Equilibrium Question 181

Question: Solubility of AgCl will be minimum in [CBSE PMT 1995]

Options:

A) 0.001,M,AgNO3

B) Pure water

C) 0.30M

D) 0.01,M,NaCl

Show Answer

Answer:

Correct Answer: C

Solution:

0.01 M CaCl2 gives maximum Cl ions to keep Ksp of AgCl constant, decrease in [Ag+] will be maximum.



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