JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
The solubility product of \(PbI _{2}\) is \(8.0 \times 10^{-9} .\) The solubility of lead iodide in \(0.1\) molar solution of lead nitrate is \(x \times 10^{-6} \,mol / L\). The value of \(x\) is ....... .(Rounded off to the nearest integer) \([\) Given \(: \sqrt{2}=1.41]\)
- A \(196\)
- B \(169\)
- C \(112\)
- D \(141\)
Answer & Solution
Correct Answer
(D) \(141\)
Step-by-step Solution
Detailed explanation
Given : \(\left[ K _{ sp }\right]_{ PbI _{2}}=8 \times 10^{-9}\) To calculate : solubility of \(PbI _{2}\) in \(0.1 \,M\) sol of \(Pb \left( NO _{3}\right)_{2}\) \((I)\) \(Pb \left( NO _{3}\right)_{2} \rightarrow Pb _{\text {(aq) }}^{+2}+2 NO _{3}^{-}( aq )\)…
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