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Q.

At 25° C, the solubility product of Mg(OH)2 is 1.0×10−11 . At which pH, will Mg2+ ions starts precipitating in the form of Mg(OH)2 from a solution of 0.001 M Mg2+ ions?

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a

11

b

8

c

10

d

9

answer is B.

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Detailed Solution

Mg(OH)2(s)→Mg2++2OH−1

Moles of Mg(OH)2(s)= moles of Mg2+

Ksp=Mg2+∗OH−121∗10−11=0.001∗OH−12OH−1=10−4M[H]+=10−14OH−1=10−1410−4=10−10MpH=−log⁡[H]+

putting values

pH=10

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