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Question

The solubility of sparingly soluble salt CdCO3 is 2×106 mol L1 at 25oC. The Ksp value for CdCO3 is:

A
8×1012 mol2L2
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B
2×1012 mol2L2
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C
4×1012 mol2L2
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D
4×1015 mol2L2
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Solution

The correct option is C 4×1012 mol2L2
Given, solubility, s=2×106 mol L1
Lets consider solid CdCO3 is in contact with its saturated aqueous solution. The equilibrium is established between undissolved CdCO3 and its ions. The equilibrium reaction is given by :
CdCO3(s)Cd2+(aq)+CO23(aq) t=0 c 0 0t=teq cs s s
Solubility product: Ksp=[Cd2+][CO23]=s×s
s2=(2×106)2Ksp=4×1012 mol2L2

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