How does the addition of a common ion affect the molar solubility of a salt?

It decreases the molar solubility. According to Le Chatelier's Principle, increasing the concentration of a product ion shifts the equilibrium toward the solid reactant, causing more precipitate to form.

What is the difference between the effect of a common ion on solubility versus its effect on the $K_{sp}$ value?

The common ion decreases the solubility (the amount of salt that dissolves), but it has no effect on the $K_{sp}$ value. $K_{sp}$ is a constant that only changes with temperature.

When comparing the solubility of $AgCl$ in pure water versus $0.1$ M $NaCl$, which is lower and why?

The solubility is lower in $0.1$ M $NaCl$. The high initial concentration of $Cl^-$ ions from $NaCl$ forces the $AgCl$ equilibrium to stay mostly on the solid side to satisfy the $K_{sp}$ constant.

What is a frequent mistake when calculating the common ion concentration from a salt like $CaCl_2$?

Forgetting the stoichiometry of the source salt. $0.1$ M $CaCl_2$ provides $0.2$ M $Cl^-$ ions because there are two moles of chloride for every one mole of calcium chloride.

Why is it an error to assume the concentration of the common ion comes solely from the sparingly soluble salt?

Because the common ion is typically provided by a highly soluble strong electrolyte. Its concentration is usually much higher than the tiny amount contributed by the sparingly soluble salt, which is often negligible.

What happens if a student forgets to square the concentration of an ion in the $K_{sp}$ expression for a salt like $PbI_2$?

The calculated solubility will be incorrect because the $K_{sp}$ expression must reflect the stoichiometry of the balanced equation ($K_{sp} = [Pb^{2+}][I^-]^2$).

Define the 'Common-Ion Effect'.

It is the reduction in the solubility of an ionic compound caused by the addition of a soluble compound that shares one of the same ions as the original compound.

What is 'Molar Solubility' ($s$)?

The number of moles of a solute that can be dissolved in one liter of solution before the solution becomes saturated.

What is the $K_{sp}$ expression for $CaF_2$?

$K_{sp} = [Ca^{2+}][F^-]^2$. This expression defines the equilibrium state for the saturated solution.

How does Le Chatelier's Principle explain the common-ion effect?

Adding a common ion increases the concentration of a product in the dissolution equilibrium. The system responds by shifting to the left (reactant side) to consume the excess ion, forming more solid.

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Common-Ion Effect

Summary

The common-ion effect describes the reduction in the solubility of an ionic compound when a soluble salt containing one of its constituent ions is added to the solution. This phenomenon is a direct application of Le Chatelier's Principle to solubility equilibria, where the increased concentration of a product ion shifts the equilibrium toward the solid reactant, resulting in precipitation.

1. Definition & Core Concepts

The Common-Ion Effect occurs when a substance is added to a solution that already contains one of the ions present in the substance's equilibrium expression. This 'common' ion suppresses the ionization or dissolution of the original substance.

In the context of solubility, it specifically refers to the decrease in the molar solubility of a sparingly soluble salt when a strong electrolyte containing a shared ion is introduced.

A saturated solution represents a dynamic equilibrium between the undissolved solid and its dissolved ions, expressed as: $MX(s) \rightleftharpoons M^+(aq) + X^-(aq)$ . Adding more $M^+$ or $X^-$ from an external source disrupts this balance.

Diagram showing a saturated solution where the addition of a common ion leads to increased solid precipitate at the bottom of the beaker.

2. Underlying Principles

3. Methods & Techniques

4. Key Distinctions

5. Exam Strategy & Tips

6. Common Pitfalls & Misconceptions

Common-Ion Effect

Summary

1. Definition & Core Concepts

In the context of solubility, it specifically refers to the decrease in the molar solubility of a sparingly soluble salt when a strong electrolyte containing a shared ion is introduced.

Diagram showing a saturated solution where the addition of a common ion leads to increased solid precipitate at the bottom of the beaker.

2. Underlying Principles

The effect is governed by Le Chatelier's Principle, which states that if a system at equilibrium is disturbed, the system will shift to counteract the disturbance. Increasing the concentration of a product ion drives the reaction toward the reactant (solid) side.

Mathematically, the Solubility Product Constant ( $K_{sp}$ ) remains constant at a fixed temperature. For the equilibrium $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$ , the expression is $K_{sp} = [Ag^+][Cl^-]$ .

If the concentration of $Cl^-$ is increased by adding $NaCl$ , the concentration of $Ag^+$ must decrease proportionally to keep the product equal to the constant $K_{sp}$ value.

3. Methods & Techniques

Step 1: Identify the Common Ion: Determine which ion is being provided by the soluble salt (the strong electrolyte).
Step 2: Set up the ICE Table: Unlike pure water calculations where initial ion concentrations are zero, the common ion will have a non-zero initial concentration ( $C_0$ ).
Step 3: Define Change: Let $s$ be the molar solubility. The change in ion concentrations will be $+s$ (or $+ns$ based on stoichiometry).
Step 4: Apply the Approximation: If $K_{sp}$ is very small relative to the common ion concentration, assume $C_0 + s \approx C_0$ to simplify the algebra.
Step 5: Solve for $s$ : Substitute the equilibrium terms into the $K_{sp}$ expression and solve for the unknown solubility.

