The solubility of `Mg (OH)_(2)` in pure water is `9.57 xx 10^(-3) gL^(-1)`. Calculate its solubility (in `gL^(-1))` in `0.02M Mg(NO_(3))_(2)` solution.

Explanation

Solubility of `Mg(OH)_(2)` in pure water in moles/litre br> `=(9.57 xx 10^(-3))/(58.31) [Mg(OH)_(2)=58.31]`
`=1.64 xx 10^(-4)`
`[Mg^(2+)]=1.64 xx 10^(-4)`
and `[OH^(-)]=2 xx 1.64 xx 10^(-4)`.
`K_(sp) [Mg(OH)_(2)]=[Mg^(2+)] [OH^(-)]^2`
`=(1.64 xx 10^(-4)) (2 xx 1.64 xx 10^(-4))^2 =1.764 xx 10^(-11)`
Let hte solubility of `Mg(OH)_(2)` in 0.02 M `Mg(NO_(3))_(2)` solution be S moles per litre. In this case,
`[Mg^(2+)]=S+0.02`
`[OH^(-)]=2S`
`K_(sp)=(S+002) (2S)^2=1.764 xx 10^(-11)`
`S=1.48 xx 10^(-5)" mole per litre"`
`=1.48 xx 10^(-5) xx 58.31" g per litre"`
`=8.63 xx 10^(-4)" g per litre"`