I assume that the reaction would be $\ce{HNO3(aq) + NH4Cl(aq) -> HCl(aq) + NH4NO3(aq)}$

But will this reaction really happen?


3 Answers 3


Be aware that all 4 compounds, dissolved in water, are dissociated to respective ions. E.g. $\ce{NH4Cl(aq)}$ is just a shortcut for $\ce{NH4+(aq) + Cl-(aq)}$.

So it is not combination of $\ce{HNO3}$, $\ce{NH4Cl}$, $\ce{HCl}$ and $\ce{NH4NO3}$, but of (hydrated) ions $\ce{H+}$, $\ce{NO3-}$, $\ce{NH4+}$ and $\ce{Cl-}$.

Ammonium nitrate cannot be formed even by crystallization of the less soluble salt:

$$\ce{NH4Cl(s,aq) + HNO3(aq) -> NH4NO3(s) + HCl(aq)},$$

because it is more soluble than ammonium chloride.

OTOH, if the mixture has minimum of water, there can be reaction like:

$$\ce{NH4Cl(s,aq) + HNO3(l,aq) <=> NH4NO3(s,aq) + HCl(aq,g)}$$

But, in such high concentrations, there would be significant oxidation of chlorides, forming nitrogen oxides, chlorine and nitrosyl chloride.

  • 1
    $\begingroup$ As a comment to Poutnik's message, it should be mentioned that the original $\ce{HNO3}$ reacts with the $\ce{HCl}$ produced in a second reaction like $\ce{HNO3 + 3 HCl -> NOCl + Cl2 + 2 H2O}$, which occurs in the solution called aqua regia. $\endgroup$
    – Maurice
    Dec 26, 2022 at 10:57
  • $\begingroup$ It was already said, by other words. $\endgroup$
    – Poutnik
    Dec 27, 2022 at 8:58

Dissolving those two compounds in water would lead to four species of ions floating around (without discussing solvation and oxidation).

To get the solid compounds back again would require removing the water, e.g., with a vacuum pump, to get the anions and cations to link (crystallize) again. Then the question arises, which of the resultant compounds are least soluble, and would precipitate first?

In this way, brines are separated into (relatively) purer substances through fractional crystallization. For example, solar evaporation ponds are used commercially to separate $\ce{NaCl}$, $\ce{K2SO4}$, and $\ce{MgCl2}$ from the mix of ions in Utah's Great Salt Lake brine.


Redox reactions aside, the reaction would tend to go the other way, precipitating ammonium chloride.

Below, we note the solubilities of a few ammonium salts in water at 20°C, translated into moles of ammonium ions per kilogtlram of water. Mass solubility from which the molar solubilities are computed are given in the linked sources.

$\ce{NH4Cl}: 6.8$ (https://en.wikipedia.org/wiki/Ammonium_chloride) (interpolated between 0 and 25°C)

$\ce{NH4HSO4}: 10.0$ (https://www.scbt.com/p/ammonium-bisulfate-7803-63-6) (it is assumed that in the acid solutions described below sulfate ions would be converted to bisulfate)

$\ce{NH4NO3}: 18.7$ (https://en.wikipedia.org/wiki/Ammonium_nitrate)

Thus given sufficient salt concentrations, the addition of $\ce{HCl}$ would precipitatethe less soluble ammonium chloride from solutions of the other salts listed:

$\ce{NH4HSO4 + HCl -> NH4Cl(s) + H2SO4}$

$\ce{NH4NO3 + HCl -> NH4Cl(s) + HNO3}$

As described by Maurice in a comment, given the high solute concentrations required plus the formation of nitric acid, the reaction with ammonium nitrate would likely be accompanied by oxidation of the ammonium and chloride ions. With sulfuric acid the oxidizing action is not as strong, so the (bi)sulfate reaction would be cleaner.

The reduced solubility of ammonium chloride may be attributed to its relatively efficiently packed lattice structure. This salt has a caesium chloride type simple cubic structure, with ammonium ions oriented so that their protic hydrogen atoms are facing alternating chloride ions. With this arrangement both ionic bonding and hydrogen bonding serve to stabilize the solid. The bisulfate and nitrate, with bulkier and nonspherical anions, do not offer as good a fit in the solid lattice and thus the solid is more easily attacked by the water solvent.


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