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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
`Mg(OH)_(2) to Mg^(2+) + 2OH^(-)`
`K_(sp)=[Mg^(2+)][OH^(-2)]`
`[OH^(-)]=sqrt(K_(sp)/([Mg^(2+)])) = sqrt((1xx10^(-11))/0.001) = 10^(-4)`
`pOH = -log[OH^(-)] = -log[10^(-4)]`
`pOH=4 therefore pH=14-pOH =14-4=10`
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