What mass of NaF must be added to 1.0 L of 0.0036 M Pb2+ to initiate precipitation of PbF2(s)? (Ksp of PbF2 is 3.3 × 10-8; assume no volume change on addition of NaF.)

Question

can you please explain

Answer 1

..................PbF2 ==> Pb^2+ + 2F^-
I...................solid...........0............0
C..................solid..........x.............2x
E...................solid..........x............2x
Ksp = (Pb^+)(F^-)^2
You know Ksp. You know (Pb^2+) from the problem. You want to know (F^-). Solve for that (in mols/L). Convert (F^-) to (NaF) in mols/L and change that to grams/L. The idea here is that PbF2 will ppt when Ksp is exceeded. So you're calculating the amount of NaF to add to the solution so (Pb^+)(F^-)^2 is greater than Ksp for PbF2. Post your work if you get stuck.