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

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

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    $\begingroup$ Search the pKa of NH3/NH2- $\endgroup$ – Waylander Nov 11 '20 at 15:10
  • $\begingroup$ @Waylander I mean at a molecular level. I know that its pKa explains this. $\endgroup$ – david david Nov 11 '20 at 15:15
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    $\begingroup$ It might, but the equilibrium lies far to the left. Hence, ammonia is not a very effective acid in water. $\endgroup$ – Zhe Nov 11 '20 at 16:06
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    $\begingroup$ 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. $\endgroup$ – Karl Nov 11 '20 at 17:12
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    $\begingroup$ @daviddavid In your second equation above, it acts as an acid. That's what your question is about. $\endgroup$ – 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.

  • $\begingroup$ @KarstenTheis Methane is CH4 ... $\endgroup$ – david david Nov 13 '20 at 12:09
  • $\begingroup$ You've only said that it is more reluctant to lose it but you haven't actually said why. $\endgroup$ – david david Nov 13 '20 at 12:11
  • $\begingroup$ @David. The only reason is the result of experiment. $\endgroup$ – Maurice Nov 13 '20 at 14:19
  • $\begingroup$ @Maurice I don't think you get it. Results have to be reasoned theoretically and checked practically (with experiments). I am asking for the theoretical explanation. $\endgroup$ – david david Nov 13 '20 at 14:23

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