# Does ammonium react with water in an ammonia solution?

In the following equilibrium:

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$

Is it possible for the ammonium to simultaneously establish another reaction with water?

$$\ce{NH4+ + H2O <=> NH3 + H3O+}$$

If so, we would enter a cycle where the additional concentration of ammonia resulting from the second reaction would, in turn, shift the first equilibrium to the right according to Le Châtelier's principle. I surmise that this system will eventually stabilize yielding hydroxide ions (with a negligible concentration of hydronium ions), hence the basic character of ammonia. Are these valid conclusions? Thanks in advance.

• Possible duplicate of Is there a difference between ‘ammonium hydroxide’ and ‘aqueous ammonia’? – Nilay Ghosh Feb 12 '19 at 18:18
• Everything is in equilibrium at the same time. I think you are confusing what happens to a single molecule, but the equilibrium process happens in bulk. – Zhe Feb 12 '19 at 18:33
• Oh, ok. I see. When the equilibrium is reached in the first reaction, there is no such thing as a second reaction because the system is already in equilibrium? Am I correct? – Zach Galois Feb 12 '19 at 18:35
• To be complete, you should also include the reaction $\ce{H3O+ + HO- <=> 2H2O}$. Perhaps then it will be more clear how everything equilibrates. – Andrew Feb 13 '19 at 1:04
• @Andrew If you add up the two reactions given by the OP, you get the auto dissociation of water. It is a bit hidden, but already there. Alternatively, you could keep the auto dissociation reaction and drop the 2nd reaction given by the OP to remove the redundancy. – Karsten Theis Feb 13 '19 at 15:12

Is it possible for the ammonium to simultaneously establish another reaction with water?

It might help to add these two reactions to see what the net effect is:

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$ $$\ce{NH4+ + H2O <=> NH3 + H3O+}$$ $$+\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$$ $$\ce{2H2O <=> H3O+ + OH-}$$

So yes, this is the autodissociation of water, and this reaction goes on at the same time. The three equilibrium constants are linked (K1 x K2 = Kw), so there is never a situation where two of these reactions are at equilibrium and the third one isn't, causing an endless change in concentrations. In fact, for reactions in solution, there is always a state where all of the reactions are at equilibrium, and if all reactions proceed, they will all reach equilibrium eventually.

You can take any reaction that includes hydronium (or hydroxide) ions and formulate another reaction that includes hydroxide (or hydronium) ions instead by combining it with the autodissociation reaction.

If so, we would enter a cycle where the additional concentration of ammonia resulting from the second reaction would, in turn, shift the first equilibrium to the right according to Le Châtelier's principle.

There is no cycle, unless you call reactants making products and products making reactants at the same time a cycle. The latter is a feature of all reactions that are at equilibrium. If you label a hydrogen, it will forever cycle between being part of ammonia, ammonium, hydroxide, water, and hydronium. There are some more complicated systems of reactions where the path toward equilibrium involves oscillations of concentrations, but in most cases, concentrations approach the equilibrium concentration in a monotonic way.

Since the equilibrium has been reached in the first reaction, the concentration of ammonium no longer varies and therefore there is no reaction with water. I was confusing this with salt hydrolysis, where the ammonium resulting from a salt does react with water because salts are strong electrolytes and there is no equilibrium involved.

• That's a misconception. If you have multiple equilibrium reactions in the same system, all have to reach equilibrium before changes in concentrations cease. Otherwise, the first equilibrium will be disturbed as the second equilibrium process changes concentrations to reach its equilibrium. – Karsten Theis Feb 13 '19 at 15:10