Question

How would you calculate the molar solubility of Mn(OH)2 (Ksp = 1.6e-13) in pure water?

Answer 1

`Mn(OH)_2(s) rightleftharpoons Mn^(2+) + 2OH^-`

`K_(sp) = 1.6xx10^(-13)`

Explanation:

`K_(sp) = 1.6xx10^(-13)=[Mn^(2+)][OH^-]^2`

If we call the solubility of manganese(II) hydroxide `S`, then by the stoichiometry of the reaction:

`K_(sp)=S(2S)^2=4S^3`.

So `S = root(3)((1.6 xx 10^(-13))/4)`