what is the molar solubility of barium fluoride in 0.15M NaF at 25 degrees C?

Answer 1

BaF2 ==> Ba^+2 + 2F^-
S = solubility
..S......S........2S

NaF ==> Na^+ + F^-
0.15....0.15....0.15

Ksp = (Ba^+2)(F^-)^2 =
(S)(S+0.15)^2 = Ksp. Solve for S.

Answer 2

This is almost correct, except it should be (s)(2s+.15)^2=Ksp

Answer 3

6.9*10-3