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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

8

B

9

C

10

D

11

Text Solution

Verified by Experts

The correct Answer is:
C
`K_(sp)=[Mg^(+2)][OH^(-)]^(2)`
`1xx10^(-11)=10^(-3)xx[OH^(-)]^(2)`
`[OH^(0)]^(2)=10^(-8)`
`OH^(-)=10^(-4)`
pOH=4 [ pH+pOH=14]
pH=10
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