Question

Please explain and show all work on how to solve this:

The solubility product of CaF2 is 4.0x10^-4. Calc. the molar solubility (s) of calcium Floride in pure water.

ksp=(Ca^2+)(F^-)^2

ksp=(s)(2s)^2 =4.0x10^-4

4s^3=4.0x10^-4

s=2.15x10^-4 molar solubility

I don't understand how to get s?

Answer 1

The solubility and Ksp are related as

CaF2 --> Ca⁺² + 2F^-

Ksp = Solubility of Ca+2 X (2 X solubility of F-)^2

Let solubility of CaF2 = s

so solubility of Ca+2 = s

Solubility of F- = s

Ksp = s X (2s)²

4.0x10^-4 = 4s^3

Let us sole the expression as

s^3 = 4.0x10^-4 / 4 = 10^-4

We can solve it by taking log on both the side

3 log (s) = -4 log 10

3log(s) = -4

so log (s) = -4/3 = -1.33

Now let us take antilog of the two side

s = 0.0467

So the answer is solubility = 0.0467

Let us cross check our results (as they are different from what you have mentioned)

We will put the solubility values and will calculat the Ksp

Ksp = 0.0467 X (2 X 0.0467) ^2 = 0.000407

So Ksp = 4 X 10^-4

So yes our answer is correct