4. Key Distinctions

Feature	Solubility in Pure Water	Solubility with Common Ion
Initial Ion Conc.	Zero for all ions	Non-zero for the common ion
Equilibrium Shift	None (standard state)	Shifts toward the solid phase
Molar Solubility	Higher	Significantly Lower
Ksp Value	Constant at fixed Temp	Constant at fixed Temp

Solubility vs. Ksp: It is critical to remember that while the solubility (the amount that dissolves) changes, the solubility product constant ( $K_{sp}$ ) does not change unless the temperature changes.
Q vs. Ksp: When the common ion is added, the reaction quotient $Q$ temporarily exceeds $K_{sp}$ , triggering precipitation until $Q$ again equals $K_{sp}$ .

5. Exam Strategy & Tips

Check Stoichiometry: Always verify the mole ratio of the common ion in the source salt. For example, $0.10$ M $MgCl_2$ provides $0.20$ M $Cl^-$ ions.
The 'Small x' Rule: In almost all common-ion problems, the solubility $s$ is so small compared to the initial concentration of the common ion that you can ignore it in addition/subtraction steps. Always state this assumption.
Reasonableness Check: Your calculated solubility in a common-ion solution should always be several orders of magnitude smaller than the solubility in pure water.
Units: Ensure all concentrations are in Molarity (mol/L) before plugging them into the $K_{sp}$ expression.

6. Common Pitfalls & Misconceptions

Misconception: Thinking that adding a common ion increases the $K_{sp}$ value. $K_{sp}$ is an equilibrium constant and is only temperature-dependent.
Calculation Error: Forgetting to square or cube the concentration of the common ion if the stoichiometry requires it (e.g., in $PbI_2$ , the $[I^-]$ term is squared).
Source Confusion: Students often confuse the solubility of the sparingly soluble salt with the concentration of the common ion from the soluble salt. The common ion concentration is usually dominated by the fully dissociated strong electrolyte.

If the concentration of $Cl^-$ is increased by adding $NaCl$ , the concentration of $Ag^+$ must decrease proportionally to keep the product equal to the constant $K_{sp}$ value.

Step 1: Identify the Common Ion: Determine which ion is being provided by the soluble salt (the strong electrolyte).
Step 2: Set up the ICE Table: Unlike pure water calculations where initial ion concentrations are zero, the common ion will have a non-zero initial concentration ( $C_0$ ).
Step 3: Define Change: Let $s$ be the molar solubility. The change in ion concentrations will be $+s$ (or $+ns$ based on stoichiometry).
Step 4: Apply the Approximation: If $K_{sp}$ is very small relative to the common ion concentration, assume $C_0 + s \approx C_0$ to simplify the algebra.
Step 5: Solve for $s$ : Substitute the equilibrium terms into the $K_{sp}$ expression and solve for the unknown solubility.

Feature	Solubility in Pure Water	Solubility with Common Ion
Initial Ion Conc.	Zero for all ions	Non-zero for the common ion
Equilibrium Shift	None (standard state)	Shifts toward the solid phase
Molar Solubility	Higher	Significantly Lower
Ksp Value	Constant at fixed Temp	Constant at fixed Temp

Solubility vs. Ksp: It is critical to remember that while the solubility (the amount that dissolves) changes, the solubility product constant ( $K_{sp}$ ) does not change unless the temperature changes.
Q vs. Ksp: When the common ion is added, the reaction quotient $Q$ temporarily exceeds $K_{sp}$ , triggering precipitation until $Q$ again equals $K_{sp}$ .

Check Stoichiometry: Always verify the mole ratio of the common ion in the source salt. For example, $0.10$ M $MgCl_2$ provides $0.20$ M $Cl^-$ ions.
The 'Small x' Rule: In almost all common-ion problems, the solubility $s$ is so small compared to the initial concentration of the common ion that you can ignore it in addition/subtraction steps. Always state this assumption.
Reasonableness Check: Your calculated solubility in a common-ion solution should always be several orders of magnitude smaller than the solubility in pure water.
Units: Ensure all concentrations are in Molarity (mol/L) before plugging them into the $K_{sp}$ expression.

Misconception: Thinking that adding a common ion increases the $K_{sp}$ value. $K_{sp}$ is an equilibrium constant and is only temperature-dependent.
Calculation Error: Forgetting to square or cube the concentration of the common ion if the stoichiometry requires it (e.g., in $PbI_2$ , the $[I^-]$ term is squared).
Source Confusion: Students often confuse the solubility of the sparingly soluble salt with the concentration of the common ion from the soluble salt. The common ion concentration is usually dominated by the fully dissociated strong electrolyte.