7.12 Common Ion Effect: Understanding and Applying Equilibrium Concepts

What Is the Common Ion Effect?

In our exploration of solubility equilibria, we have primarily focused on how compounds dissolve in pure water. However, what if a solute is added to a solution already containing one of the ions the solute dissociates into? This is where the common ion effect comes into play.

The Common Ion Effect Defined

The common ion effect occurs when the solubility of a substance is suppressed by the presence of a common ion. This ion is already present in the solution from another source and influences the equilibrium, reducing the overall solubility of the solute. This effect can be explained using Le Chatelier’s Principle, which states that a system at equilibrium will adjust to counteract any imposed changes.

For example, consider dissolving AgBr in a solution of NaBr. The Br⁻ ion, which is present in both AgBr and NaBr, is the common ion. Adding NaBr introduces an excess of Br⁻, shifting the equilibrium to favor the formation of more solid AgBr to reduce the concentration of free ions.

Mathematical Explanation of the Common Ion Effect

Understanding the Equilibrium

For a dissolution reaction:

$$\text{AgBr (s)} \leftrightharpoons \text{Ag⁺ (aq)} + \text{Br⁻ (aq)}$$

The solubility product constant (Ksp) is defined as:

$$K_{sp} = [\text{Ag⁺}][\text{Br⁻}]$$

When a common ion is present (e.g., Br⁻ from NaBr), the initial concentration of Br⁻ is not zero. This affects the equilibrium position, reducing the solubility of AgBr.

Practice Problem: Calculating the Effect of a Common Ion

Problem: Calculate the molar solubility of AgBr (Ksp = 7.7 x 10⁻¹³) in:

1. Pure water
2. A 0.0010 M NaBr solution

1. Molar Solubility in Pure Water

1. Write the dissociation reaction:

   $$\text{AgBr (s)} \leftrightharpoons \text{Ag⁺ (aq)} + \text{Br⁻ (aq)}$$
   

2. Set up an ICE table:

   Reaction	AgBr	Ag⁺	Br⁻
   Initial	—-	0.00 M	0.00 M
   Change	—-	+x	+x
   Equilibrium	—-	x	x

3. Substitute into the Ksp expression:

   $$K_{sp} = [\text{Ag⁺}][\text{Br⁻}] = x^2$$
   

   Given Ksp = 7.7 x 10⁻¹³:

   $$x^2 = 7.7 \times 10^{-13} \Rightarrow x = 8.8 \times 10^{-7} \text{ M}$$
   

2. Molar Solubility in a 0.0010 M NaBr Solution

1. Initial concentrations: [Br⁻] = 0.0010 M

2. Adjust the ICE table:

   Reaction	AgBr	Ag⁺	Br⁻
   Initial	—-	0.00 M	0.0010 M
   Change	—-	+x	+x
   Equilibrium	—-	x	0.0010 + x

3. Substitute into the Ksp expression:

   $$K_{sp} = [\text{Ag⁺}][\text{Br⁻}] = [x][0.0010 + x]$$
   

   Since x is very small, approximate 0.0010 + x ≈ 0.0010:

   $$K_{sp} = [x][0.0010] = 7.7 \times 10^{-13}$$
   

   Solving for x:

   $$x = \frac{7.7 \times 10^{-13}}{0.0010} = 7.7 \times 10^{-10} \text{ M}$$
   

Conclusion

The molar solubility of AgBr in 0.0010 M NaBr is 7.7 x 10⁻¹⁰ M, which is significantly lower than its solubility in pure water (8.8 x 10⁻⁷ M).

Justifying the Common Ion Effect

The reduction in solubility due to the common ion effect is explained by Le Chatelier’s Principle. When a common ion is present, it acts as an external stress on the equilibrium system, causing the reaction to shift towards the formation of the undissolved solid to reduce the concentration of the excess ion.

Example: Dissolving CaSO₄ in a CuSO₄ solution (where the common ion is SO₄²⁻) results in decreased solubility of CaSO₄ as the equilibrium shifts to form more solid CaSO₄.

Key Takeaways

• The common ion effect decreases the solubility of a compound when a solution already contains one of the ions present in the compound.
• Le Chatelier’s Principle explains this effect as the system adjusting to minimize stress by shifting equilibrium.
• Practical calculations using Ksp help quantify changes in solubility due to common ions.