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.



NCERT Chapter Video Solution

Dual Pane