Equilibrium Question 181
Question: Solubility of $ AgCl $ will be minimum in [CBSE PMT 1995]
Options:
A) $ 0.001,M,AgNO_3 $
B) Pure water
C) $ 0.30M $
D) $ 0.01,M,NaCl $
Show Answer
Answer:
Correct Answer: C
Solution:
0.01 M $ CaCl_2 $ gives maximum $ C{l^{-}} $ ions to keep $ K _{sp} $ of AgCl constant, decrease in $ [A{g^{+}}] $ will be maximum.