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Question

The Ksp of Mg(OH)2 is 1×1012. 0.01 m Mg(OH)2 will precipitate at the limiting pH equal to:

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Solution

The solubility product expression is:

KSP=[Mg2+][OH]2

Rearrange the above expression:

[OH]=Ksp[Mg2+]

Substitute values in the above expression and calculate hydroxide ion concentration-

[OH]=1×10120.01

[OH]=1×105 M

Calculate pOH from the hydroxide ion concentration:

pOH=log[OH]=log(1×105)

pOH=5

Calculate pH from pOH-

pH=14pOH

pH=145

pH=9

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