Common ion effect — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: This chapter covers the definition of the common ion effect, Le Chatelier’s principle applications to acid-base and solubility equilibria, pH calculation of common ion buffer systems, and calculation of solubility in the presence of a common ion for AP Chemistry exam questions.

You should already know: Le Chatelier’s principle for dynamic chemical equilibrium, equilibrium constant expressions for $K_a$, $K_{sp}$, and $K_b$, ICE table construction for equilibrium problems.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Common ion effect?

The common ion effect describes the suppression of ionization or dissolution of a weak electrolyte (weak acid, weak base, or sparingly soluble ionic salt) when a soluble compound containing an ion already present in the equilibrium (the "common ion") is added to the solution. It is sometimes called common ion suppression, and is a direct application of Le Chatelier’s principle to equilibrium systems. Unlike minor ionic strength effects from adding non-common ion salts (which are not tested on AP Chemistry), the common ion effect produces a large, easily calculable change in equilibrium concentrations that is explicitly tested on the exam.

Per the AP Chemistry CED, this topic falls under Unit 7: Equilibrium, which accounts for 7-9% of total AP exam score. Common ion questions appear in both MCQ and FRQ sections: MCQ typically test conceptual understanding of pH or solubility changes, while FRQ require quantitative calculation of pH or molar solubility. The common ion effect is also the foundational principle behind buffer solutions, one of the most heavily tested topics in Unit 7. (Word count: 248)

2. Common Ion Effect on Weak Acid and Weak Base Ionization

When a weak electrolyte (e.g., weak acid HA) ionizes in solution, it establishes the equilibrium: $$HA(aq)\rightleftharpoons H^{+}(aq)+A^{-}(aq)$$ If a soluble salt of the conjugate base NaA is added, it dissociates completely to release $A^{-}$, the common ion. Adding $A^{-}$ increases the product concentration, shifting equilibrium left toward undissociated HA. This reduces the percent ionization of HA, lowers $[H^{+}]$, and increases pH compared to a solution of HA alone. The same logic applies to weak bases: adding a soluble salt of the conjugate acid (the common ion) suppresses base ionization, lowers $[O{H}^{-}]$, and decreases pH.

We can calculate pH using the $K_a$ expression and the small x (5%) approximation, which assumes that the amount of HA that dissociates is negligible compared to the initial concentrations of HA and added $A^{-}$: $$K_a=\frac{[H^{+}][A^{-}]}{[HA]}\approx \frac{[H^{+}][A^{-}{]}_0}{[HA{]}_0}$$ Rearranging gives the Henderson-Hasselbalch equation, which is a direct result of the common ion effect: $$pH=p{K}_a+\log \left(\frac{[A^{-}{]}_0}{[HA{]}_0}\right)$$

Worked Example

Calculate the pH of a solution that is 0.15 M hydrofluoric acid (HF, $K_a=6.6\times 10^{-4}$) and 0.25 M sodium fluoride (NaF).

1. Identify the common ion: $F^{-}$ is common to both HF and NaF. NaF dissociates completely, so initial $[F^{-}{]}_0=0.25$ M, and initial $[HF{]}_0=0.15$ M.
2. Let $x=[H^{+}]$ from HF dissociation. The equilibrium concentrations are $[HF]=0.15-x$, $[H^{+}]=x$, $[F^{-}]=0.25+x$.
3. Apply the 5% approximation: $x<<0.15$ and $x<<0.25$, so $0.15-x\approx 0.15$ and $0.25+x\approx 0.25$. Substitute into $K_a$: $$6.6\times 10^{-4}=\frac{(x)(0.25)}{0.15}$$
4. Solve for $x=[H^{+}]=3.96\times 10^{-4}$ M. Check approximation: $x=0.16\%$ of 0.25 M, so approximation is valid.
5. Calculate pH: $pH=-\log (3.96\times 10^{-4})\approx 3.40$.

Exam tip: On the AP exam, you can directly use the Henderson-Hasselbalch equation for common ion weak acid/base systems to save time, as the 5% approximation almost always holds for these problems.

