7.12 Common Ion Effect

The common ion effect means a salt dissolves less in a solution that already contains one of its ions. You can explain this with Le Chatelier's principle, because the extra ion pushes the dissolution equilibrium back toward the solid, or calculate it by plugging the common ion's concentration into the $K_{sp}$ expression. For AP Chemistry, put the common ion's starting concentration in the initial row of the ICE table.

Why This Matters for the AP Chemistry Exam

This topic connects solubility equilibria, Ksp, and Le Chatelier's principle into one reasoning skill. On the AP Chemistry exam, you may be asked to find the solubility of a salt or the value of Ksp when a common ion is already present, and to explain how changing the solution changes the result. Because the skill here focuses on explaining how a change to conditions alters an outcome, expect to both calculate molar solubility with an ICE table and justify the direction of the shift in words. Getting comfortable with Qsp versus Ksp also strengthens your equilibrium reasoning across the whole unit.

Key Takeaways

• Adding a common ion lowers the molar solubility of a salt compared to its solubility in pure water.
• The common-ion effect is just Le Chatelier's principle applied to a dissolution equilibrium: extra product ion shifts the system back toward the solid.
• Ksp itself does not change when you add a common ion; only the solubility (how much dissolves) changes.
• Set up the ICE table with the common ion's starting concentration already in the "initial" row.
• When the common ion concentration is much larger than x, you can approximate (0.0010 + x) as 0.0010 to avoid the quadratic.
• You can confirm the effect by comparing Qsp to Ksp: extra common ion makes Qsp greater than Ksp, so the system shifts toward forming more solid.

The Common-Ion Effect Explained

The common-ion effect describes how an ion already in solution suppresses the solubility of a salt that produces that same ion. A "common ion" is just an ion shared between two compounds. When you dissolve a salt into a solution that already contains one of its ions, the dissolution equilibrium adjusts, and less of the salt dissolves than it would in pure water.

For example, if you dissolve AgBr in a solution of NaBr, there is already bromide ion (Br-) present in the system. The solubility equilibrium responds by shifting toward the solid (the reactant side). This is Le Chatelier's principle in action. Like other situations where a stress is introduced to a chemical system, the common-ion effect can be justified by comparing Q to K.

Here is the reasoning with symbols:

For the dissolution $\text{AB}(s) \rightleftharpoons \text{A}^+(aq) + \text{B}^-(aq)$, the solubility product is $K_{sp} = [\text{A}^+][\text{B}^-]$. Suppose the solution already contains some initial concentration $X$ of $\text{B}^-$. Right when you add the solid, $Q_{sp} = [\text{A}^+][\text{B}^- + X]$. Because that extra $\text{B}^-$ raises the ion product, $Q_{sp} > K_{sp}$, so the system shifts back toward the solid and dissolution is inhibited. The example below shows what this looks like with real numbers.

Common-Ion Effect Practice Problem

Start with the molar solubility in pure water, the same way you would in solubility equilibria.

The dissolution is:

$\text{AgBr}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Br}^-(aq)$

Set up an ICE table:

$K_{sp} = [\text{Ag}^+][\text{Br}^-] = x^2 = 7.7 \times 10^{-13}$

So $x = 8.8 \times 10^{-7} \text{ M}$.

Now find the molar solubility in 0.0010 M NaBr. The NaBr gives you a starting 0.0010 M concentration of Br-, so put that in the initial row:

So:

$K_{sp} = [\text{Ag}^+][\text{Br}^-] = [x][0.0010 + x] \approx [x][0.0010] = 7.7 \times 10^{-13}$

Here you can approximate $(0.0010 + x)$ as 0.0010 because $x$ is tiny compared to 0.0010. Solving gives $x = 7.7 \times 10^{-10} \text{ M}$.

Notice this is much smaller than the 8.8 * 10^(-7) M you found in pure water. The added bromide ion suppressed the solubility, exactly as the common-ion effect predicts.

Justifying the Common-Ion Effect

You can calculate molar solubilities when common ions are present, but it helps to know why the effect happens. The answer is Le Chatelier's principle.

As a reminder, Le Chatelier's principle says that when an external stress is placed on a system at equilibrium, the system shifts toward reactants or products to relieve that stress and reach a new equilibrium. For instance, adding more of a reactant pushes the system toward making more products.

Apply that to dissolution. The common-ion effect happens when an ion is already present in a solution where the same ion is being released by a dissolving salt. For example, dissolving $\text{CaSO}_4$ ($K_{sp} = 2.4 \times 10^{-5}$) in a solution of $\text{CuSO}_4$ brings in a common ion: sulfate ($\text{SO}_4^{2-}$). The dissolution equilibrium is:

$\text{CaSO}_4(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{SO}_4^{2-}(aq)$

The already-present sulfate acts as a stress on the right side of the equilibrium. Le Chatelier's principle predicts the system shifts toward the reactant (the solid), which lowers the molar solubility of $\text{CaSO}_4$.

So Le Chatelier's principle gives you the qualitative justification for the common-ion effect. You can confirm it either by comparing Qsp to Ksp or by calculating the new molar solubility and seeing that it is lower than in pure water.

How to Use This on the AP Chemistry Exam

Problem Solving

• Write the balanced dissolution equation first, then the Ksp expression with correct stoichiometric exponents.
• In the ICE table, put the common ion's starting concentration in the initial row before adding +x.
• Use the approximation that x is negligible compared to the common ion concentration when it is reasonable. This avoids the quadratic and usually matches the real answer closely.
• Watch your units. Molar solubility is in mol/L, and you may need to convert to g/L if the question asks.

Free Response

• If asked to explain why solubility drops, name Le Chatelier's principle and identify the specific common ion as the stress.
• You can also justify the shift by comparing Qsp to Ksp: the added ion makes Qsp greater than Ksp, so the system shifts toward the solid.
• State clearly that Ksp stays the same and that it is the solubility that decreases.

Common Trap

• Do not change the value of Ksp when a common ion is added. Ksp depends only on temperature.
• Do not forget the stoichiometric coefficients as exponents in the Ksp expression, especially for salts like CaF2 where one ion appears with a coefficient of 2.

Common Misconceptions

• "Adding a common ion changes Ksp." Ksp is constant at a given temperature. The common ion changes how much salt dissolves, not the value of Ksp.
• "The common-ion effect makes salts dissolve more." It does the opposite. The shared ion shifts the equilibrium toward the solid and lowers solubility.
• "The +x in the ICE table is the same as the common ion concentration." The common ion's starting amount goes in the initial row. The +x is the additional amount that dissolves, which is the molar solubility you are solving for.
• "You always have to solve the quadratic." When the common ion concentration is much larger than x, the approximation (initial + x) ≈ initial is valid and faster. Just be ready to check that the approximation holds.
• "Qsp > Ksp means nothing is happening." It means the solution is past saturation for those conditions, so the system shifts toward forming solid until Qsp equals Ksp again.

Related AP Chemistry Guides

• 7.1 Introduction to Equilibrium
• Unit 7 Overview: Equilibrium
• 7.2 Direction of Reversible Reactions
• 7.3 Reaction Quotient and Equilibrium Constant
• 7.4 Calculating the Equilibrium Constant
• 7.11 Introduction to Solubility Equilibria