The common ion effect

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.

• 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 M_mX_x.
  
  M_mX_{x (s)} $\, \rightleftharpoons \,$ M^x+_(aq) + X^m-_(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:
  
  M_mX_{x (s)} $\,$→$\,$ M^x+_(aq) + $\,$ X^m-_(aq)
  The reverse reaction is the crystallization process that changes the substance from aqueous back to the solid state, as a precipitate:
  
  M^x+ _(aq) + $\,$ X^m-_(aq) $\,$→$\,$ M_m X_{x (s)}
  
  This is useful for when we want to reduce or increase the solubility of some compounds. Even though K_sp 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:
  
  Ca(OH)_{2 (s)} $\, \rightleftharpoons \,$ 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)} $\, \rightleftharpoons \,$ Ca²⁺ _(aq) + 2OH^- _(aq)
  To maintain the K_sp 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.
  
  
  • 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 CaCO_{3 (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.
  
  
• 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:

Ca(OH)_{2 (s)} $\, \rightleftharpoons \,$ Ca²⁺_(aq) + 2OH^-_(aq)
Where K_sp = 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.

H⁺ _(aq) + OH^- _(aq) $\, \rightleftharpoons \,$ H₂O _(l)
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.

Ca(OH)_{2 (s)} + 2H⁺ _(aq) + 2OH^- _(aq) $\, \rightleftharpoons \,$ Ca²⁺ _(aq) + 2OH^- _(aq) + 2H₂O _(l)
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.

• 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:

CaCl_{2 (s)} $\, \rightleftharpoons \,$ Ca²⁺ _(aq) + 2Cl^- _(aq)
Where K_sp = [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.


• (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:

$\Delta G$ = $\Delta H$ - $T \Delta S$


• $\Delta H$ is the enthalpy change and $\Delta S$ is the entropy change. A process is feasible provided that $\Delta G$ is negative overall.

There are many examples where a species dissolves ($\Delta G$ is negative) but the temperature of the solution decreases ($\Delta H$ 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:
• 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.

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.


• In short, the common ion effect and increasing solubility works by these principles:
• When a solution is saturated and the K_sp 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.