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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(OH)_(2)hArr Mg^(2+) +2OH^(-)`
`K_(sp) = [Mg^(2+)][OH^(-)]^(2)`
` 1 xx 10^(-12) = 0.01 [OH^-)]^(2)`
`[OH^(-)]^(2) = 1 xx 10^(-10) rArr [OH^(-)] = 10^(-5)`
`[H^(+)]=10^(-14)//10^(-5) = 10^(-9)`
`pH = - log [H^(+)] = - log [10^(-9) ] = 9 `
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