MHT CET · Chemistry · Ionic Equilibrium
Find solubility of \(\mathrm{PbI}_2\) if its solubility product is \(7.0 \times 10^{-9}\).
- A \(1.21 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\)
- B \(3.228 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\)
- C \(2.831 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\)
- D \(1.811 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\)
Answer & Solution
Correct Answer
(A) \(1.21 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\text { For } \mathrm{PbI}_2, \)
\( \mathrm{PbI}_{2(\mathrm{~s})} \rightleftharpoons \mathrm{Pb}_{\text {(aq) }}^{2+}+2 \mathrm{I}_{(\mathrm{aq})}^{-} \)
\( x=1, \mathrm{y}=2 \)
\( \therefore \mathrm{K}_{\mathrm{sp}}=x^x \mathrm{y}^{\mathrm{y}} \mathrm{S}^{x+\mathrm{y}}=(1)^1(2)^2 \mathrm{~S}^{1+2}=4 \mathrm{~S}^3 \)
\( \therefore \mathrm{S}=\sqrt[3]{\frac{\mathrm{Ksp}}{4}}=\sqrt[3]{\frac{7.0 \times 10^{-9}}{4}}=1.21 \times 10^{-3}\)\(\mathrm{~mol} \mathrm{~L}^{-1}\)
\( \mathrm{PbI}_{2(\mathrm{~s})} \rightleftharpoons \mathrm{Pb}_{\text {(aq) }}^{2+}+2 \mathrm{I}_{(\mathrm{aq})}^{-} \)
\( x=1, \mathrm{y}=2 \)
\( \therefore \mathrm{K}_{\mathrm{sp}}=x^x \mathrm{y}^{\mathrm{y}} \mathrm{S}^{x+\mathrm{y}}=(1)^1(2)^2 \mathrm{~S}^{1+2}=4 \mathrm{~S}^3 \)
\( \therefore \mathrm{S}=\sqrt[3]{\frac{\mathrm{Ksp}}{4}}=\sqrt[3]{\frac{7.0 \times 10^{-9}}{4}}=1.21 \times 10^{-3}\)\(\mathrm{~mol} \mathrm{~L}^{-1}\)
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