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Question

Ksp of Mg(OH)2 is 4.0×106. At what minimum pH, Mg2+ ions starts precipitating 0.01 MgCI.

A
2+log2
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B
2log2
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C
12+log2
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D
12log2
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Solution

The correct option is C 12+log2
According to given data, [OH] needed for precipitation of
Mg(OH)2=(4×106102)1/2=(4×104)1/2M=2×102M
pOH=log[OH]=2log2
So pH=14pOH=12+log2

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