E - Common Ion Effect (Lesson)
Common Ion Effect
All of the solubility equilibria investigated in the previous lesson were examples of solutions made in pure water. In other words, there were no additional ions present before the solid dissolved. However, what if there were other ions present ? What effect would this have on the solubility of the sparingly soluble salt? The equilibrium shown below involving Al(OH)3 can be used to illustrate this point.
Al(OH)3 (s) ⇌ Al+3 (aq) + 3 OH- (aq)
Ksp = 1.3 x 10-33
In pure water, the solubility of Al(OH)3 (s) could be calculated as follows:
Al(OH)3 (s) | Al+3 (aq) | 3 OH- (aq) | |
---|---|---|---|
Initial | -- | 0 | 0 |
Change | -- | +x | +3x |
Equilibrium | -- | x | 3x |
Plugging the values from the ICE table allows the calculation of the molar solubility of Al(OH)3 (s):
Ksp = [Al+3][OH-]3 = (x)(3x)3 = 27x4 = 1.3 x 10-33
x = 2.6 x 10-9; therefore the molar solubility of Al(OH)3 is 2.6 x 10-9 M
But, what if Al(OH)3 is not dissolved in pure water? For example, what if Al(OH)3 is added to a solution that already contains hydroxide (OH-)? Using Le Chatelier's principle, it is possible to make predictions regarding this scenario.
Take for example a scenario where Al(OH)3 is added to a solution of 0.15 M solution of NaOH (aq) instead of pure water. Both Al(OH)3 and NaOH contain the common ion of hydroxide. This results in a lowering of the solubility because the presence of the common ion causes an equilibrium shift in the direction of the solid Al(OH)3 thereby lowering the solubility. This can be demonstrated in the calculations shown below:
Al(OH)3 (s) | Al+3 (aq) | 3 OH- (aq) | |
---|---|---|---|
Initial | -- | 0 | 0.15 |
Change | -- | +x | +3x |
Equilibrium | -- | x | 0.15 + 3x |
Ksp = [Al+3][OH-]3 = (x)(0.15 +3x)3 = 1.3 x 10-33
Expanding the equation above out would result in a somewhat complicated cubic equation which is not a trivial equation to solve. In order to avoid this, an assumption can be made. The assumption is that the amount of hydroxide produced by the dissolution of Al(OH)3 is minuscule compared to the 0.15 M hydroxide originally present. The equation above could then be simplified as follows:
Ksp = [Al+3][OH-]3 = (x)(0.15)3 = 3.3 x 10-8x = 1.3 x 10-33
x = 3.9 x 10-26
Notice that the solubility of Al(OH)3 has been significantly decreased by the presence of the common ion of hydroxide. The solubility has been reduced from 2.6 x 10-9 M to 3.9 x 10-26 which is over 17 orders of magnitude lower!
Predicting Precipitation Using Ksp Values
Because solubility equilibria are simply specific examples of equilibria situations, it is possible to use the concentrations of ions present in the solution to determine if the equilibrium will shift to the right (dissolve) or shift to the left (precipitate). The video shown below offers an example of how these calculations can be performed. Be sure your volume is turned up!
You Try It!
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