In which of the given solutions with EQUAL concentrations will silver bromide be LEAST soluble?

The solubility of silver halides in water is another equilibrium reaction, one that we treat in the normal way:

`AgBr(s) rightleftharpoons Ag^+ + Br^(-); K_(sp) = ?`

Note that the reactant, `AgBr(s)`, does not appear in the equilibrium expression because as a solid it has no concentration. The equilibrium constant, `K_(sp),` is simply a number and is considered dimensionless. We don't know what it is, but it is very small. Given that it is an equilibrium constant, we can write:

`K_(sp) = ? = [Ag^+][Br^-]`

This relationship governs the solubility of silver bromide in water. If `[Ag^+][Br^-]` > `K_(sp)`, then silver bromide, `AgBr`, will precipitate until `[Ag^+][Br^-]` = `K_(sp)`.

Please do not be intimidated by the length of this answer. The crucial bit of information governing the reaction is the rxn,

`AgBr(s) rightleftharpoons Ag^+ + Br^(-); K_(sp) = ?`

If the ion product `[Ag^+][Br^-]` is above a certain number, precipitation of the silver salt will occur. If we introduce ammonia, `NH_3` to the reaction mixture, which can form the soluble ion `Ag(NH_3)_2^+`, do you think that some of the `AgBr` precipitate will go up?

I take it this is a 1st year chemistry question?