The pH of an aqueous solution of `Mg(OH)_(2)` is 9.0. If the solubility product of `Mg(OH)_(2)` is `1xx10^(-11)`, what is `[Mg^(2+)]` ?

Options

A. `1xx10^(-5)`
B. `1.0xx10^(-4)`
C. `1xx10^(-2)`
D. `0.1`

Answer

D

Explanation

`pH=9, " " therefore [H^(+)]=10^(-9)`
`[OH^(-)]=(1xx10^(-14))/(10^(-9))=10^(-5)`
`Mg(OH)_(2)hArr Mg^(2+)+2OH^(-)`
`K_(sp)=[Mg^(2+)][OH^(-)]^(2)`
`1xx10^(-11)=[Mg^(2+)][10^(-5)]^(2)`
`[Mg^(2+)]=(1xx10^(-11))/((10^(-5))^(2))=10^(-1)=0.1`