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K(sp) of Mg(OH)(2) is 1xx10^(-12), 0.01M...

`K_(sp) of Mg(OH)_(2)` is `1xx10^(-12), 0.01M MgCI_(2)` will be precipitating at the limiting `pH`:

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The correct Answer is:
T
`M(OH)_(3) `is precipitate
`[M^(3+)][OH^(Theta)]^(3) gt K_(sp) [OH^(Theta)]^(3)gt 1 xx 10^(-9)`
`[OH^(Theta)] gt 1 xx 10^(-3)`
Maximum `pOH = 3` Minimum `pH = 9`
Hence true.
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