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The K(sp) of Mg(OH)(2) is 1xx10^(-12). 0...

The `K_(sp)` of `Mg(OH)_(2)` is `1xx10^(-12). 0.01M Mg^(2+)` will precipitate at the limiting pH of

A

3

B

9

C

12

D

8

Text Solution

Verified by Experts

The correct Answer is:
B
`[Mg^(2+][OH^(-)]^(2)= 1xx10^(-12)`,
`:. [OH^(-)]= sqrt((10^(-12))/(0.01))=10^(-5)`
or `pOH=5`
and thus `pH=9`
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