Question

What is the molar solubility of `Cd(OH)_2` given `K_"sp"=5.3xx10^-15`?

Answer 1

`[Cd(OH)_2]=1.10xx10^-5*mol*L^-1`

Explanation:

For the reaction:

`Cd(OH)_2(s) rightleftharpoons Cd^(2+) + 2HO^-`

We write `K_"sp"=[Cd^(2+)][HO^-]^2`

And if we call the solubility of `Cd(OH)_2`, `S`, then.........

`K_"sp"=Sxx(2S)^2=4S^3`

And so `S=""^3sqrt(K_"sp"/4)=""^3sqrt((5.3xx10^-15)/4)=1.10xx10^-5*mol*L^-1`, with respect to `Cd(OH)_2`..........

If the `pH` of the solution were buffered to HIGH or LOW `pH` how would the solubility evolve?