The minimum volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution (\(K_{sp}\) of \(\mathrm{PbCl_2} = 3.2 \times 10^{-8}\), atomic mass of Pb = 207 u) is :
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- 0.36 L
- 17.98 L
- 0.18 L
- 1.798 L
The Correct Option is C
Solution and Explanation
Moles in 0.1 g = 0.1/278 = \(3.6 \times 10^{-4}\) mol.
\(\mathrm{PbCl_2 \rightleftharpoons Pb^{2+} + 2Cl^-}\), so \(K_{sp} = s(2s)^2 = 4s^3\).
\(s^3 = 3.2\times10^{-8}/4 = 8\times10^{-9}\), so \(s = 2\times10^{-3}\) mol/L.
Volume of water needed = moles/s = \(3.6\times10^{-4}/(2\times10^{-3}) \approx 0.18\) L.
Answer: (3) 0.18 L
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