<span>Answer:
MgF2 ====⇒ Mg+2 + 2F-
Ksp = [Mg+2][F-]^2
Let S = molar solubility of MgF2
[Mg+2] = S ; [F-] = 2s
Since the [F-] is initially 0.40 M, then [F-] = 0.40 + 2S
6.4 x 10^-9 = (S) (0.40 + 2S)^2 ; one can neglect the 2S in the 0.40 + 2S expression since it is very, very small compared to the 0.40 already present.
6.4 x 10^-9 = S(0.40)^2
S = 4.0 x 10^-8
Molar solubility = 4.0 x 10^-8</span>