3. Common Ion Effect on Solubility of Sparingly Soluble Salts

The common ion effect significantly reduces the molar solubility (s, moles of solid that dissolve per liter of solution) of sparingly soluble ionic compounds. For the general dissolution equilibrium: $$M_x{X}_y(s)\rightleftharpoons x{M}^{y+}(aq)+y{X}^{x-}(aq)$$ the solubility product constant is $K_{sp}=[M^{y+}{]}^x[X^{x-}{]}^y$. If a common ion (e.g., $M^{y+}$ from a soluble nitrate salt) is added, equilibrium shifts left toward the solid, reducing the amount of $M_x{X}_y$ that dissolves. For sparingly soluble salts, the solubility s is already very small, so the contribution of dissolved $M_x{X}_y$ to the common ion concentration is negligible, so we can approximate the common ion concentration as equal to its initial added concentration. This allows us to solve directly for s.

For example, if we add $M^{y+}$ at initial concentration $[M^{y+}{]}_0$, the approximation gives $[M^{y+}]\approx [M^{y+}{]}_0$, and $[X^{x-}]=ys$, so $K_{sp}=([M^{y+}{]}_0{)}^x(ys{)}^y$, which can be solved for s.

Worked Example

Calculate the molar solubility of calcium oxalate ($Ca{C}_2{O}_4$, $K_{sp}=2.3\times 10^{-9}$) in a 0.15 M solution of calcium chloride ($CaC{l}_2$).

1. Write the balanced dissolution equilibrium: $Ca{C}_2{O}_4(s)\rightleftharpoons C{a}^{2+}(aq)+C_2{O}_4^{2-}(aq)$. The common ion is $C{a}^{2+}$, from complete dissociation of $CaC{l}_2$, so initial $[C{a}^{2+}{]}_0=0.15$ M.
2. Let $s$ = molar solubility of $Ca{C}_2{O}_4$. Each mole of dissolved $Ca{C}_2{O}_4$ gives 1 mole of $C{a}^{2+}$ and 1 mole of $C_2{O}_4^{2-}$, so equilibrium concentrations are $[C{a}^{2+}]=0.15+s$, $[C_2{O}_4^{2-}]=s$.
3. Apply the small s approximation: $s<<0.15$, so $0.15+s\approx 0.15$. Substitute into $K_{sp}$: $$2.3\times 10^{-9}=(0.15)(s)$$
4. Solve for s: $s=\frac{2.3\times 10^{-9}}{0.15}\approx 1.5\times 10^{-8}$ M. Check approximation: $1.5\times 10^{-8}$ M is orders of magnitude smaller than 0.15 M, so approximation is valid. For comparison, solubility in pure water is ~$4.8\times 10^{-5}$ M, 3000x higher than in the common ion solution, confirming the large suppression of solubility.

Exam tip: Always write the balanced dissolution reaction before plugging concentrations into $K_{sp}$; do not assume all salts are 1:1 and use the 1:1 formula for salts with different stoichiometry.

4. Qualitative Predictions of Common Ion Effect Changes

A very common AP exam question asks for qualitative predictions of how adding a common ion changes pH, percent ionization, or solubility, without requiring full calculation. These questions rely on applying Le Chatelier’s principle correctly to the equilibrium system. The core rules are:

1. Adding a common product ion always shifts equilibrium toward the reactant side (undissociated weak electrolyte or undissolved solid).
2. No common ion = no common ion effect (minor ionic strength effects are not tested on AP, so you can assume no change).
3. For weak acids: adding common conjugate base → lower $[H^{+}]$ → higher pH → lower percent ionization.
4. For weak bases: adding common conjugate acid → lower $[O{H}^{-}]$ → lower pH → lower percent ionization.
5. For sparingly soluble salts: adding any common ion (cation or anion) → lower solubility.

Worked Example

For each change to a 0.10 M acetic acid solution (initial pH = 2.87), state whether pH increases, decreases, or stays the same, and justify: (a) Solid sodium acetate is added to the solution. (b) Solid sodium chloride is added to the solution.

