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Question

The solubility of lead (II) iodide is 0.064 g/100 mL at 25o C. What is its solubility product? 


Your Answer
A
2.08×108 mol3 L3
Correct Answer
B
1.07×108 mol3 L3
Your Answer
C
0.17×108 mol3 L3
Your Answer
D
3.5×108 mol3 L3

Solution

The correct option is D 1.07×108 mol3 L3
Given:
Solubility=0.064 g/100mL
Solubility=0.64 g/L
So, Calculating Solubility in mol/L,
s=0.64461 mol/L   ,as molar mass ofPbI2=461g
[PbI2]=1.39×103 M

The equilibrium established will be,
Pbl2(s)Pb2+(aq)+2I(aq)[c]i1.39×10300[c]e1.39×1032.78×103

Ksp=[Pb2+]×[Cl]2

Putting the values,
Ksp=1.39×103×(2.78×103)2=1.07×108 mol3 L3

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