.........BaF2 ==> Ba^2+ + 2F^-
I.......solid.....0........0
C........-x.......x........2x
E.......solid.....x........2x
Ksp = (Ba^2+)(F^-)^2
You know (Ba^2+) = 2E-3M
Solve for F^-
..........ScF3 ==> Sc^2+ + 3F^-
I.........solid.....0.......0
C..........-x.......x.......3x
E..........solid....x.......3x
Ksp = (Sc^3+)(F^-)^3
You know (Sc^3+) = 1E-2
Solve for F^-.
Since you are adding F^- drop by drop, the one that ppts first will be the one with the smallest F^- in your calculations. That will be the one that reaches Ksp first.
A solution contains 2.0 × 10^–3 M Ba^2+ and 1.0 × 10^–2 M Sc^3+. A source of F^– is slowly added to the solution. Will BaF2 (Ksp = 1.84 × 10^–7) or ScF3 (Ksp = 5.81 × 10^–24) precipitate first?
I believe that ScF3 would precipitate first but I don't know how to prove this mathematically.
1 answer
1 answer