Calculate the concentration of `Mg^(2+)` ions and `OH^(-)` ions in a saturated soluton of `Mg(OH)_(2)`.The solubility product of `Mg(OH)_(2) " is " 9.0 xx 10^(-12)`

Answer

`Mg^(2+) =1.31 xx 10^(-4) "mo" L^(-1) , OH^(-1) =2.62 xx 10(-4) "mol "L^(-1)`

Explanation

Let the solubility of Mg `(OH)_(2) =S " moles " L^(-1)`
The solubility product of `Mg(OH)_(2) =9.0 xx 10^(-12)`
The solubility equilibrium of `Mg(OH)_(2)` may be represented
`Mg(OH)_(2) ,hArr, underset(S)(Mg^(2+)(aq)),+,underset(2S)(2OH^(-) (aq))`
`K_(sp) =[Mg^(2+) (aq)]^(2) [OH^(-)(aq)]^(2)=S xx (2S)^(2) =4S^(3)`
`" or " " "S=((K_(sp))/(4)) ^(1//3) =((9.0 xx 10^(-12))/(4))^(1//3) =1.31 xx 10^(-4) " mol" L^(-1)`
`:.` concentration of `Mg^(2+)` ions is solution `=1.31 xx 10^(-4) " mol " L^(-1)`
concentration of `OH^(-)` ions in solution `=2xx 1.31 xx 10^(-4) =2.62 xx 10^(-4) " mol " L^(-1)`