MCQ
If the solubility product of lead iodide $(Pb{l_2})$ is $3.2 \times {10^{ - 8}},$ then its solubility in moles/litre will be
- ✓$2 \times {10^{ - 3}}$
- B$4 \times {10^{ - 4}}$
- C$1.6 \times {10^{ - 5}}$
- D$1.8 \times {10^{ - 5}}$
✓
Answer
Correct option: A.
$2 \times {10^{ - 3}}$
(a) ${K_{sp}} = 4{S^3}$
$4{S^3} = 3.2 \times {10^{ - 8}}$ ; $S = 2 \times {10^{ - 3}}M$.
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