Would the solubility of #"Cu"_2"S"# increase, decrease or stay the same if #"AgNO"_3# is added to the solution?

1 Answer
Jun 3, 2016

Answer:

It will increase.

Explanation:

Copper(I) sulfide, #"Cu"_2"S"#, is practically insoluble in aqueous solution.

This means that very, very small amounts of the salt will dissociate to form copper(I) cations, #"Cu"^(+)#, and sulfide anions, #"S"^(2-)#, in solution, and thus an equilibrium will be established between the undissolved salt and the dissolved ions

#"Cu"_ 2"S"_ ((s)) rightleftharpoons 2"Cu"_ ((aq))^(+) + "S"_ ((aq))^(2-)#

Silver nitrate, #"AgNO"_3#, on the other hand, is soluble in aqueous solution.

This means that it will dissociate completely to form silver cations ,#"Ag"^(+)#, and nitrate anions, #"NO"_3^(-)#, in solution.

#"AgNO"_ (3(aq)) -> "Ag"_ ((aq))^(+) + "NO"_ (3(aq))^(-)#

Now, when a solution of silver nitrate is added to a solution of copper(I) sulfide, the silver cations will combine with the sulfide anions to form silver sulfide, #"Ag"_2"S"#, an insoluble solid that precipitates out of solution.

#2"Ag"_ ((aq))^(+) + "S"_ ((aq))^(2-) -> "Ag"_ 2"S"_((s)) darr#

This reaction will consume some of the sulfide anions that were delivered to the solution by the insoluble copper(I) sulfide.

As a result, the equilibrium reaction that governs the salt's dissociation will shift to the right in order to compensate for the fact that the sulfide anions are now combining with the silver cations to form silver sulfide #-># think Le Chatelier's Principle here.

#color(white)()#

#"Cu"_ 2"S"_ ((s)) rightleftharpoons 2"Cu"_ ((aq))^(+) + "S"_ ((aq))^(2-)#

#stackrel(color(blue)(rarr))(color(white)(aacolor(green)("shift to the right")aaa))#

Since the forward reaction aims at increasing the concentration of sulfide anions in solution, a shift to the right implies that more of the salt will dissociate.

Therefore, the solubility of copper(I) sulfide will increase upon the addition of a solution of silver nitrate.