Common-ion effect Knowledge-Based Study

Study notes

1. In aqueous solution chemistry, the common-ion effect refers to the suppression of the ionization or dissolution of a compound when an additional soluble substance containing a shared ion is introduced into the mixture.

   When applied to sparingly soluble salts, this phenomenon is directly governed by Le Chatelier's principle. Consider the dynamic dissolution equilibrium of a generic sparingly soluble ionic solid, $\text{M}_p\text{X}_q(s)$:

   $$\text{M}_p\text{X}_q(s) \rightleftharpoons p\,\text{M}^{n+}(aq) + q\,\text{X}^{m-}(aq)$$

   The corresponding solubility product constant expression is:

   $$K_{\text{sp}} = [\text{M}^{n+}]^p [\text{X}^{m-}]^q$$

   Because $K_{\text{sp}}$ depends solely on temperature, the product of the ionic concentrations (raised to their stoichiometric powers) in a saturated solution remains constant. If a strong electrolyte providing either $\text{M}^{n+}$ or $\text{X}^{m-}$ is added, the system shifts toward the reactants (the solid phase) to reduce the concentration of the added ion, resulting in decreased molar solubility.

   Understanding the common-ion effect qualitatively requires tracking which ion is introduced and identifying the direction in which the solubility equilibrium shifts.

   Example: Silver Chloride

   Silver chloride ($\text{AgCl}$) is a sparingly soluble salt that dissolves according to the equilibrium:

   $$\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$$

   • In pure water: Dissolution yields equimolar amounts of silver ions and chloride ions, such that $[\text{Ag}^+] = [\text{Cl}^-] = s$, where $s$ is the molar solubility.
   • In a solution containing $\text{NaCl}$: Sodium chloride completely dissociates into $\text{Na}^+(aq)$ and $\text{Cl}^-(aq)$. The extra $\text{Cl}^-$ acts as a common ion.
   • Equilibrium response: The presence of additional $\text{Cl}^-$ increases the reaction quotient $Q$ above $K_{\text{sp}}$. To restore equilibrium, $\text{Ag}^+$ combines with excess $\text{Cl}^-$ to precipitate as solid $\text{AgCl}$, driving the equilibrium to the left.

   Consequently, the concentration of free $\text{Ag}^+$ in solution decreases, and the molar solubility of $\text{AgCl}$ is noticeably lower in a solution containing a chloride source than in pure water.

   To determine the molar solubility quantitatively in the presence of a common ion, an equilibrium table (ICE approach) is constructed, taking the initial concentration of the common ion into account.

   Worked Example

   Calculate the molar solubility of calcium fluoride, $\text{CaF}_2$ ($K_{\text{sp}} = 3.9 \times 10^{-11}$), in a 0.10 M solution of sodium fluoride ($\text{NaF}$).

   First, write the dissolution equilibrium:

   $$\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\,\text{F}^-(aq)$$

   Sodium fluoride dissociates completely:

   $$\text{NaF}(aq) \rightarrow \text{Na}^+(aq) + \text{F}^-(aq)$$

   Therefore, the initial fluoride concentration from the strong electrolyte is $[\text{F}^-]_0 = 0.10\text{ M}$.

   • Initial: $[\text{Ca}^{2+}] = 0$, $[\text{F}^-] = 0.10\text{ M}$
   • Change: $[\text{Ca}^{2+}] = +s'$, $[\text{F}^-] = +2s'$
   • Equilibrium: $[\text{Ca}^{2+}] = s'$, $[\text{F}^-] = 0.10 + 2s'$

   Substitute these values into the solubility product expression:

   $$K_{\text{sp}} = [\text{Ca}^{2+}][\text{F}^-]^2 = (s')(0.10 + 2s')^2 = 3.9 \times 10^{-11}$$

   Because $K_{\text{sp}}$ is very small, we make the approximation that $2s' \ll 0.10$, which simplifies the expression:

   $$(s')(0.10)^2 \approx 3.9 \times 10^{-11}$$

   $$s'(0.010) = 3.9 \times 10^{-11} \implies s' = 3.9 \times 10^{-9}\text{ M}$$

   For comparison, in pure water, the solubility of $\text{CaF}_2$ is given by $4s^3 = 3.9 \times 10^{-11}$, yielding $s \approx 2.1 \times 10^{-4}\text{ M}$. The presence of 0.10 M $\text{F}^-$ reduces the solubility by several orders of magnitude.

   The common-ion effect plays a crucial role in analytical chemistry, wastewater remediation, and chemical synthesis.

   Selective Precipitation

   When a solution contains multiple metal cations, adding a common counterion gradually allows chemists to precipitate one ion while leaving others in solution. Precipitation occurs when the ion product quotient $Q$ exceeds $K_{\text{sp}}$:

   • The compound with the smaller concentration threshold for precipitation precipitates first.
   • By carefully controlling the common-ion concentration (such as hydroxide, sulfate, or sulfide ions), distinct separation of cations can be achieved.

   Gravimetric Analysis

   In gravimetric determinations, an analyte is precipitated as an insoluble compound. To minimize losses caused by incomplete precipitation, an excess of the precipitating agent (a source of the common ion) is added. This shifts the solubility equilibrium toward the solid phase, ensuring that nearly all target ions leave the solution phase and are captured quantitatively on the filter.