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The solubility product of Mg(OH)(2) is 1...

The solubility product of `Mg(OH)_(2)` is `1.0 xx 10^(-12)`. Concentrated aqueous `NaOH` solution is added to a `0.01 M` aqueous solution of `MgCl_(2)`. The pH at which precipitation occur is -

A

7.2

B

7.8

C

`8.0`

D

`9.0`

Text Solution

Verified by Experts

The correct Answer is:
D
`{:(MgCl_(2)rarr,Mg^(+2),2Cl^(-)),(,?),(,0.01M):}`
`K_(sp) = Q = [Mg^(+2)][OH^(-)]^(2)`
`10^(-12)=[0.01][OH^(-)]^(2)`
`[OH^(-)]^(2) = 10^(-10)`
`[OH^(-)] = 10^(-5)`
`pOH = 5 :. pH = 9`
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