Equilibrium Question 183

Question: At 298 K, the solubility product of $ PbCl_2 $ is $ 1.0\times {10^{-6}} $ . What will be the solubility of $ PbCl_2 $ in moles/litre [MP PMT 1990; CPMT 1985, 96]

Options:

A) $ 6.3\times {10^{-3}} $

B) $ 1.0\times {10^{-3}} $

C) $ 3.0\times {10^{-3}} $

D) $ 4.6\times {10^{-14}} $

Show Answer

Answer:

Correct Answer: A

Solution:

$ K _{sp}=4s^{3} $

$ S=\sqrt[3]{\frac{K _{sp}}{4}}=\sqrt[3]{\frac{1.0\times {10^{-6}}}{4}} $

$ =6.3\times {10^{-3}} $ .



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