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

#a.# #NH_3(aq)#
#b.# #NaNO_3(aq)(aq)#
#c.# #NaBr(aq)#
#d.# #Ca(NO_3)_2(aq)#

1 Answer
Sep 14, 2015

Answer:

The answer is indeed (c), #NaBr#, due to the common ion effect. The common ion is bromide, which restricts solubility of the silver ion.

Explanation:

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?