In this lesson, we will learn:
- How to use the common ion effect to decrease the solubility of a saturated solution.
- How to use a solubility table to suggest ways to increase the solubility of a saturated solution.
- Whenever we talk about a compound with low solubility or a saturated solution, we should always write the equilibrium that has been created. For any salt MmXx.
MmXx (s) Mx+(aq) + Xm-(aq)
This can be broken up into the individual forward and reverse reactions. The forward reaction is the dissolving process that changes the substance from solid to aqueous state:
MmXx (s) → Mx+(aq) + Xm-(aq)
The reverse reaction is the crystallization process that changes the substance from aqueous back to the solid state, as a precipitate:
Mx+ (aq) + Xm-(aq) → Mm Xx (s)
This is useful for when we want to reduce or increase the solubility of some compounds. Even though Ksp cannot change, there are ways to change solubility in a given solution without changing the temperature!
- The common ion effect is a way to change the solubility of a compound by adding a soluble salt that has an ion in common with the compound you are trying to change the solubility of. For example, the salt calcium hydroxide, Ca(OH)2, when saturated has the equilibrium:
Ca(OH)2 (s) Ca2+ (aq) + 2OH- (aq)
- To decrease the solubility of Ca(OH)2, we need the equilibrium to shift to the left and favor the precipitate. To do this, we add a compound that will dissolve to produce more Ca2+ or OH-ions in solution.
Using a solubility table and our Predicting the solubility of salts lesson recall that compounds with a nitrate (NO3-) anion are highly soluble in water, If we added some calcium nitrate (Ca(NO3)2) to the solution, the following happens:
Ca(NO3)2 (s) → Ca2+ (aq) + 2NO3- (aq)
As it is highly soluble, this is not an equilibrium, it is a straightforward dissolving process. But the extra Ca2+ (aq) will now disturb the Ca(OH)2 equilibrium:
Ca(OH)2 (s) Ca2+ (aq) + 2OH- (aq)
To maintain the Ksp concentration of Ca(OH)2 at equilibrium, the equilibrium must shift to the left (favoring the crystallization reaction) and makes more Ca(OH)2 (s). In this way, we have decreased the solubility of Ca(OH)2 as more of it is in precipitate form now.
- To increase the solubility of Ca(OH)2, we need to do the opposite; the equilibrium must shift to the right and favor the dissolving reaction. To do this, we need to add a compound that will reduce the amount of Ca2+ or OH- ions in solution by precipitating one of the ions out of solution.
Using a solubility table, we can see that compounds containing sodium ions (Na+) will dissolve in water, and that calcium carbonate, CaCO3, has low solubility. If we added sodium carbonate, Na2CO3, we would begin to precipitate CaCO3 (s) while reducing the Ca2+ (aq) concentration. The Ca(OH)2 equilibrium will respond by shifting to the right to produce more Ca2+ ions. Remember that Na+ like NO3- is a spectator ion and will not form a precipitate!
The same could be done with the OH- ions dissolved; adding Pb(NO3)2 to the solution would cause Pb(OH)2 (s) to precipitate and the equilibrium will shift to the right. This would produce more OH- ions to rebalance those that were lost when Pb(OH)2 (s) started forming.
- In short, the common ion effect and increasing solubility works by these principles:
- When a solution is saturated and the Ksp equilibrium is established, changing the ion concentrations will change the equilibrium position which, here, is the compound’s solubility.
- To decrease the solubility of a saturated solution of a compound, add a soluble salts with an ion in common to it. Use a solubility table to find a salt, and remember spectator ions make soluble salts!
- To increase the solubility, adding soluble salts with an ion that will form compounds of low solubility with one of the aqueous ions at equilibrium. For example, to a saturated AgCl solution, adding soluble Pb(NO3)2 which will precipitate PbCl2 with the Cl- ions in solution.