In which of the following would the molar solubility of Ag2CrO4(s) be the highest

Question

In which of the following would the molar solubility of Ag2CrO4(s) be the highest.

a. 0.10 M AgNO3

b. 0.20 M AgNO3

c. 0.10 M K2CrO4

d. 0.20 M K2CrO4

e. 0.20 M KNO3

i know the answer is E but why

Answer 1

Answer e . when the solution contains other salts  formed with the cation or anion of the Ag2CrO4 i. e. Ag+ or CrO4-- ,the solubilty of ag2CrO4-- decreases

Answer 2

Because it does not contain any of the ions present in silver chromate(VI) and so will not shift the equilibrium -

Ag2CrO4(s) <=> 2Ag+(aq) + [(CrO4)2-](aq)

to the left, thereby decreasing solubility.

Answer 3

the answer is E because it's the only solution in the list that doesn't contain any of the ions Ag or CrO that can disrupt the solubility of Ag2CrO4 (Silver Chromate). The rest of the solutions have either Ag or CrO in them that may decrease the molar solubility of Ag2CrO4.

In addition, according to Le Chatelier's Law concerning equilibrium, any stress put on a system at equilibrium will cause the system to shift to counteract that change. So when Ag2CrO4 is added to a solution and dissociates, the free Ag or CrO4 ions in the other solutions added to the product side will cause the system to shift to the left, using up the ions in the process and producing more of the reactant Ag2CrO4 which is why free ions can disturb the molar solubility of a substance.

Ag2CrO4 <~~> 2Ag(+) + CrO4(2-)