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

In pure water, solubility = `9.57/58 xx 10^(-3)M = 1.65 xx 10^(-4)M`
`K_(sp) = 4S^(3) = 4 (1.65 xx 10^(-4))^(3) = 1.8 xx 10^(-11)`
In 0.02 M `Mg(NO_(3))_(2),`
Solubility of `Mg(OH_(2)) = sqrt(K_(sp)/[Mg^(2+)]) xx 1/2 = 1.5 xx 10^(-5) molL^(-1)`
` = 1.5 xx 10^(-5) xx 58 gL^(-1) = 8.7 xx 10^(-4) gL^(-1)`