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Multiple Choice
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
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).
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.
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.
Substitute the expressions for [Ba^2+] and [F^-] into the Ksp expression: Ksp = (s)(2s)^2 = 4s^3.
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.