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Question

Find moles of NH4Cl required to prevent Mg(OH)2 from precipitating in a litre of solution which contains 0.02 mole of NH3 and 0.001 mole of Mg2+ ions.
Given : Kb(NH3)=105;Ksp[Mg(OH)2]=1011

A
104
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B
2×103
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C
0.02
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D
0.1
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Solution

The correct option is B 2×103
[OH]=Ksp[Mg2+]=1011103=104
pOH=pKb+log[NH4Cl][NH3]
4=5+log[NH4Cl][NH3]
0.10=[NH4Cl][NH3]
moles of NH4Cl required =0.1×0.02
=2×103
Below this concentration, Mg(OH)2 will precipiate.

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