Equilibrium Question 293
Question: If the solubility products of $ AgCl $ and $ AgBr $ are $ 1.0\times {10^{-8}},M $ and $ 3.5\times {10^{-13}} $ respectively, then the relation between the solubilities (denoted by the symbol $ ‘S’ $ ) of these salts can correctly be represented as [MP PET 1994]
Options:
A) $ S $ of $ AgBr $ is less than that of $ AgCl $
B) $ S $ of $ {{(CH_3)}_3}{C^{+}} $ is greater than that of $ AgCl $
C) $ {10^{-11}}M $ of $ BOH+{H^{+}} $ is equal to that of $ AgCl $
D) $ S $ of $ AgBr $ is $ 10^{6} $ times greater than that of $ AgCl $
Show Answer
Answer:
Correct Answer: A
Solution:
AgCl $ K _{sp}=1.2\times {10^{-10}} $
$ S=\sqrt{1.2\times {10^{-10}}} $ ; $ S=1.09\times {10^{-5}} $ AgBr $ K _{sp}=3.5\times {10^{-13}} $
$ S=\sqrt{3.5\times {10^{-13}}} $
$ =5.91\times {10^{-6}} $ So that S of AgBr is less than that of AgCl.
