Question

Question #cc857

Answer 1

Here's what's going on here.

Explanation:

Ammonium carbonate, `("NH"_4)_2"CO"_3`, is a soluble ionic compound that dissociates completely to form ammonium cations, `"NH"_4^(+)`, and carbonate anions, `"CO"_3^(2-)`, when dissolved in water.

`("NH"_4)_2"CO"_text(3(aq]) -> 2"NH"_text(4(aq])^(+) + "CO"_text(3(aq])^(2-)`

So far, so good.

Now, both of these ions will react with water. Two distinct equilibrium reactions will be established

`"NH"_text(4(aq])^(+) + "H"_2"O"_text((l]) rightleftharpoons "NH"_text(3(aq]) + "H"_3"O"_text((aq])^(+)`

The ammonium ion will hydrolize to form ammonia, `"NH"_3`, and hydronium ions, `"H"_3"O"^(+)`. This tells you that the ammonium ions is actually acidic in aqueous solution.

`"CO"_text(3(aq])^(2-) + "H"_2"O"_text((l]) rightleftharpoons "HCO"_text(3(aq])^(-) + "OH"_text((aq])^(-)`

The cabonate anion will hydrolize to form bicarbonate anions, `"HCO"_3^(-)`, and hydroxide anions, `"OH"^(-)`. This tells you that the carbonate anion is actually basic in aqueous solution.

So, what is the overall reaction for when ammonium carbonate dissolves in water?

Simply add these three equations to see what you get - remember to cancel out the species that are found on both sides of the overall chemical equation

`("NH"_4)_2"CO"_text(3(aq]) + color(red)(cancel(color(black)("NH"_text(4(aq])^(+)))) + "H"_2"O"_text((l]) + "H"_2"O"_text((l]) + color(red)(cancel(color(black)("CO"_text(3(aq])^(2-)))) -> color(red)(cancel(color(black)(2)))"NH"_text(4(aq])^(+) + color(red)(cancel(color(black)("CO"_text(3(aq])^(2-)))) + "NH"_text(3(aq]) + "H"_3"O"_text((aq])^(+) + "HCO"_text(3(aq])^(-) + "OH"_text((aq])^(-)`

Now, this reaction still needs some work. Notice that you have hydronium and hydroxide ions in the same solution. As you know, these ions will neutralize each other to produce water

`"H"_3"O"_text((aq])^(+) + "OH"_text((aq])^(-) -> 2"H"_2"O"_text((l])`

Since you have two water molecules on the left-hand side as well, these will cancel each other out. The overall reaction will thus be

`("NH"_4)_2"CO"_text(3(aq]) + color(red)(cancel(color(black)(2"H"_2"O"_text((l]) ))) -> "NH"_text(4(aq])^(+) + "NH"_text(3(aq]) + "HCO"_text(3(aq])^(-) + color(red)(cancel(color(black)(2"H"_2"O"_text((l]))))`

`("NH"_4)_2"CO"_text(3(aq]) -> "NH"_text(3(aq]) + "NH"_text(4(aq])^(+) + "HCO"_text(3(aq])^(-)`

So, you can say that the ionic equation, which includes all ions, will be

`2"NH"_text(4(aq])^(+) + "CO"_text(3(aq])^(2-) -> "NH"_text(3(aq]) + "NH"_text(4(aq])^(+) + "HCO"_text(3(aq])^(-)`

The net ionic equation, for which spectator ions are excluded, will be

`"NH"_text(4(aq])^(+) + "CO"_text(3(Aq])^(2-) -> "NH"_text(3(aq]) + "HCO"_text(3(aq])^(-)`