The solubility of `Mg(OH)_(2)` is increased by the addition of `overset(o+)NH_(4)` ion. Calculate
a. `Kc` for the reaction:
`Mg(OH)_(2) +2overset(o+)NH_(4) hArr 2NH_(3) +2H_(2)O + Mg^(+2)`
`K_(sp) of Mg(OH)_(@) = 6 xx 10^(-12), K_(b) of NH_(3) = 1.8 xx 10^(-5)`.
b. Find the solubility of `Mg(OH)_(2)` in a solution containing `0.5M NH_(4)C1` before addition of `Mg(OH)_(2)`.b

Explanation

`K_(c) = ([NH_(3)]^(2)[Mg^(+2)])/([overset(o+)NH_(4)]^(2)) = ([NH_(4)OH]^(2)[Mg^(+2)])/([overset(o+)NH_(4)]^(2)) …(i)`
Also,
`NH_(4)OH hArr overset(o+)NH_(4) + overset(Θ)OH`
`K_(b) = ([overset(o+)NH_(4)][overset(Θ)OH])/([NH_(4)OH]) ..(ii)`
`:. K_(c) xx K_(b)^(2) = [Mg^(2+)] [overset(Θ)OH]^(2) = K_(sp) of Mg (OH)_(2)`
`:. K_(c) = (K_(sp))/(k_(b)^(2)) = (6 xx 10^(-12))/((1.8xx10^(-5))^(2)) = 1.85 xx 10^(-2)`
Let a `molL^(-1)` of `Mg(OH)_(2)` be dissolved in presence of `0.05 M NH_(4)C1`
`{:(Mg(OH)_(2)+,2NH_(4)^(o+)hArr,2NH_(3)+,2NH_(2)O+,Mg^(2+)),("Moles before reaction",rArr0.5,0,-,0),("Moles after reaction",rArr(0.5-0.2a),2a,-,a):}`
`:. K_(c) =(axx(2a)^(2))/((0.5-2a)^(2))`
`1.85 xx 10^(-12) = (4a^(3))/((0.5-2a)^(2)) ..(iii)`
Solving cubic equation,
`:. a= 0.081M`
Note: If cubic equaiton is not solved, then neglecting `2a` in comparison to `0.5`, them 'a' becomes equal to `0.1049M`.