Asked by jade
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.)
can you please explain
can you please explain
Answers
Answered by
DrBob222
..................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.
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.