Question

The solubility products of `Fe(OH)_2` and `Fe(OH)_3` are `10^(-17)` and `10^(-38)` respectively. If the concentrations of `Fe^(2+)` and `Fe^(3+)` are each `10^(-5)`, what is the `[OH^-]` when each just begin to precipitate?

Answer 1

`[OH^-]=1 xx 10^-6`
and`[OH^-]=1 xx 10^-11`

Explanation:

for the reactions
`Fe(OH)_2 = Fe^(2+) + 2 OH^-` and
`Fe(OH)_3 = Fe^(3+) + 3 OH^-` we have the espressions of Kps that are
`Kps = [Fe^(2+)] xx [OH^-]^2 = 1 xx 10^-17` and
`Kps = [Fe^(3+)] xx [OH^-]^3 = 1 xx 10^-38`
from wich you have
`[OH^-]= sqrt ((Kps)/[Fe^(2+)])= sqrt ((1 xx 10^-17)/(1 xx 10^-5))=1 xx 10^-6`
pOH = 6; pH = 8
and
`[OH^-]= root(3)((Kps)/[Fe^(3+)])= root(3) ((1 xx 10^-38)/(1 xx 10^-5))=1 xx 10^-11`
pOH = 11; pH = 3