The solubility product of $\mathrm{PbCl}_2$ at $298 \mathrm{~K}$ is $3.2 \times 10^{-5}$. What is its…

The solubility product of $\mathrm{PbCl}_2$ at $298 \mathrm{~K}$ is $3.2 \times 10^{-5}$. What is its solubility in mol dm${ }^{-3}$ ?
  1. $8 \times 10^{-6}$
  2. $2 \times 10^{-2}$
  3. $5.6 \times 10^{-3}$
  4. $5.0 \times 10^{-2}$

Solution

$\begin{aligned} & \mathrm{K}_{\text {sp }}=3.2 \times 10^{-5} \\ & \mathrm{PbCl}_{2(\mathrm{~s})} \rightleftharpoons \mathrm{Pb}_{(\mathrm{aq})}^{2+}+2 \mathrm{Cl}_{(\mathrm{aq})}^{-} \\ & \therefore \quad \mathrm{K}_{\mathrm{sp}}=4 \mathrm{~S}^3 \\ & \therefore \quad 4 \mathrm{~S}^3=3.2 \times 10^{-5} \\ & \mathrm{~S}=\sqrt[3]{\frac{3.2 \times 10^{-5}}{4}} \\ & =\sqrt[3]{8 \times 10^{-6}} \\ & =2 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3} \\ & \end{aligned}$

Asked in: MHT CET 2023 (14 May Shift 2)

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