Common Ion Effect

The Common Ion Effect is a significant principle in chemistry that describes the decrease in solubility of a compound when an ion that is already part of the compound is added to the solution. This phenomenon is important in various contexts, such as water treatment, pharmaceutical formulation, and industrial processes.

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Introductory Overview

• Consider the issue of lead contamination in water systems. The common ion effect can address this by causing precipitations of harmful ions like lead, thus lowering their concentration.
• A comprehensive understanding of this effect is essential for both preventing undesirable chemical reactions and enhancing beneficial ones.

Definition and Explanation

• Common Ion Effect: Decreased solubility of a compound in a solution when it contains an ion present in the compound.
• Dynamic Equilibrium: Ongoing balance between dissolution and precipitation.
• Sparingly Soluble Salt: Compounds with minimal solubility in water.
• Solubility Product Constant (K$_{sp}$): Indicates solubility under equilibrium conditions.

Real-world Application with Silver Chloride (AgCl)

• Equation:
  $\text{AgCl(s)} \rightleftharpoons \text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$
• Practical Insight: Introducing table salt (NaCl) to a saltwater solution adds more Cl⁻ ions. This shifts the equilibrium to precipitate additional solid AgCl, thereby reducing its solubility.
• Visualisation: Particles continuously exchange between states, illustrating ongoing equilibrium.

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Solubility Calculations with Common Ions

Overview

• The influence of common ions on solubility is measured by the solubility product constant (K$_{sp}$), which is crucial in predicting solubility within solutions.

Mathematical Framework

• K$_{sp}$ Formula:
  • For a salt AB:
    $K_{sp} = [A^+]^{m}[B^-]^{n}$
• Denotes the ion concentrations, each raised to the power of their stoichiometric coefficients.

Example Problems

Example 1: Lead(II) Bromide

• Chemical Equilibrium: $\text{PbBr}_2 (s) \leftrightarrow \text{Pb}^{2+} (aq) + 2\text{Br}^- (aq)$ $K_{sp} = [\text{Pb}^{2+}][\text{Br}^-]^2$
• Common Ion Effect:
  • Adding more bromide ions reduces the solubility of PbBr₂ by shifting the equilibrium to the left.

Example 2: Silver Chloride in Sodium Chloride Solution

• Influential Ion: Cl⁻ from NaCl affects the solubility of AgCl.
• K$_{sp}$ Expression: $K_{sp} = [\text{Ag}^+][\text{Cl}^-]$
• Step-by-Step Solution:
  1. Write the K$_{sp}$ expression: K$_{sp}$ = [Ag⁺][Cl⁻]
  2. If [Cl⁻] is known (e.g., from NaCl solution), rearrange to find [Ag⁺]: [Ag⁺] = K$_{sp}$/[Cl⁻]
  3. The molar solubility of AgCl equals [Ag⁺] in this case

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Practical Applications

Water Treatment

• Method: Introducing sodium chloride (NaCl) reduces calcium's solubility, prompting its precipitation.
• Outcome: Results in cleaner water, as depicted in the diagram.

Pharmaceutical Formulation

• Effect on Bioavailability: Common ions influence drug dissolution rates, affecting effectiveness.
• Example: Ibuprofen's bioavailability can be impacted by ions present in the stomach.

Environmental Implications

• Example: Discharge of ions by mining industries influences water solubility.
• Preventive Measures: Utilise alternative substances to decrease ion release.

Industrial Chemistry

• Process Optimisation: Controlling ion effects improves production efficiency and product quality.
• Case Study: The Bayer process demonstrates efficient alumina extraction.

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Problem with Solution

Determine the solubility of AgCl in a 0.1 M NaCl solution, given K$_{sp}$ of AgCl = 1.8 × 10$^{-10}$.

Solution:

1. Identify the equilibrium: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

2. Write the K$_{sp}$ expression: K$_{sp}$ = [Ag⁺][Cl⁻] = 1.8 × 10$^{-10}$

3. Analyse the Cl⁻ concentration: [Cl⁻] = 0.1 M (from NaCl) + x (from AgCl dissolution) Since x will be very small compared to 0.1 M, we can approximate: [Cl⁻] ≈ 0.1 M

4. Calculate [Ag⁺]: [Ag⁺] = K$_{sp}$/[Cl⁻] = (1.8 × 10$^{-10}$)/(0.1) = 1.8 × 10$^{-9}$ M

5. Determine solubility: The molar solubility of AgCl equals [Ag⁺] = 1.8 × 10$^{-9}$ M

Therefore, the solubility of AgCl in 0.1 M NaCl is 1.8 × 10$^{-9}$ M, significantly lower than its solubility in pure water (1.3 × 10$^{-5}$ M), demonstrating the common ion effect.