MHT CET · Chemistry · Ionic Equilibrium
The solubility product of \(\mathrm{PbCl}_2\) at \(298 \mathrm{~K}\) is \(3.2 \times 10^{-5}\). What is its solubility in mol dm\({ }^{-3}\) ?
- A \(8 \times 10^{-6}\)
- B \(2 \times 10^{-2}\)
- C \(5.6 \times 10^{-3}\)
- D \(5.0 \times 10^{-2}\)
Answer & Solution
Correct Answer
(B) \(2 \times 10^{-2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{K}_{\text {sp }}=3.2 \times 10^{-5} \\ & \mathrm{PbCl}_{2(\mathrm{~s})} \rightleftharpoons \mathrm{Pb}_{(\mathrm{aq})}^{2+}+2 \mathrm{Cl}_{(\mathrm{aq})}^{-} \\ & \therefore \quad \mathrm{K}_{\mathrm{sp}}=4 \mathrm{~S}^3 \\ & \therefore \quad 4 \mathrm{~S}^3=3.2 \times 10^{-5} \\ & \mathrm{~S}=\sqrt[3]{\frac{3.2 \times 10^{-5}}{4}} \\ & =\sqrt[3]{8 \times 10^{-6}} \\ & =2 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3} \\ & \end{aligned}\)
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