| 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-]
= 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 Therefore Ksp = [Ag+][Br-]
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. |
