`Mg(OH)_(2)` is precipitated when NaOH is added to a solution of `Mg^(2+)` . If the final concentration of `Mg^(2+)` is `10^(-10)` .M, the concetration of `OH^(-)` (M) is the solution is b
[Solubility product for `Mg(OH)_(2)=5.6xx10^(-12)`]

Options

A. 0.056
B. 0.12
C. 0.24
D. 0.025

Answer

C

Explanation

`K_(sp) Mg(OH)_(2)=[Mg^(+2)][OH^(-)]^(2)`
`5.6xx10^(-12)=[10^(-10)][OH^(-)]^(2)`
`[OH^(-)]=sqrt(5.6xx10^(-2))=0.24 M`