Equilibrium Question 888
Question: Solubility of $ PbCl_2 $ in 0.5 M NaCl is $ S_1 $ and that in 0.5 M $ Pb{{(NO_3)}_2} $ is $ S_2 $ then which of 5 the following is correct:
Options:
A) $ \frac{S_1}{S_2}=1 $
B) $ S_1-S_2>0 $
C) $ S_1-S_2<0 $
D) $ \frac{S_1}{S_2}>1 $
Show Answer
Answer:
Correct Answer: C
Solution:
$ S_1=\frac{K _{sp}}{{{0.5}^{2}}},,S_2=\sqrt{\frac{K _{sp}}{0.5}} $
$ \frac{S_1}{S_2}=\frac{K _{sp}}{{{0.5}^{2}}}\times \sqrt{\frac{0.5}{K _{sp}}}=\frac{\sqrt{K _{sp}}}{0.5\times \sqrt{0.5}}=\frac{\sqrt{K _{sp}}}{0.707\times 0.5} $
But $ \because $ $ K _{sp}«<1 $
$ \therefore $ $\frac{S_1}{S_2}<1 $
$ \therefore $ $S_1<S_2 $
$ \therefore $ $S_1-S_2<0 $
BETA
