Question

`2NH3 rightleftharpoons (NH_4)^+ + (NH_2)^-`. `(NH_2)^-` acts as?

Answer 1

I think you are describing the equilibrium that is conceived to occur in liquid ammonia....

Explanation:

All acid base behaviour is modified by the identity of the solvent...in water we conceive of the autoprotolysis reaction, that has been carefully, and quantitatively studied, and is so well-developed that it now appears as a component of chemistry education:

`2H_2O(l) rightleftharpoonsH_3O^+ + HO^-`

And at `298*K`, `K_w=10^-14=[H_3O^+][HO^-]`

But there are other solvent systems, for instance liquid `SO_2` or acetic acid, or liquid ammonia, which is widely used, and accessible in a lab....for which an analogous ammonolysis reaction operates...

`underbrace(2NH_3(l))_"the solvent" rightleftharpoonsunderbrace(NH_2^(-) + NH_4^+)_"the base and the acid"`

`K_"ammonolysis"=[NH_4^+][NH_2^(-)]=10^(-33)`

This equilibrium does NOT occur in aqueous solution, where ammonia behaves as a weak Bronsted base.... And so here ammonia acts as another solvent....the which allows more basic reagents to be employed as compared to water.