The common ion effect
All in One PlaceEverything you need for better grades in university, high school and elementary. | Learn with EaseMade in Canada with help for all provincial curriculums, so you can study in confidence. | Instant and Unlimited HelpGet the best tips, walkthroughs, and practice questions. |
Make math click 🤔 and get better grades! 💯Join for Free
0/6
Intros
Lessons
0/2
Examples
Lessons
- Recall the use of the common-ion effect to change the solubility of a saturated solution.
A student has a saturated solution of AgCl, which is in equilibrium with its ions.AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
What will adding the following salts to the saturated solution do to the solubility equilibrium?
Explain your answer.- Pb(NO3)2
- Na2S
- AgNO3
- KCl
- Apply the common ion effect to suggest ways of changing the solubility of a saturated solution.
A student has a saturated solution of lead chloride, with the precipitate in equilibrium with its aqueous ions.PbCl2 ⇌ Pb2+ (aq) + 2Cl- (aq)
What compounds could be added to have the following effects on the PbCl2 solution?
Free to Join!
StudyPug is a learning help platform covering math and science from grade 4 all the way to second year university. Our video tutorials, unlimited practice problems, and step-by-step explanations provide you or your child with all the help you need to master concepts. On top of that, it's fun — with achievements, customizable avatars, and awards to keep you motivated.
Easily See Your Progress
We track the progress you've made on a topic so you know what you've done. From the course view you can easily see what topics have what and the progress you've made on them. Fill the rings to completely master that section or mouse over the icon to see more details.Make Use of Our Learning Aids
Earn Achievements as You Learn
Make the most of your time as you use StudyPug to help you achieve your goals. Earn fun little badges the more you watch, practice, and use our service.Create and Customize Your Avatar
Play with our fun little avatar builder to create and customize your own avatar on StudyPug. Choose your face, eye colour, hair colour and style, and background. Unlock more options the more you use StudyPug.
Topic Notes
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.
- How pH and thermodynamics influence the solubility of salts.
Notes:
- 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.
Like any process at equilibrium, the common ion effect is governed by Le Chatelier’s principle. This is important in predicting how the solubility will change.- For example, the salt calcium hydroxide, Ca(OH)2, when saturated has the equilibrium:
- 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.
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. Applying Le Chatelier’s principle, increasing [Ca2+] or [OH-] will cause the system to shift away from them, which is towards more Ca(OH)2 (s).
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). When this happens, we will have decreased the solubility of Ca(OH)2 as more of it is in precipitate form now. - As the above hopefully shows, the common ion effect is governed by Le Chatelier’s principle.
Le Chatelier’s principle also applies to acid-base equilibria so the solubility of many salts is pH sensitive; if one of the aqueous ions is a conjugate pair with a weak acid or base, a change in pH will affect its concentration and the solubility equilibrium of the salt. - For example, Ca(OH)2 has the solubility equilibrium:
- If you have a solubility equilibrium with an aqueous ion that is the conjugate base of a strong acid (chlorides, bromides, sulfates, nitrates), pH has no effect on solubility.
This is because adding aqueous acid or base will not lead to the strong acid/base being formed; strong acids and bases have 100% dissociation.
For example, calcium chloride, CaCl2, has the following solubility equilibrium: - (AP CHEMISTRY ONLY)
When a substance dissolves, its solubility is influenced by several thermodynamic factors. In Entropy and Gibbs free energy, we looked at the thermodynamics behind a chemical change – is it feasible or not? Recall that the Gibbs free energy equation has two thermodynamic factors involved: - ΔH is the enthalpy change and ΔS is the entropy change. A process is feasible provided that ΔG is negative overall.
- The NaCl lattice breaks up, so the Na+ Cl- ionic bonds are overcome.
- Some solvent-solvent interactions are broken up, as ions will occupy some space that solvent molecules previously did. In a water solvent, this will be some hydrogen bonds being disrupted and reorganized.
- As the ionic lattice is broken up, the resulting aqueous Na+ and Cl- ions will make many new polar interactions with the water solvent.
- 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 salt 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.
Where Ksp = 5.02 * 10-6 = [Ca2+(aq)][OH-(aq)]2
Adding aqueous acid (H+) to a saturated solution of Ca(OH)2 to lower the pH will form water, using the hydroxide ions from Ca(OH)2 that dissolved.
This removal of OH- (aq) as water forms means the solubility equilibrium is disturbed. There is less OH (aq), so according to Le Chatelier’s principle the equilibrium will shift towards the aqueous products to restore the original OH- that was lost. To do this, more Ca(OH)2 (s) must dissolve.
You can combine the two equations above which shows the process in one step.
With the equations combine you can see the process clearly: making the solution more acidic will increase the solubility of Ca(OH)2 by forming water.
Where Ksp = [Ca2+ (aq)][Cl-]2
Adding H+ (aq) to a saturated solution of CaCl2 containing Cl- ions will not cause HCl to form because as a strong acid, it stays (virtually) 100% dissociated. So the Cl- (aq) will remain aqueous chloride ions, and the solubility equilibrium is undisturbed.
There are many examples where a species dissolves (ΔG is negative) but the temperature of the solution decreases (ΔH is positive).
The explanation for this is that a large enough positive entropy change offsets a positive enthalpy change to make an endothermic process feasible. When a substance dissolves, make a note of the chemical changes occurring.
With sodium chloride dissolving in water, the following three chemical changes are occurring:
A positive enthalpy change tells us that overall, more energy was needed to break the existing bonds/interactions than was released by forming new ion-solvent interactions.
Does this surprise you? Dissolving an ionic substance breaks strong ionic bonds!
Still, the large entropy increase of a lattice breaking into aqueous ions offset this.
Every ionic species will have a different balance of ionic bonds broken, solvent-ion interactions formed and an associated entropy change. This is what gives rise to the variety in solubility we find in ionic substances.
2
videos
remaining today
remaining today
5
practice questions
remaining today
remaining today