1. Write the acetic acid ionization equilibrium: $C{H}_{3}COOH(aq)\rightleftharpoons H^{+}(aq)+C{H}_{3}CO{O}^{-}(aq)$.
2. For (a): Sodium acetate dissociates completely to release $C{H}_{3}CO{O}^{-}$, which is the common product ion. Adding $C{H}_{3}CO{O}^{-}$ shifts equilibrium left toward undissociated acetic acid. This reduces $[H^{+}]$, so pH increases above 2.87.
3. For (b): Sodium chloride dissociates to $N{a}^{+}$ and $C{l}^{-}$, neither of which are present in the acetic acid equilibrium. There is no common ion, so the common ion effect does not change $[H^{+}]$, so pH stays the same (for AP purposes).

Exam tip: Always write the balanced equilibrium reaction in your justification for qualitative questions; AP graders require explicit reference to the shift direction to award full credit.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Calculating the common ion concentration by adding the full contribution of the dissolved weak electrolyte/sparingly soluble salt to the added common ion concentration, leading to an incorrect low value for s or pH. Why: Students forget that dissociation/dissolution is already suppressed, so the contribution from the weak electrolyte is negligible. Correct move: Always apply the small x approximation first; only add the contribution of the weak electrolyte if the approximation fails (which almost never happens on AP problems).
• Wrong move: Predicting that adding a common conjugate base to a weak acid decreases pH. Why: Students confuse lower $[H^{+}]$ with lower pH, or mix up the direction of the equilibrium shift. Correct move: Always write the equilibrium before predicting: adding product ion shifts left, so $[H^{+}]$ decreases, and pH (the negative log of $[H^{+}]$) increases.
• Wrong move: For $Ca{F}_2$ in a NaF solution, writing $K_{sp}=[C{a}^{2+}][F^{-}]=(s)(0.10)=K_{sp}$, forgetting the stoichiometric coefficient for $F^{-}$. Why: Students memorize the 1:1 salt formula and apply it to all salts regardless of stoichiometry. Correct move: Always balance the dissolution reaction and map equilibrium concentrations to solubility before plugging into $K_{sp}$.
• Wrong move: Claims adding any salt to a weak electrolyte solution triggers the common ion effect. Why: Students assume all salts contain a common ion, but only salts with one ion matching the equilibrium produce the effect. Correct move: Check both ions of the added salt against the equilibrium to confirm one is common before invoking the common ion effect.
• Wrong move: Swapping the concentrations of weak acid and conjugate base in the Henderson-Hasselbalch equation for common ion systems. Why: Students misremember the order of terms in the formula. Correct move: Derive $[H^{+}]$ directly from the $K_a$ expression if you cannot remember the order of terms to avoid this mistake.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Which of the following changes will result in a decrease in the percent ionization of 0.20 M ammonia (NH₃, a weak base) due exclusively to the common ion effect? A) Adding pure water to dilute the solution B) Adding 0.10 mol of solid ammonium nitrate (NH₄NO₃) to the solution C) Adding 0.10 mol of solid sodium chloride (NaCl) to the solution D) Adding 0.10 mol of solid potassium nitrate (KNO₃) to the solution

Worked Solution: First, write the ionization equilibrium for ammonia: $N{H}_3(aq)+H_{2}O(l)\rightleftharpoons N{H}_4^{+}(aq)+O{H}^{-}(aq)$. The common ion effect requires adding a soluble salt that contains an ion already present in this equilibrium. Ammonium nitrate dissociates into $N{H}_4^{+}$ and $N{O}_3^{-}$, and $N{H}_4^{+}$ is a product of ammonia ionization. Adding $N{H}_4^{+}$ shifts equilibrium left, reducing the amount of ammonia that ionizes, so percent ionization decreases. Option A is incorrect because dilution increases percent ionization and does not act via the common ion effect. Options C and D have no common ions with the ammonia equilibrium, so the common ion effect does not apply. Correct answer: B

Question 2 (Free Response)

A student prepares a solution containing 0.30 M propanoic acid ($C{H}_{3}C{H}_{2}COOH$, $K_a=1.3\times 10^{-5}$) and 0.20 M calcium propanoate ($(C{H}_{3}C{H}_{2}COO{)}_{2}Ca$). (a) Calculate the pH of the resulting solution. (b) Explain why the pH of this solution is higher than the pH of a 0.30 M solution of propanoic acid with no added calcium propanoate. (c) The student adds a small amount of strong acid to the solution. Does the ratio $\frac{[C{H}_{3}C{H}_{2}CO{O}^{-}]}{[C{H}_{3}C{H}_{2}COOH]}$ increase, decrease, or stay the same? Justify your answer in terms of the common ion effect and Le Chatelier’s principle.

