ExamBro
ExamBro
MHT CET · Chemistry · Ionic Equilibrium

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

  1. A \(1.21 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\)
  2. B \(3.228 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\)
  3. C \(2.831 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\)
  4. D \(1.811 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(1.21 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\)

Step-by-step Solution

Detailed explanation

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