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Question

Chemistry Question on Equilibrium

If the solubility product of lead iodide (PbI2)(PbI_2) is 3.2×10−83.2 \times 10^{-8}, its solubility will be

A

2×10−3 M2 \times 10^{-3} \,M

B

4×10−4 M4 \times 10^{-4} \,M

C

1.6×10−5 M1.6 \times 10^{-5} \,M

D

1.8×10−5 M1.8 \times 10^{-5} \,M

Answer

2×10−3 M2 \times 10^{-3} \,M

Explanation

Solution

PbI2⇌Pb2+S+2I− 2S?;Ksp=4S3PbI_{2} {\rightleftharpoons} \underset{\text{S}}{{Pb^{2+}}}+\underset{\text{ 2S}}{{2I^{-}}} ? ; K_{sp}=4S^{3} ∴ 4S3=3.2×10−8\therefore\, 4S^{3}=3.2\times 10^{-8} ⇒S=2×10−3 M\Rightarrow S=2\times10^{-3}\,M