AP Chemistry Quiz: Common Ion Effect

What this quiz covers

This quiz focuses on Common Ion Effect, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. It increases because the added salt acts as a buffer for $\text{F}^-$.
2. It decreases because the equilibrium shifts toward $\text{CaF}_2(s)$. (correct answer)
3. It increases because extra ions help pull solid into solution.
4. It stays the same because adding a common ion does not affect $K_{sp}$.
5. It stays the same because the solution is already saturated.

Explanation: This question tests the common ion effect on the solubility of ionic compounds. Adding a soluble fluoride salt increases [F⁻], which is already present from the CaF₂ equilibrium. According to Le Chatelier's principle, the increased fluoride concentration shifts the equilibrium left toward solid CaF₂, decreasing the amount of CaF₂ that can dissolve (its solubility). Students who choose E incorrectly recognize that Ksp remains constant but fail to understand that constant Ksp with increased [F⁻] requires decreased [Ca²⁺], which means less CaF₂ dissolves. When a common ion is added to a saturated solution, always expect the solubility of the sparingly soluble salt to decrease due to the equilibrium shift.

Question 2

1. It shifts left, decreasing $[\text{OH}^-]$. (correct answer)
2. It shifts right, increasing $[\text{OH}^-]$.
3. It does not shift because water is a pure liquid.
4. It shifts right because adding a salt always increases ion formation.
5. It does not shift because $\text{NH}_4^+$ makes the solution a buffer.

Explanation: This question tests the common ion effect on weak base equilibria. Adding NH₄Cl increases the concentration of NH₄⁺ ions, which are already present as a product in the ammonia equilibrium. By Le Chatelier's principle, increasing [NH₄⁺] causes the equilibrium to shift left, consuming OH⁻ ions and forming more NH₃. Students who choose C incorrectly think that because water is a pure liquid, the equilibrium cannot shift, but water's constant activity doesn't prevent the equilibrium from responding to changes in ion concentrations. To predict the effect of adding a common ion to a weak base system, identify which side of the equilibrium contains the common ion and apply Le Chatelier's principle.

Question 3

1. The equilibrium shifts left because $\text{HCO}_3^-$ is a common ion (product). (correct answer)
2. The equilibrium shifts right because adding a salt always increases $\text{OH}^-$ production.
3. The equilibrium position does not change because $K_b$ is constant.
4. The equilibrium shifts right because $\text{HCO}_3^-$ is a strong acid that neutralizes $\text{OH}^-$.
5. The equilibrium is unchanged because adding $\text{HCO}_3^-$ only creates a buffer, not a shift.

Explanation: This question tests the common-ion effect on hydrolysis equilibria of ions like CO3^2-. Adding sodium bicarbonate introduces HCO3- ions, a common ion (product) in the equilibrium CO3^2-(aq) + H2O(l) ⇌ HCO3-(aq) + OH-(aq). By Le Chatelier's principle, the increased HCO3- shifts the equilibrium to the left to consume the excess HCO3-, decreasing OH- production. This reduces the extent of CO3^2- hydrolysis. A tempting distractor is choice C, which claims no change because Kb is constant, based on the misconception that equilibrium constants prevent positional shifts from concentration changes. For hydrolysis reactions, detect common ions and apply Le Chatelier's principle to evaluate equilibrium responses.

Question 4

1. HF ionizes more because adding $\text{F}^-$ increases conductivity.
2. HF ionizes less because the equilibrium shifts toward HF. (correct answer)
3. HF ionization is unchanged because $K_a$ changes when salt is added.
4. HF ionizes more because $\text{NaF}$ acts as a buffer that forces dissociation.
5. HF ionization is unchanged because the added ion is a spectator.

