Question

How to find solubility of a salt based on Ksp?

The problem says, 'If the Ksp value for `CdF_2` is `6.44\times10^(-3)`, what is the solubility of `CdF_2`?'

I'm not entirely sure where to start with this. Thanks!

Answer 1

`K_"sp"` values express equilibrium concentrations...

Explanation:

Cadmium fluoride is moderately soluble...and in contact with water participates in the following solubility equilibrium...

`CdF_2(s) stackrel(H_2O)rightleftharpoonsCd^(2+) + 2F^-`

And this equilibrium can be quantified...

`K_"sp"=[Ca^(2+)][F^-]^2=6.44xx10^-3`...

...`CaF_2(s)` does not appear in the solubility expression because as a solid, it CANNOT express a concentration....

Now if we let the `"solubility of cadmium fluoride"-=S`...then clearly...`[Cd^(2+)]=S`, and `[F^-]=2S`. Agreed? This is probably the only part of the solution requiring any thinking.... And of course we assume that cadmium fluoride is the only source of cadmium and fluoride ions....in other problems, this condition might not hold....for instance if the solution were acidified....

`K_"sp"=[Ca^(2+)][F^-]^2=6.44xx10^-3-=Sxx(2S)^2=4S^3`...

`S=""^(3)sqrt{(K_"sp")/4}=""^(3)sqrt{(6.44xx10^-3)/4}=0.117*mol*L^-1`

...and this converts into a mass solubility of `0.117*mol*L^-1xx150.41*g*mol^-1=17.6*g*L^-1`...

For more on `K_"sp"`...see this old answer.