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.