The Common Ion Effect and Solubility
Le Châtelier's Principle indicates that it should be possible to vary the solubility of a solid by varying the concentration of the ions at equilibrium.
 
i.e. To decrease the [Ag+] in a solution of AgBr, we can add excess Br- ions in the form of a soluble salt such as NaBr. This excess puts a stress on the equilibrium system which adjusts itself by shifting left. As it consumes Br- ions and forms AgBr(s).
The Ksp is a constant. If one ion gets larger then the other must get smaller.
 
Ksp = [Ag+][Br-]
 
Ksp = [Ag+][Br-]
 
Each Br- ion that reacts ties up a Ag+ ion which removes it from the solution.

Ksp = [Ag+][Br-] = 4.8 x 10-13
 

The value of Ksp for KBr is so small that even a little Br- from the NaBr will cause a ppt. The overall effect is that the solubility of AgBr is decreased. AgBr is less soluble in a solution of NaBr then it would be in pure water.
 
Compare the molar solubility of AgBr in pure water and in a 0.10 M NaBr solution.

AgBr(s) <====> Ag+(aq) + Br-(aq)

In pure water

Let 'x' be the [Ag+] and the [Br-] in a saturated solution of AgBr.

Therefore   Ksp = [Ag+][Br-]
                         = (x) (x)
        4.8 x 10-13 = x2

                         = 6.9 x 10-7 mol/L
 

Therefore in a solution made with pure water the [Ag+] = [Br-] = 6.9 x 10-7 mol/L.

In a 0.10 M NaBr soln.

In a 0.10 mol/L solution, the [Br-] = 0.10 mol/Lalready
 

The small quantity of Br- furnished from the dissolving of the AgBr is insignificant since in pure water it was only 6.9 x 10-7 M. Since dissolution is repressed by the presence of the common Br- ion, it will be even less. Therefore we can forget about it.

Since [Br-] = x + 0.10 mol/L
                 = 6.9 x 10-7 mol/L + 0.10 mol/L
                 = 0.10000069 mol/L

Therefore Ksp = [Ag+][Br-]
     4.8 x 10-13 = (x)(0.10 mol/L)
                    x = 4.9 x 10-12 mol/L

Since 0.10 + 4.8 x 10-12 0.10 our assumption is valid. AgBr is about 100000 times more soluble in pure water than it would be in a 0.10 M NaBr solution.
 

Follow-Up Problem
Compare the molar solubility of PbI2 in pure water and in 0.10 NaI.