Calculate the solubility of a sparingly soluble compund A2X3 at 25∘C. The solubility product Ksp of A2X3 is 1.08×10−23 at the given temperature.
A
1×10−7molL−1
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B
1.08×10−7molL−1
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C
1×10−5molL−1
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D
1.08×10−3molL−1
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Solution
The correct option is C1×10−5molL−1 The equilibrium reaction is: A2X3(s)⇌2A3+(aq)+3X2−(aq)c00c−s2s3s 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=(1.08×10−23108)15=(10−25)15Ksp=1×10−5molL−1