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What is the minimum pH of a solution of 0.1 M in `Mg^(2+)` from which `Mg(OH)_(2)` will not precipitate `K_(sp) =1.2 xx 10^(-11) M^(3).`

Answer» `K_(sp) =[Mg^(2+)][OH^(-)]^(2)`
`1.2 xx 10^(11) =[0.1][OH^(-)]^(2)`
`[OH^(-)]^(2) =(1.2 xx10^(-11))/(0.1) =1.2 xx 10^(-10)`
`[OH^(-)]=(1.2 xx 10^(-10))^(1//2) = 1.0954 xx 10^(-5) M`
`pOH =- log (OH^(-)) =- log (1.0954 xx 10^(-5)) = (5 -0.04) =4.96`
`pH =14 -4.96 = 9.04`


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