The salt `ZN (OH)_(2)` is involved in the following two equilibria:
`Zn(OH)_(2)(s) hArr Zn^(3+)(aq) +2 overset(Theta)OH(aq) K_(sp) = 1.2 xx 10^(-17)`
`Zn(IH)_(2)(s)+2 overset(Theta)OH hArr Zn(OH)_(4)^(2-)(aq)K_(f) = 0.12`
Calculate `|overset(Theta)OH|` at which solubility of `Zn(OH)_(2)` be a minimum. Also find the solubility of `Zn(OH)_(2)` at this `pH`.

Explanation

a. `Zn(OH)_(2)(s) hArr Zn^(2+) +2 overset(Theta)OH`
`K_(sp) = [Zn^(2+)] [overset(Theta)OH]^(2)`
b. `Zn(OH)_(2)(s) + 2 overset(Theta)OH hArr (Zn(OH)_(4))^(2-)`
`K_(f) = ([(Zn(OH)_(4))^(2-)])/([overset(Theta)OH]^(2))`
Let the solubility be `S` is charge (a) and thus solubility will also be `S` in change (b). By coupling (a) and(b), `2Zn(OH)_(2)(s) hArr Zn^(2+) (aq) + (Zn(OH)_(4))^(2-)(aq)`.
`K_(sp) xx K_(f) = S xx S`
or `S = sqrt(K_(sp) xx K_(f)) = sqrt(2xx10^(-17)xx0.2) = 1.2 xx 10^(-9)M`
Thus, total solubility will be
`2S = 2 xx 1.2 xx 10^(-9) = 2.4 xx 10^(-9)M`
This solubility is minimum when
`[Zn^(2+)] [overset(Theta)OH]^(2) = 1.2 xx 10^(-17)`
`1.2 xx 10^(-9) [overset(Theta)OH]^(2) = 1.2 xx 10^(-17)`
`:. [overset(Theta)OH] = 10^(-4)M`
Note: If only (a) change than solubility of
`Zn(OH)_(2) =3sqrt((K_(sp))/(4)) =3sqrt((1.2xx10^(-17))/(4)) = 1.44 xx 10^(-6)M` and at this concentration `[overset(Theta)OH] = 2 xx 1.44 xx 10^(-6)M`
`= 2.88 xx 10^(-6)M`