# Basic behavior of ammonia in aqueous medium [closed]

Thanks to the reversible reaction

$$\ce{NH3 + H2O <=> NH4+ + OH-},\tag{R1}$$

we know that ammonia acts as a weak base. However, I was wondering why it ends up forming ammonium $$\ce{NH4+}$$ and not azanide $$\ce{NH2-}$$ according to

$$\ce{NH3 + H2O <=> NH2- + H3O+}.\tag{R2}$$

• Search the pKa of NH3/NH2- – Waylander Nov 11 '20 at 15:10
• @Waylander I mean at a molecular level. I know that its pKa explains this. – david david Nov 11 '20 at 15:15
• It might, but the equilibrium lies far to the left. Hence, ammonia is not a very effective acid in water. – Zhe Nov 11 '20 at 16:06
• The conjugate acid of a strong base (and NH2- is a strong base) is very weak. And a very weak acid like NH3 doesn't protonate water. – Karl Nov 11 '20 at 17:12
• @daviddavid In your second equation above, it acts as an acid. That's what your question is about. – Karl Nov 11 '20 at 19:07

All what Karl and Poutnik are saying is right. But the real question is : Why ? Why is $$\ce{NH3}$$ more reluctant to loose one proton than $$\ce{H2O}$$, at the atomic level ? Why is $$\ce{NH2^-}$$ such a strong base ?
Let's compare this behaviour in the series $$\ce{N-O-F}$$. Starting with $$\ce{F}$$, we can observe that $$\ce{F}$$ holds one $$\ce{H}$$, only one $$\ce{H}$$, and it does not hesitate to give its unique $$\ce{H}$$ atom (as a proton) to a great variety of bases. The next in line, the oxygen atom $$\ce{O}$$ can hold two $$\ce{H}$$ atoms. This is one more than what does $$\ce{F}$$. Well, $$\ce{O}$$ in $$\ce{H2O}$$ is also more reluctant to loose one $$\ce{H}$$, and accepts to give it (as a proton) only to strong bases. The third in line is the nitrogen atom $$\ce{N}$$ which can hold three $$\ce{H}$$ atoms. It's still more than what does Oxygen. But $$\ce{N}$$ in $$\ce{NH3}$$ is still more reluctant to loose one of its three $$\ce{H}$$ atoms as a proton $$\ce{H+}$$. It even prefers stealing one $$\ce{H+}$$ ion from the neighborhood rather than giving it. Apparently the more $$\ce{H}$$ atoms the central atom has, the more it wants to increase this number. It looks like the human society : the more money you've got, the more you want to increase your fortune.
Edit : Of course it would be interesting to see if this tendency $$\ce{F-O-N}$$ can be extrapolated to the carbon atom. If my reasoning is correct, the Carbon atom in $$\ce{CH4}$$ should be extremely reluctant to loose one $$\ce{H+}$$ ion for producing $$\ce{CH3^-}$$. Indeed I am not sure that the ion $$\ce{CH3^-}$$ does exist, and has a measured $$p\mathrm{K_a}$$ value. It should be extremely high. On the other hand, $$\ce{CH4}$$ should be eager to get one more Hydrogen. However the electronic situation is not similar to $$\ce{F, O}$$ and $$\ce{N}$$. In $$\ce{HF, H2O, NH3, F^-, OH^-, NH2^-, H3O+, NH4^+}$$, the central atom has always the electronic state of Neon. It is known that $$\ce{CH4}$$ does produce $$\ce{CH5^+}$$ ion in high pressure mass spectrometry. But the electronic structure of the carbon atom in $$\ce{CH5^+}$$ is unknown, to my knowledge. So I am not sure whether the tendency, valid for $$\ce{F-O-N}$$, can be extrapolated to the Carbon atom.