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}} $ .
