`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)`
` 5.6 xx 10^(-12) = [10^(-10)] [OH^(-)]^(2)`
`[OH^(-)]= sqrt(5.6 xx10^(-2)) = 0.24 M `.