Water ($\ce{H2O}$) can be added to ammonia ($\ce{NH3}$) in order to form ammonium hydroxide ($\ce{NH4OH}$): $$\ce{NH3 + H2O -> NH4OH}$$

But we know that ammonia is gaseous at STP so writing $\ce{NH3 + H2O}$ equals writing $\ce{NH3(aq)}$ right?

If yes, then the equation will look like: $$\ce{NH3(aq) -> NH4OH}$$

So my question is: is ammonium hydroxide $\ce{NH4OH}$ simply aqueous ammonia $\ce{NH3(aq)}$? Or am I missing something?
Can I resume the first equation like this? $$\ce{NH3(g) + H2O(l) -> NH3(aq)}$$


Ammonia is a weak base, ammonium is a weak acid. Water is a very weak acid and hydroxide is a very strong base. Taken together, this tells us that equilibrium $(1)$ is pushed towards the reactant side.

$$\ce{NH3 (aq) + H2O (l) <<=> NH4+ (aq) + OH- (aq)}\tag{1}$$

Thus, it is practically impossible to form stoichiometric ammonium hydroxide; the $\mathrm{p}K_\mathrm{a}$ values tell us that ammonium ions and hydroxide ions put together will generate water and ammonia.

However, it is also true that equation $(1)$ is an equilibrium — everybody has seen a solution of ammonia in water display basic properties so hydroxide ions (and thus ammonium ions) are unquestionably there. The predominant species is dissolved ammonia, though.

$$\ce{NH3 (g) + H2O (l) <=>> NH3 (aq) + H2O (l)}\tag{2}$$

It is correct that the most important process when dissolving ammonia in water is $(2)$ — $(1)$ plays a more minor role. Thus, it is also more correct to speak of solutions of ammonia — but at the same time ammonium hydroxide is commonly understood even if not formally correct.

By the way beware: occasionally students are sent off to fetch ‘solid ammonium hydroxide’ as a joke — similar to students getting sent off to fetch ‘vacuum gas’, ‘three phenol rings’ etc.

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