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Question

Calculate the solubility of a sparingly soluble compund A2X3 at 25C.
The solubility product Ksp of A2X3 is 1.08×1023 at the given temperature.

A
1×107 mol L1
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B
1.08×107 mol L1
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C
1×105 mol L1
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D
1.08×103 mol L1
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Solution

The correct option is C 1×105 mol L1
The equilibrium reaction is:
A2X3 (s)2A3+(aq)+3X2(aq)c00cs2s3s
Here s is the solubility of A2X3
Since,
Ksp=[A3+]2[X2]3
Ksp=(2s)2×(3s)3
Ksp=4s2×27s3
108s5=Ksp
s=(Ksp108)15

Putting the value of Ksp
s=(1.08×1023108)15=(1025)15Ksp=1×105 mol L1

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