The solubility of `Mg( OH)_2` in pure water is ` 9.57 xx 10 ^(-3) g L^(-1) ` .Calculate its solubility I ` (g L^(-1))` in 0.02 M Mg `(NO_3)_2` solutions.

Explanation

Molecular mass of `Mg(OH)_(2) = 58 g mol^(-1)`
Solubility of `Mg(OH)_(2)` in pure water
`= 9.57 xx 10^(-3) gL^(-1) = (9. 57xx10^(-3))/(58)M`
`Mg (OH)_(2) (s) hArr Mg^(2+) (aq) + 2overset(Θ)OH (aq)`
If `S` is the solubility of `Mg(OH)_(2) "in mol" L^(-1)`, then
solubility `= [Mg^(2+)] = S` and `[OH^(Θ)] = 2S`
`K_(sp) = [Mg^(2+)] [overset(Θ)OH]^(2) - S xx (2S)^(2) = 4S^(3)`
`= 4 xx ((9.57 xx 10^(-3))/(58))^(3) = 1.8 xx 10^(-11)`
In `0.02M Mg (NO_(3))_(2)`
`{:(,Mg(NO_(3))_(2)(s)rarr,Mg^(2+)(aq),+,2Noverset(Θ)O_(3)(aq)),("Final",0.02M,0,,),("Final",0,0.02M,,):}`
`[Mg^(2+)] = [Mg^(2+)]` from `Mg(NO_(3))_(2) added + [Mg^(2+)]`
from `Mg(OH)_(2) = 0.02M + S ~~ 0.02 M, [OH^(Θ)] = 2S`
`K_(sp) = [Mg^(2+)] [overset(Θ)OH]^(2) = (0.02) xx (2S)^(2) = 0.08S^(2)`
`0.08S^(2) = 1.8 xx 10^(-11)` and `S^(2) = (1.8 xx 10(-11))/(0.08) = 2.25 xx 10^(-10)`
`S = sqrt(2.25 xx 10^(-10)) = 1.5 xx 10^(-5)M`
`= 1.5 xx 10^(-5) xx 58 = 8.7 xx 10^(-4) gL^(-1)`