Compare the solubility of `Fe(OH)_(3) (K_(sp) = 4 xx 10^(-38))` and `Ni(OH)_(2).(K_(sp) = 2 xx 10^(-16))` at `pH = 5.0`

Explanation

`pH = 5.0, pOH = 14 - 5 = 9`
`[overset(Θ)OH] =- Antilog 9.0 = 1 xx 10^(-9)M`
`Fe(OH)_(3) hArr Fe^(3+) (aq) +3 overset(Θ)OH (aq)`
`[Fe^(3+)] [overset(Θ)OH]^(3) = K_(sp) = 4 xx 10^(-38)`
`[Fe^(3+)] = S = (4xx10^(-38))/((1.8xx10^(-9))^(3)) = 4 xx 10^(-11)M`
`Ni^(2+)` ions can be present in high concentration in solution and do not precipiate. Let `[Ni^(2+)]`be `1M`
`[Ni^(2+)] [overset(Θ)OH]^(2) = (1)^(2) (+- 10^(-9))^(2) lt 1xx10^(-18) lt K_(sp)`