# 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: $$ \mathrm{NaF}_{(\mathrm{s})} \rightarrow \mathrm{Na}^{+}_{(\mathrm{aq})} + \mathrm{F}^{-}_{(\mathrm{aq})} $$ 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 ($\mathrm{KClO_4}$) 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: $$ \mathrm{KCl}_{(\mathrm{s})} \rightarrow \mathrm{K}^{+}_{(\mathrm{aq})} + \mathrm{Cl}^{-}_{(\mathrm{aq})} $$ 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 $\mathrm{ClO_4}^-$ ions, leading to the precipitation of solid $\mathrm{KClO_4}$ 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:** $$ \mathrm{HCl}_{(\mathrm{aq})} \rightarrow \mathrm{H}^{+}_{(\mathrm{aq})} + \mathrm{Cl}^{-}_{(\mathrm{aq})} $$ 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, $\mathrm{NH_4Cl}$, is a water-soluble salt. What will happen if this salt is added to a solution containing ammonium hydroxide, $\mathrm{NH_4OH}$? $$ \mathrm{NH}_{4}\mathrm{OH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{NH}_{4(\mathrm{aq})}^{+} + \mathrm{OH}_{(\mathrm{aq})}^{-} $$ **A:** $\mathrm{NH_4Cl}$ is a strong electrolyte that dissociates completely into $\mathrm{NH_4}^+$ and $\mathrm{Cl}^-$. The added $\mathrm{NH_4}^+$ is the common ion. According to Le Chatelier's principle, the equilibrium will shift to the left, suppressing the ionization of the weak base $\mathrm{NH_4OH}$. The concentration of $\mathrm{OH}^-$ will decrease. **Q:** Carbonic acid, $\mathrm{H_2CO_3}$, is a weak acid. What will happen if a strong electrolyte such as $\mathrm{Na_2CO_3}$ 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:** $\mathrm{Na_2CO_3}$ is a strong electrolyte that provides a high concentration of the common ion, $\mathrm{CO_3}^{2-}$. This will shift the carbonic acid equilibrium to the left, reducing its dissociation and lowering the concentration of $\mathrm{H}^+$ ions in the solution.