What products result from the addition of aqueous solutions of #Cu(NO_3)_2# and #(NH_4)_2S#?

1 Answer
Jan 4, 2016

Answer:

Solid copper(II) sulfide and aqueous ammonium nitrate.

Explanation:

Copper(II) nitrate, #"Cu"("NO"_3)_2#, and ammonium sulfide, #("NH"_4)_2"S"#, are both soluble ionic compounds, which means that the dissociate completely in aqueous solution and exist as cations and anions

#"Cu"("NO"_3)_text(2(aq]) -> "Cu"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-)#

#("NH"_4)_2"S"_text((aq]) -> 2"NH"_text(4(aq])^(+) + "S"_text((aq])^(2-)#

Now, when these two solutions are mixed, the copper(II) cations, #"Cu"^(2+)#, and the sulfide anions, #"S"^(2-)#, will react to form copper(II) sulfide, an insoluble solid that precipitates out of solution.

The other product of the reaction will be ammonium nitrate, #"NH"_4"NO"_3#, a soluble compound that will exist as cations and anions in solution.

This means that you can write

#"Cu"("NO"_3)_text(2(aq]) + ("NH"_4)_2"S"_text((aq]) -> "CuS"_text((s]) darr + 2"NH"_4"NO"_text(3(aq])#

The complete ionic equation will look like this

#"Cu"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-) + 2"NH"_text(4(aq])^(+) + "S"_text((aq])^(2-) -> "CuS"_text((s]) darr + 2"NH"_text(4(aq])^(+) + 2"NO"_text(3(aq])^(-)#

Notice that some ions are present on both sides of the equation - these ions are called spectator ions.

Removing these ions will give you the net ionic equation

#"Cu"_text((aq])^(2+) + color(red)(cancel(color(black)(2"NO"_text(3(aq])^(-)))) + color(red)(cancel(color(black)(2"NH"_text(4(aq])^(+)))) + "S"_text((aq])^(2-) -> "CuS"_text((s]) darr + color(red)(cancel(color(black)(2"NH"_text(4(aq])^(+)))) + color(red)(cancel(color(black)(2"NO"_text(3(aq])^(-))))#

which looks like this

#"Cu"_text((aq])^(2+) + "S"_text((aq])^(2-) -> "CuS"_text((s]) darr#