MHT CET · Chemistry · Ionic Equilibrium
The solubility product of \(\mathrm{PbI}_2\) is \(1.08 \times 10^{-7}\). Calculate its solubility in \(\mathrm{mol~} \mathrm{dm}^{-3}\) at 298 K ?
- A \(2.018 \times 10^{-3}\)
- B \(2.011 \times 10^{-9}\)
- C \(1.259 \times 10^{-9}\)
- D \(3.0 \times 10^{-3}\)
Answer & Solution
Correct Answer
(D) \(3.0 \times 10^{-3}\)
Step-by-step Solution
Detailed explanation
\( K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{I}^{-}]^2 \) \( K_{sp} = (s)(2s)^2 = 4s^3 \)
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