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

## $a .$ $N {H}_{3} \left(a q\right)$ $b .$ $N a N {O}_{3} \left(a q\right) \left(a q\right)$ $c .$ $N a B r \left(a q\right)$ $d .$ $C a {\left(N {O}_{3}\right)}_{2} \left(a q\right)$

Sep 14, 2015

The answer is indeed (c), $N a B r$, 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, $A g B r \left(s\right)$, does not appear in the equilibrium expression because as a solid it has no concentration. The equilibrium constant, ${K}_{s p} ,$ 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 $\left[A {g}^{+}\right] \left[B {r}^{-}\right]$ > ${K}_{s p}$, then silver bromide, $A g B r$, will precipitate until $\left[A {g}^{+}\right] \left[B {r}^{-}\right]$ = ${K}_{s p}$.

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 $\left[A {g}^{+}\right] \left[B {r}^{-}\right]$ is above a certain number, precipitation of the silver salt will occur. If we introduce ammonia, $N {H}_{3}$ to the reaction mixture, which can form the soluble ion $A g {\left(N {H}_{3}\right)}_{2}^{+}$, do you think that some of the $A g B r$ precipitate will go up?

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