If PbF2 is poured into a solution of 0.012 molar F-, how much will dissolve at equilibrium?

PbF2(s) ==> Pb^+2 + 2F^-

Ksp = (Pb^+2)(F^-)^2.
Look up Ksp for PbF2, substitute 0.012 M for F^- in Ksp and solve for (Pb^+2). That will be the solubility of PbF2 in that solution.

Is the answer 0.211 mg/L PbF2?

You made an error somewhere and since you didn't show your work I can't help you find it. And I don't know what you used for Ksp. I found the value of 3.65 x 10^-8 on the Internet.

Ksp = (Pb^+2)(F^-)^2
(F^-) = 0.012 from the problem.
(Pb^+2) = 3.65 x 10^-8/(0.012)^2 = 2.53 x 10^-4 M
That x 245.2 for molar mass PbF2 = 0.062 g/L or 62 mg/L.

To determine how much PbF2 will dissolve at equilibrium, we need to consider the solubility product constant (Ksp). The Ksp of PbF2 represents the equilibrium between the solid PbF2 and its dissolved ions, Pb2+ and F-.

The balanced equation for the dissolution of PbF2 is:

PbF2(s) ⇌ Pb2+(aq) + 2F-(aq)

The Ksp expression for this reaction is:

Ksp = [Pb2+][F-]^2

Given that the concentration of F- in the solution is 0.012 M, we can substitute this value into the Ksp expression:

Ksp = [Pb2+](0.012)^2

To solve for [Pb2+], we need to know the value of the Ksp for PbF2. According to the solubility product database, the Ksp of PbF2 is approximately 4.0 x 10^-8.

Hence, we can set up the equation:

4.0 x 10^-8 = [Pb2+](0.012)^2

Solving for [Pb2+], we can rearrange the equation:

[Pb2+] = 4.0 x 10^-8 / (0.012)^2

Calculating this expression will give us the concentration of Pb2+ ions at equilibrium when PbF2 is poured into the solution of 0.012 M F- ions.