Question

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)`

Answer 1

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?