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Question

Ionic product of Ni(OH)2 is 2.0×1015. Molar solubility of Ni(OH)2 in 0.10M NaOH will be ________.

A
1.0×1013M
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B
2.0×1013M
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C
4.0×1013M
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D
8.0×1013M
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Solution

The correct option is D 2.0×1013M
Let the solubility of Ni(OH)2=S

Ni(OH)2+H2ONi2+(aq)+2OH(aq)
S S 2S

Ksp=2.0×1016=[Ni2+][OH]2

=[S][0.10+2S]2

As Ksp is small 2S<<0.10

therefore (0.10+2S)=0.10

Hence, 2.0×1015=S(0.10)2

S=2.0×1013M=[Ni2+]=[Ni(OH)2]

Hence, the correct option is B

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