3.4 Common Ion Effect

The common ion effect is a crucial concept in chemistry, particularly when dealing with the equilibrium of weak electrolytes and sparingly soluble salts. It describes how the presence of a common ion in a solution can suppress the ionization or solubility of another electrolyte.

1. Definition of the Common Ion Effect

The common ion effect describes the behavior of a solution in which the same ion is produced by two different compounds. Specifically, it is defined as:

This effect is a direct consequence of Le Chatelier's principle.

2. Mechanism: Weak Electrolyte and its Salt

Consider a weak electrolyte, such as hydrofluoric acid ($\mathrm{HF}$), which partially dissociates in water, and its salt, sodium fluoride ($\mathrm{NaF}$), which is a strong electrolyte.

Dissociation of Hydrofluoric Acid (Weak Electrolyte): ${\mathrm{HF}}_{(\mathrm{aq})}\rightleftharpoons {\mathrm{H}}_{(\mathrm{aq})}^{+}+{\mathrm{F}}_{(\mathrm{aq})}^{-}$ (Here, $K_a=7.2\times 10^{-4}$, indicating weak dissociation.)

Dissociation of Sodium Fluoride (Strong Electrolyte): ${\mathrm{NaF}}_{(\mathrm{s})}\to {\mathrm{Na}}_{(\mathrm{aq})}^{+}+{\mathrm{F}}_{(\mathrm{aq})}^{-}$ (NaF dissociates completely.)

In this scenario, ${\mathrm{F}}^{-}$ is the common ion produced by both compounds. When NaF is added to a solution of HF, the concentration of ${\mathrm{F}}^{-}$ ions significantly increases. According to Le Chatelier's principle, the equilibrium of the weak acid (HF) will shift to the left to consume the excess ${\mathrm{F}}^{-}$ ions. This shift results in:

• A decrease in the dissociation of HF.
• An increase in the concentration of undissociated HF.
• A decrease in the concentration of ${\mathrm{H}}^{+}$ ions.

Thus, the presence of the common ion suppresses the ionization of the weak electrolyte. This principle is directly related to buffer solutions.

3. Mechanism: Sparingly Soluble Salt and Highly Soluble Salt

Similarly, the common ion effect can decrease the solubility of a sparingly soluble salt. When a highly soluble salt is added to a saturated solution of a less soluble salt containing a common ion, the degree of dissociation of the less soluble salt decreases, causing a decrease in its solubility and often leading to precipitation.

4. Applications of the Common Ion Effect

The common ion effect has significant applications in various chemical processes:

• Separation and Identification of Cations: In qualitative chemical analysis, the separation and identification of cations into analytical groups are often based on the solubility product principle and the common ion effect. By carefully controlling the concentration of a common ion, specific ions can be selectively precipitated.
• Controlled Precipitation: Any procedure that involves controlled precipitation (e.g., in industrial processes or laboratory synthesis) often utilizes these principles to maximize the yield of a desired product or remove impurities.

Worked Examples

Example i: Potassium Perchlorate Precipitation

1. Given: Potassium perchlorate ($\mathrm{KCl}{\mathrm{O}}_4$) is moderately soluble in water. Potassium chloride ($\mathrm{KCl}$) is highly soluble.
2. Scenario: Highly soluble $\mathrm{KCl}$ is added to a saturated solution of $\mathrm{KCl}{\mathrm{O}}_4$.
3. Chemical Reactions:
   • Dissociation of potassium perchlorate (sparingly soluble): ${\mathrm{KClO}}_{4(\mathrm{s})}\rightleftharpoons {\mathrm{K}}_{(\mathrm{aq})}^{+}+{\mathrm{ClO}}_{4(\mathrm{aq})}^{-}$
   • Dissociation of potassium chloride (highly soluble): ${\mathrm{KCl}}_{(\mathrm{s})}\to {\mathrm{K}}_{(\mathrm{aq})}^{+}+{\mathrm{Cl}}_{(\mathrm{aq})}^{-}$
4. Explanation: The addition of $\mathrm{KCl}$ significantly increases the concentration of ${\mathrm{K}}^{+}$ ions, which is a common ion with $\mathrm{KCl}{\mathrm{O}}_4$. According to Le Chatelier's principle, the equilibrium of $\mathrm{KCl}{\mathrm{O}}_4$ shifts to the left, favoring the formation of solid $\mathrm{KCl}{\mathrm{O}}_4$. This suppresses the ionization (and thus the solubility) of $\mathrm{KCl}{\mathrm{O}}_4$, causing it to precipitate out of the solution.

