Find solubility of $\mathrm{PbI}_2$ if its solubility product is $7.0 \times 10^{-9}$.

Find solubility of $\mathrm{PbI}_2$ if its solubility product is $7.0 \times 10^{-9}$.
  1. $1.21 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}$
  2. $3.228 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}$
  3. $2.831 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}$
  4. $1.811 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}$

Solution

$\begin{array}{ll} & \text { For } \mathrm{PbI}_2, \\ & \mathrm{PbI}_{2(\mathrm{~s})} \rightleftharpoons \mathrm{Pb}_{\text {(aq) }}^{2+}+2 \mathrm{I}_{(\mathrm{aq})}^{-} \\ & x=1, \mathrm{y}=2 \\ \therefore \quad & \mathrm{K}_{\mathrm{sp}}=x^x \mathrm{y}^{\mathrm{y}} \mathrm{S}^{x+\mathrm{y}}=(1)^1(2)^2 \mathrm{~S}^{1+2}=4 \mathrm{~S}^3 \\ \therefore \quad & \mathrm{S}=\sqrt[3]{\frac{\mathrm{Ksp}}{4}}=\sqrt[3]{\frac{7.0 \times 10^{-9}}{4}}=1.21 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\end{array}$

Asked in: MHT CET 2023 (09 May Shift 2)

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