The solubility of the fictitious compound, administratium fluoride (AdF3) in water is 3.091×10−4 M. Calculate the value of the solubility product Ksp.

Answer 1

........AdF3 ==> Ad^3+ + 3F^-
I.......solid.....0.......0
C.....x dissolves..x......3x
E.......solid......x......3x

Ksp = (Ad^3+)(F^-)^3
Ksp = (x)(3x)^3
You know x = 3.091E-4. Substitute and solve for Ksp.

Answer 2

2.46x10^-13