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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
`[Mg^(2+)][OH^(-)^(2)=10^(-12)`
`0.01xx[OH-]^(2)=10^(-12)`
`[OH-]=10^(-5)M`
`:. [H^(+)]=10^(-9)M` and `pH= -log[10^(-9)]=9`
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