Question

The solubility of scandium (III) fluoride, `ScF_3`, in pure water is `2.0 x 10^-5` moles per liter. How do you calculate the value of `K_(sp)` for scandium (III) fluoride from this data?

Answer 1

`K_"sp"(ScF_3)=4.32xx10^-18`

We interrogate the reaction.....

Explanation:

We interrogate the reaction.....

`ScF_3(s) rightleftharpoons Sc^(3+) + 3F^-`

And as written, we can directly write the solubility expression, `K_"sp"`, where `K_"sp"=[Sc^(3+)][F^-]^3`

And we are given the solubility of `ScF_3` in water, `2.0xx10^-5*mol*L^-1`....

And thus `[Sc^(3+)]=2.0xx10^-5*mol*L^-1`

And `[F^-]=3xx2.0xx10^-5*mol*L^-1`

Capisce?

And then we input these numbers into the `K_"sp"` expression.....

`K_"sp"=(2.0xx10^-5)xx(6.0xx10^-5)^3=4.32xx10^-18`

PS I have not got a good text at hand, but the values I read for `K_"sp"` for scandium fluoride seem to vary widely. I would check the source of these data.