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Question

The solubility product of AgCl at 25C is 1×1010. A solution of Ag+ at a concentration of 4×103 M just fails to yield a precipitate of AgCl with a concentration of 1×103 M of Cl when the concentration of NH3 in the solution is 2×102 M. Calculate the equilibrium constant for
[Ag(NH3)2]+Ag++2NH3

A
108
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B
1010
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C
1012
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D
102
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Solution

The correct option is A 108
Ksp=[Ag+][Cl]1010=[Ag+]×103[Ag+]=107 M

[Ag(NH3)2]+Ag++2NH3Initially: 4×103 107 2×102Equilibrium:4×103x x+107 2x+2×102
Since,the above forward reaction is the dissociation of a complex, it will hardly occur
4×103x4×103x+1071072x+2×1022×102

Thus the equilibrium constant for the above dissociation equilibria of complex is given as:
K=[Ag+][NH3]2[Ag(NH3)2]+]K=107×(2×102)24×103K=108

Since K has a very low value, our assumption is verified that forward reaction will hardly occur.

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