9.11 Common Ion Effect

The common ion effect describes the decrease in the ionization or solubility of an electrolyte when a soluble compound containing a common ion is added to the solution. This phenomenon is a direct consequence of Le Chatelier's principle.

Principle of the Common Ion Effect

When a solution contains a weak electrolyte in equilibrium, adding a strong electrolyte that provides an ion in common with the weak electrolyte disturbs the equilibrium.

• Weak electrolyte: Dissociates partially, establishing an equilibrium.
• Strong electrolyte: Dissociates completely, significantly increasing the concentration of one of the ions (the "common ion").

According to Le Chatelier's principle, the system will counteract this disturbance. The equilibrium will shift to the left (towards the reactants) to consume the excess common ion, thereby reducing the dissociation of the weak electrolyte.

Effect on Weak Acids and Bases

The dissociation of a weak acid or weak base is suppressed by the addition of a salt containing its conjugate base or conjugate acid, respectively.

Example: Hydrofluoric Acid and Sodium Fluoride

Consider a solution of hydrofluoric acid ($\mathrm{HF}$), a weak acid with an acid dissociation constant, $K_a=7.2\times 10^{-4}$.

The dissociation equilibrium is: ${\mathrm{HF}}_{(\mathrm{aq})}\rightleftharpoons {\mathrm{H}}_{(\mathrm{aq})}^{+}+{\mathrm{F}}_{(\mathrm{aq})}^{-}$

If sodium fluoride ($\mathrm{NaF}$), a strong electrolyte, is added, it dissociates completely: ${\mathrm{NaF}}_{(\mathrm{s})}\to {\mathrm{Na}}_{(\mathrm{aq})}^{+}+{\mathrm{F}}_{(\mathrm{aq})}^{-}$

The addition of $\mathrm{NaF}$ increases the concentration of the fluoride ion (${\mathrm{F}}^{-}$), which is the common ion. To restore equilibrium, the system shifts to the left, consuming ${\mathrm{H}}^{+}$ and ${\mathrm{F}}^{-}$ to form more undissociated $\mathrm{HF}$. As a result, the dissociation of $\mathrm{HF}$ is suppressed.

Effect on Solubility

The solubility of a sparingly soluble salt is decreased by the addition of a soluble salt that has an ion in common with it. This concept is directly related to the solubility product principle.

Example: Potassium Perchlorate and Potassium Chloride

Potassium perchlorate ($\mathrm{KCl}{\mathrm{O}}_4$) is a moderately soluble salt. In a saturated solution, the following equilibrium exists: ${\mathrm{KClO}}_{4(\mathrm{s})}\rightleftharpoons {\mathrm{K}}_{(\mathrm{aq})}^{+}+{\mathrm{ClO}}_{4(\mathrm{aq})}^{-}$

If a highly soluble salt like potassium chloride ($\mathrm{KCl}$) is added, it dissociates completely: ${\mathrm{KCl}}_{(\mathrm{s})}\to {\mathrm{K}}_{(\mathrm{aq})}^{+}+{\mathrm{Cl}}_{(\mathrm{aq})}^{-}$

The concentration of the common ion, ${\mathrm{K}}^{+}$, increases. According to Le Chatelier's principle, the equilibrium shifts to the left. This causes ${\mathrm{K}}^{+}$ ions to combine with ${\mathrm{Cl}{\mathrm{O}}_4}^{-}$ ions, leading to the precipitation of solid $\mathrm{KCl}{\mathrm{O}}_4$ and thus reducing its solubility.

Purifying Sodium Chloride

When concentrated hydrochloric acid ($\mathrm{HCl}$) gas is passed through a saturated solution of sodium chloride ($\mathrm{NaCl}$), the concentration of the chloride ion (${\mathrm{Cl}}^{-}$) increases significantly.

• Initial Equilibrium: ${\mathrm{NaCl}}_{(\mathrm{s})}\rightleftharpoons {\mathrm{Na}}_{(\mathrm{aq})}^{+}+{\mathrm{Cl}}_{(\mathrm{aq})}^{-}$
• Added Strong Electrolyte: ${\mathrm{HCl}}_{(\mathrm{aq})}\to {\mathrm{H}}_{(\mathrm{aq})}^{+}+{\mathrm{Cl}}_{(\mathrm{aq})}^{-}$

The high concentration of the common ion, ${\mathrm{Cl}}^{-}$, from $\mathrm{HCl}$ forces the equilibrium of the $\mathrm{NaCl}$ dissociation to shift to the left. This causes sodium ions (${\mathrm{Na}}^{+}$) to combine with chloride ions (${\mathrm{Cl}}^{-}$) to form solid, pure $\mathrm{NaCl}$, which precipitates out of the solution.

Possible Questions and Answers

Q: Ammonium chloride, $\mathrm{N}{\mathrm{H}}_4\mathrm{Cl}$, is a water-soluble salt. What will happen if this salt is added to a solution containing ammonium hydroxide, $\mathrm{N}{\mathrm{H}}_4\mathrm{OH}$? ${\mathrm{NH}}_4{\mathrm{OH}}_{(\mathrm{aq})}\rightleftharpoons {\mathrm{NH}}_{4(\mathrm{aq})}^{+}+{\mathrm{OH}}_{(\mathrm{aq})}^{-}$

A: $\mathrm{N}{\mathrm{H}}_4\mathrm{Cl}$ is a strong electrolyte that dissociates completely into ${\mathrm{N}{\mathrm{H}}_4}^{+}$ and ${\mathrm{Cl}}^{-}$. The added ${\mathrm{N}{\mathrm{H}}_4}^{+}$ is the common ion. According to Le Chatelier's principle, the equilibrium will shift to the left, suppressing the ionization of the weak base $\mathrm{N}{\mathrm{H}}_4\mathrm{OH}$. The concentration of ${\mathrm{OH}}^{-}$ will decrease.

Q: Carbonic acid, ${\mathrm{H}}_2\mathrm{C}{\mathrm{O}}_3$, is a weak acid. What will happen if a strong electrolyte such as $\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3$ is added to a solution containing carbonic acid? ${\mathrm{H}}_2{\mathrm{CO}}_{3(\mathrm{aq})}\rightleftharpoons 2{\mathrm{H}}_{(\mathrm{aq})}^{+}+{\mathrm{CO}}_{3(\mathrm{aq})}^{2-}$

A: $\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3$ is a strong electrolyte that provides a high concentration of the common ion, ${\mathrm{C}{\mathrm{O}}_3}^{2-}$. This will shift the carbonic acid equilibrium to the left, reducing its dissociation and lowering the concentration of ${\mathrm{H}}^{+}$ ions in the solution.