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Question

The solubility of AgCl in water at 298 K is 1.06×105 mole per litre. Calculate its solubility product at this temperature.

A
1.12×1010 mol2L2
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B
1.12×1010 mol2L2
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C
1.12×1010 mol2L2
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D
1.12×1010 mol2L2
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Solution

The correct option is B 1.12×1010 mol2L2
The equilibrium in the saturated solution will be as :
AgCl(s)Ag++Cl1001sss
Ksp=[Ag+(aq)][Cl(aq)]
=s×s=s2

Given: the solubility of AgCl is 1.06×105 mole per litre.
s=1.06×105 mol L1

putting the values,
Ksp=(1.06×105)2
=1.12×1010 mol2 L2

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