Worked Solution: (a) Calcium propanoate dissociates completely, with 1 mole of salt producing 2 moles of propanoate ion, so initial $[C{H}_{3}C{H}_{2}CO{O}^{-}]=2\times 0.20=0.40$ M. Calculate $p{K}_a=-\log (1.3\times 10^{-5})\approx 4.89$. Substitute into the Henderson-Hasselbalch equation: $$pH=4.89+\log \left(\frac{0.40}{0.30}\right)=4.89+0.12=5.01$$ (b) Propanoic acid ionizes via $C{H}_{3}C{H}_{2}COOH(aq)\rightleftharpoons H^{+}(aq)+C{H}_{3}C{H}_{2}CO{O}^{-}(aq)$. Adding calcium propanoate introduces the common ion $C{H}_{3}C{H}_{2}CO{O}^{-}$, which shifts equilibrium left. This reduces $[H^{+}]$, so pH is higher than in pure propanoic acid solution. (c) The ratio decreases. Adding strong acid introduces $H^{+}$, a product, to the equilibrium. Per Le Chatelier’s principle, the system shifts left to consume the added $H^{+}$, converting $C{H}_{3}C{H}_{2}CO{O}^{-}$ to $C{H}_{3}C{H}_{2}COOH$. This decreases $[C{H}_{3}C{H}_{2}CO{O}^{-}]$ and increases $[C{H}_{3}C{H}_{2}COOH]$, so the ratio decreases.

Question 3 (Application / Real-World Style)

In gravimetric analysis of chloride ion, a chemist precipitates chloride as silver chloride (AgCl, $K_{sp}=1.8\times 10^{-10}$) by adding a small excess of silver nitrate to the sample solution. The final solution after precipitation has a concentration of 0.010 M Ag⁺ from the excess silver nitrate. What mass of AgCl remains dissolved in 100. mL of this final solution? Comment on why adding excess Ag⁺ is good practice for this analysis. The molar mass of AgCl is 143.32 g/mol.

Worked Solution: The dissolution equilibrium is $AgCl(s)\rightleftharpoons A{g}^{+}(aq)+C{l}^{-}(aq)$. Let s = molar solubility of AgCl. With excess Ag⁺ as the common ion, $[A{g}^{+}]\approx 0.010$ M, so $K_{sp}=[A{g}^{+}][C{l}^{-}]=(0.010)(s)=1.8\times 10^{-10}$. Solving gives $s=1.8\times 10^{-8}$ mol/L. For 0.100 L of solution, moles of dissolved AgCl = $1.8\times 10^{-8}\times 0.100=1.8\times 10^{-9}$ mol. Mass = $1.8\times 10^{-9}$ mol × 143.32 g/mol ≈ $2.6\times 10^{-7}$ g. This extremely small mass means almost no AgCl stays dissolved in solution. Adding excess Ag⁺ common ion reduces the solubility of AgCl, minimizing mass loss of the precipitate, which leads to more accurate measurement of the original chloride concentration in the sample.

7. Quick Reference Cheatsheet

8. What's Next

The common ion effect is the foundational prerequisite for buffer solutions, the next major topic in Unit 7 Equilibrium. Buffers are literally common ion systems (weak acid + common conjugate ion, or weak base + common conjugate ion) designed to resist pH change, so without understanding how common ions suppress ionization and shift equilibrium, you will not be able to calculate buffer pH or predict buffer capacity. Beyond buffers, the common ion effect is also a core concept for understanding solubility equilibria, selective precipitation, and titration pH curves, all of which are tested heavily on the AP Chemistry exam. It also connects to acid-base homeostasis in biological systems, a common AP experimental context.

• Buffer solutions
• Solubility equilibria and Ksp
• Selective precipitation
• Acid-base titration curves