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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

1

B

8

C

9

D

10

Text Solution

Verified by Experts

The correct Answer is:
D
`Mg(OH)_(2)` dissociates as :
`Mg(OH)_(2)hArr Mg^(2+)+2OH^(-)`
`K_(sp)=[Mg^(2+)][OH^(-)]^(2)`
`[OH^(-)]=sqrt((K_(sp))/([Mg^(2+)]))`
`[OH^(-)]=sqrt((1xx10^(-11))/(0.001))=1xx10^(-4)`
`therefore pOH = 4` and pH = 10.
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