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At 25^(@)C, the solubility product of Mg...

At `25^(@)C`, the solubility product of `Mg(OH)_(2)` is `1.0xx10^(-11)`. At which `pH`, will `Mg^(2+)` ions start precipitating in the form of `Mg(OH)_(2)` from a solution of `0.001 M Mg^(2+)` ions ?

A

`9`

B

`10`

C

`11`

D

`8`

Text Solution

Verified by Experts

The correct Answer is:
B

`K_(SP) = [Mg^(+2)][OH^(-)]^(2)`
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