Question

The molar solubility of MgF2 at 25°C is 1.2 10-3 mol/L. Calculate Ksp.

i got the euqation ksp= [F-]^2[Mg]2+ and plugged in 1.2e-3 and got 1.7e-9 but it is wrong.

also this question

If 0.0968 g of CaF2 dissolves per liter of solution, calculate the solubility-product constant.

how do i do that?
......MgF2 ==> Mg^2+ + 2F^-
I.....solid.....0.......0
C..x dissolves..x.......2x
E......solid....x.......2x

Ksp = (Mg^2+)(F^-)^2
Ksp = (x)(2x)^2
Solve for Ksp. (I'm sure the problem you have is that you didn't double x before you squared.
You can save some time by
Ksp = 4x^3 = 4(1.2E-3)^3 = ?

The CaF2 is the same process.

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