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Mg(OH)(2) is precipitated when NaOH is a...

`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)`]

A

`0.056`

B

`0.12`

C

`0.24`

D

`0.025`

Text Solution

Verified by Experts

The correct Answer is:
C
`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 `.
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