Example ii: Sodium Chloride Precipitation from Brine

1. Given: A saturated solution of sodium chloride (brine). Hydrogen chloride gas ($\mathrm{HCl}$) is passed through it.
2. Scenario: $\mathrm{HCl}$ gas is passed through a saturated solution of $\mathrm{NaCl}$.
3. Chemical Reactions:
   • Dissociation of sodium chloride (in saturated solution): ${\mathrm{NaCl}}_{(\mathrm{s})}\rightleftharpoons {\mathrm{Na}}_{(\mathrm{aq})}^{+}+{\mathrm{Cl}}_{(\mathrm{aq})}^{-}$
   • Dissociation of hydrogen chloride (strong acid): ${\mathrm{HCl}}_{(\mathrm{aq})}\to {\mathrm{H}}_{(\mathrm{aq})}^{+}+{\mathrm{Cl}}_{(\mathrm{aq})}^{-}$
4. Explanation: When $\mathrm{HCl}$ gas is passed into the solution, it dissolves and dissociates completely, increasing the concentration of $\mathrm{C}{\mathrm{l}}^{-}$ ions. $\mathrm{C}{\mathrm{l}}^{-}$ is a common ion with $\mathrm{NaCl}$. According to Le Chatelier's principle, the increased $\mathrm{C}{\mathrm{l}}^{-}$ concentration shifts the equilibrium of $\mathrm{NaCl}$ to the left, causing pure $\mathrm{NaCl}$ to precipitate out. This method is used for purifying common salt.

Possible Questions/Answers

Q1: 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 ($\mathrm{N}{\mathrm{H}}_4\mathrm{OH}$)? ${\mathrm{N}{\mathrm{H}}_4\mathrm{OH}}_{(\mathrm{aq})}\rightleftharpoons {\mathrm{NH}}_{4(\mathrm{aq})}^{+}+{\mathrm{OH}}_{(\mathrm{aq})}^{-}$

A1: Ammonium hydroxide ($\mathrm{N}{\mathrm{H}}_4\mathrm{OH}$) is a weak base that partially dissociates in water. Ammonium chloride ($\mathrm{N}{\mathrm{H}}_4\mathrm{Cl}$) is a strong electrolyte that completely dissociates, producing ${\mathrm{N}{\mathrm{H}}_4}^{+}$ ions. These ${\mathrm{N}{\mathrm{H}}_4}^{+}$ ions are common ions. According to Le Chatelier's principle, the addition of ${\mathrm{N}{\mathrm{H}}_4}^{+}$ ions will shift the equilibrium of $\mathrm{N}{\mathrm{H}}_4\mathrm{OH}$ to the left, suppressing its ionization and decreasing the concentration of $\mathrm{O}{\mathrm{H}}^{-}$ ions, making the solution less basic.

Q2: Carbonic acid is a weak acid. It ionizes in water as follows: ${{\mathrm{H}}_2\mathrm{CO}}_{3(\mathrm{aq})}\rightleftharpoons 2{\mathrm{H}}_{(\mathrm{aq})}^{+}+{\mathrm{CO}}_{3(\mathrm{aq})}^{2-}$ 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?

A2: Carbonic acid (${\mathrm{H}}_2\mathrm{C}{\mathrm{O}}_3$) is a weak acid that partially ionizes. Sodium carbonate ($\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3$) is a strong electrolyte that completely dissociates, producing ${\mathrm{C}{\mathrm{O}}_3}^{2-}$ ions. These ${\mathrm{C}{\mathrm{O}}_3}^{2-}$ ions are common ions. According to Le Chatelier's principle, the addition of ${\mathrm{C}{\mathrm{O}}_3}^{2-}$ ions will shift the equilibrium of ${\mathrm{H}}_2\mathrm{C}{\mathrm{O}}_3$ to the left, suppressing its ionization and decreasing the concentration of ${\mathrm{H}}^{+}$ ions, making the solution less acidic.