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If I were to combine solutions of $\ce{(NH4)2CO3}$ and $\ce{HNO3}$ in water, the only ionic compounds that could form are $\ce{H2CO3}$ and $\ce{NH4NO3}$ right? But since both are soluble no precipitate would form correct? So does that mean that no reaction occurred? Does it also mean that an ion like $\ce{NO3-}$ will never react in an aqueous solution because it is always soluble in water?

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You picked a difficult question because this question involves acid-base reactions. It is no longer a simple "double displacement" reaction where the cations and anions "swap partners".

If I were to combine solutions of $\ce{(NH4)2CO3}$ and $\ce{HNO3}$ in water, the only ionic compounds that could form are $\ce{H2CO3}$ and $\ce{NH4NO3}$ right?

No, not really. You have to realise that the protonation of $\ce{CO3^2-}$ occurs in two stages: firstly, to $\ce{HCO3-}$, and then only to $\ce{H2CO3}$. Also, the $\ce{NH4+}$ ion is acidic, and can react with water to form $\ce{NH3}$.

On top of that, $\ce{H2CO3}$ will spontaneously decompose into $\ce{H2O}$ and $\ce{CO2}$, the latter of which is released as a gas.

Since these species are all soluble in water, and soluble ionic compounds exist as ions in water, a more appropriate way to think about this question is "what species are present", not "what ionic compounds are present". Species is just a general term that includes ions as well as molecules (since, for example, $\ce{CO2}$ and $\ce{NH3}$ are not "ions").

But since both are soluble no precipitate would form correct?

That is correct.

So does that mean that no reaction occurred?

No, there is a reaction happening because of the $\ce{CO2}$ gas formed. You will observe a lot of effervescence if you were to mix these two together.

However, if this question did not involve $\ce{CO2}$... Let's say you are reacting $\ce{NaCl}$ with $\ce{KBr}$. Since all the possible products ($\ce{NaBr}$ and $\ce{KCl}$) are soluble, you could say that in this case, no reaction would occur.

Does it also mean that an ion like $\ce{NO3-}$ will never react in an aqueous solution because it is always soluble in water?

No. You could say it will never be precipitated out. However, $\ce{NO3-}$ can also react in other ways apart from precipitation. For example, it can be used as an oxidising agent. One famous example is the dissolution of a copper coin in concentrated nitric acid.

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  • $\begingroup$ Great answer, I'd just add one fine point. $\ce{NH4NO3}$ does have a finite solubility in water. The reason that no ppt is formed is because $\ce{NH4NO3}$ is much more soluble than $\ce{NH4CO3}$. $\endgroup$ – MaxW Nov 2 '15 at 18:07
  • $\begingroup$ Would the formation of $\ce{CO2}$ form bubbles? Or would it happen very slowly because the $\ce{H2CO3}$ has to decay first, and the bubbles wouldn't be visible? $\endgroup$ – Lubed Up Slug Nov 2 '15 at 21:37
  • $\begingroup$ It is very fast. Watch this: youtube.com/watch?v=sHhT9tmePbk $\endgroup$ – orthocresol Nov 2 '15 at 21:45

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