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The K(sp) of Mg(OH)(2) is 1xx10^(-12). 0...

The `K_(sp)` of `Mg(OH)_(2)` is `1xx10^(-12)`. `0.01 M Mg(OH)_(2)` will precipitate at the limiting `pH`

A

3

B

9

C

5

D

8

Text Solution

Verified by Experts

The correct Answer is:
B
`K_(sp)` for `Mg(OH)_(2)=[Mg^(2+)][OH^(-)]^(2)=10^(-12)`
`(0.01)(OH^(-))^(2)=10^(-12)` or `[OH^(-)] =10^(-5)`
`[H^(+)]=(K_(w))/([OH^(-)])=(10^(-14))/(10^(-5))=10^(-9)`
`pH=9`
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