Chemistry question!! Don't understand

Question

The molar solubility of chromium(III) fluoride is 1.3 x 10^{-3} M. What is the Ksp for this compound?

I tried attempting this problem but don't understand what information to take out from the equation to multiply it by the solubility.

Answer 1

Molar solubility means moles of CrF₃ soluble in 1 liter of solution = 1.3x10^-3 mol/L = 1.3x10^-3 M

When this dissolves you can write the equilibrium expression as follows:

CrF₃(s) <==> Cr³⁺(aq) + 3F^-(aq)

Ksp = [Cr³⁺][F^-]3

From the molar solubility we know [Cr³⁺] = 1.3x10^-3 M and [F^-] = 3 x 1.3x10^-3 M = 3.9x10^-3 M

Plugging these values into the Ksp expression, we can solve for Ksp...

Ksp = [1.3x10^-3][3.9x10^-3]³

Ksp = 7.71x10^-11

Answer 2

Cr(III)F→Cr³⁺+3F^-

Ksp=[Cr³⁺][F^-]³

[Cr³⁺] [F^-]

intial 0 0

change 1x10^-3 3(1x10^-3)

equilibrium 1x10^-3 3x10^-3

Ksp=(1x10^-3)(3x10^-3)³=2.7x10^-11