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

Jan 4, 2016

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

${\text{Cu"("NO"_3)_text(2(aq]) -> "Cu"_text((aq])^(2+) + 2"NO}}_{\textrm{3 \left(a q\right]}}^{-}$

("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, ${\text{Cu}}^{2 +}$, and the sulfide anions, ${\text{S}}^{2 -}$, 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, ${\text{NH"_4"NO}}_{3}$, a soluble compound that will exist as cations and anions in solution.

This means that you can write

${\text{Cu"("NO"_3)_text(2(aq]) + ("NH"_4)_2"S"_text((aq]) -> "CuS"_text((s]) darr + 2"NH"_4"NO}}_{\textrm{3 \left(a q\right]}}$

The complete ionic equation will look like this

${\text{Cu"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-) + 2"NH"_text(4(aq])^(+) + "S"_text((aq])^(2-) -> "CuS"_text((s]) darr + 2"NH"_text(4(aq])^(+) + 2"NO}}_{\textrm{3 \left(a q\right]}}^{-}$

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

${\text{Cu"_text((aq])^(2+) + "S"_text((aq])^(2-) -> "CuS}}_{\textrm{\left(s\right]}} \downarrow$