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

At `25^@C`, the solubility product of `Mg(OH)_(2) is 1.0 xx 10^(-11).` At what 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(OH)_2)=[Mg^(2+)] [OH^(-)]^2`
`1 xx 10^(-11)=[0.001] [OH^(-)]^2 rArr [OH^(-)]^2=(10^(-11))/(10^(-3))=10^(-8)`
`rArr [OH^(-)]=10^(-4) rArr pOH =4`
Thus, pH =14-4=10
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