Ksp = [Ba^2+][F^-]^2. Here, [Ba^2+] is the concentration of barium ions and [F^-] is the concentration of fluoride ions.
3. Let the solubility of BaF2 be 's' mol/L. This means [Ba^2+] = s and [F^-] = 2s because each formula unit of BaF2 produces one Ba^2+ ion and two F^- ions.
4. Substitute the expressions for [Ba^2+] and [F^-] into the Ksp expression: Ksp = (s)(2s)^2 = 4s^3.
5. Solve for 's' using the given Ksp value: 1.7 x 10^-6 = 4s^3. Rearrange to find s, which represents the solubility of BaF2, and then calculate [F^-] = 2s to find the concentration of fluoride ions.