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