What concentration of the lead ion, Pb2+, must be exceeded to precipitate PbF2 from a solution that is 1.00×10−2M in th...
What concentration of the lead ion, Pb2+, must be exceeded to precipitate PbF2 from a solution that is 1.00×10−2M in the fluoride ion, F−? Ksp for lead(II) fluoride is 3.3×10−8 .
Homework Answers
Q > K
Q > 3.3×10−8 .
Q = [Pb+2][F-]^2
if F- = 10^-2
Solve for Pb+2
Q = [Pb+2][F-]^2
3.3×10−8 . = (Pb+2) ( 10^-2 .)^2
[Pb+2] = 3.3*10^-4
Anything more of that concnetration will make a precipitate
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