help! calculate calcium ions in a saturated solution of calcium fluoride = 2.15 x 10^-14. ksp?

Question

In an experiment, you determined the concentration fo calcium ions in a saturated solution of calcium fuoride to be 2.15 x 10^-4 M. ksp?

Answer 1

The equilibrium is
CaF2 <---> Ca2+ + 2F-
the requirement for this equilibrium is that
Ksp = [Ca2+] [ F-]^2
Let x = moles/l of CaF2 that dissolve : we get x moles/l Ca2+ and 2x moles/l F-
Ksp = (x)(2x)^2 = 4x^3 = 2.15 x 10^-14
x = 1.75 x 10^-5 M = [ Ca2+]

Answer 2

CaF2<-->Ca++ +2F-
ksp = [x]*[2x]^2 = 4x^3 , where x is the concentration

so 4x^3= 2.15*10^-14
x^3=5.375*10^-15

x=1.75*10^-5M

Answer 3

Do you own homework.