The solubility products of `AI(OH)_(3)` and `Zn(OH)_(2)` are `8.5xx10^(-23)` and `1.8xx10^(-14)` respectively. If `NH_(4)OH` is added to a solution containing `AI^(3+)` and `Zn^(2+)` ions, then substance precipitated first is:

Options

A. `AI(OH)_(3)`
B. `Zn(OH)_(2)`
C. Both (a) and (b)
D. None of these

Answer

A

Explanation

(Assume `[AI^(3+)] =[Zn^(2+)])`
The compound which have less `K_(sp)` value will be precipitated first. Since the `K_(sp)` of `AI(OH)_(3)` is less than `K_(sp)` of `Zn(OH)_(2)`. So former is precipitated.
Note: In such questions no need to calcuting the exact value only approximation is requires.
`[overset(Theta)OH]_(min)` for `AI^(3+) = ((K_(sp) "of" AI(OH)_(3))/([AI^(3+)]))^(1//3)`
`[overset(Theta)(O)H]_(min) "for" Zn^(2+) = ((K_(sp) "of" Zn(OH)_(2))/([Zn^(2+)]))^(1//2)`