Explanation: This question tests the common ion effect on weak acid ionization. Adding NaF increases the concentration of F⁻ ions, which are already present as a product in the HF equilibrium. By Le Chatelier's principle, the increased [F⁻] causes the equilibrium to shift left, forming more undissociated HF and decreasing the extent of ionization. Students who choose E incorrectly think F⁻ is a spectator ion, but F⁻ is actually a participant in the HF equilibrium and directly affects the position of equilibrium. To analyze common ion effects on weak acids, identify the conjugate base as the common ion and recognize that its addition always suppresses acid ionization.

Question 5

1. It shifts right, increasing $[\text{OH}^-]$.
2. It does not shift because $\text{NH}_4^+$ makes the solution a buffer.
3. It shifts right because adding a salt always increases ion formation.
4. It shifts left, decreasing $[\text{OH}^-]$. (correct answer)
5. It does not shift because water is a pure liquid.

Explanation: This question tests the common ion effect on weak base equilibria. Adding NH₄Cl increases the concentration of NH₄⁺ ions, which are already present as a product in the ammonia equilibrium. By Le Chatelier's principle, increasing [NH₄⁺] causes the equilibrium to shift left, consuming OH⁻ ions and forming more NH₃. Students who choose C incorrectly think that because water is a pure liquid, the equilibrium cannot shift, but water's constant activity doesn't prevent the equilibrium from responding to changes in ion concentrations. To predict the effect of adding a common ion to a weak base system, identify which side of the equilibrium contains the common ion and apply Le Chatelier's principle.

Question 6

1. The equilibrium shifts right and HF dissociates more.
2. The equilibrium shifts left and HF dissociates less. (correct answer)
3. The equilibrium position does not change because $K_a$ is constant.
4. The equilibrium shifts right because adding ions increases conductivity.
5. The solution becomes buffered, so HF must dissociate more.

Explanation: This question tests understanding of the common ion effect on weak acid dissociation. When NaF is added to the HF solution, it provides F⁻ ions (the common ion) to the equilibrium HF(aq) ⇌ H⁺(aq) + F⁻(aq). Le Chatelier's principle predicts that the increased F⁻ concentration will cause the equilibrium to shift to the left, forming more undissociated HF molecules. This leftward shift means HF dissociates less, resulting in lower concentrations of both H⁺ and F⁻ ions from the HF dissociation. Students might incorrectly choose option A thinking that adding ions always promotes dissociation, but this misconception ignores how the common ion suppresses the ionization of weak acids. When analyzing common ion effects on weak electrolytes, remember that the presence of a common ion always decreases the extent of dissociation.

Question 7

1. The equilibrium shifts right and the solubility of $\text{AgCl}$ increases.
2. The equilibrium shifts left and the solubility of $\text{AgCl}$ decreases. (correct answer)
3. The equilibrium shifts left and the solubility of $\text{AgCl}$ increases.
4. The equilibrium position does not change and the solubility is unchanged.
5. The added chloride acts as a buffer, so the solubility remains the same.

Explanation: This question tests understanding of the common ion effect on the solubility of sparingly soluble salts. When NaCl is added to the saturated AgCl solution, it provides additional Cl⁻ ions (the common ion) to the equilibrium system AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq). Le Chatelier's principle predicts that the equilibrium will shift to the left to reduce the stress of increased Cl⁻ concentration, forming more solid AgCl. This leftward shift decreases the solubility of AgCl because fewer Ag⁺ and Cl⁻ ions remain in solution at the new equilibrium position. Students might incorrectly choose option A thinking that adding a salt always increases solubility, but this ignores the specific effect of the common ion on the equilibrium. When solving common ion problems, always identify the shared ion between the original equilibrium and the added compound, then predict the shift direction.

Question 8

1. The solubility increases because more ions are present to dissolve the solid.
2. The solubility decreases because the equilibrium shifts left. (correct answer)
3. The solubility is unchanged because $K_{sp}$ does not change.
4. The solubility increases because $\text{I}^-$ is a spectator ion.
5. The solubility is unchanged because the added salt creates a buffer.

Explanation: This question tests understanding of the common ion effect on the solubility of sparingly soluble salts. When KI is added to the saturated PbI₂ solution, it provides additional I⁻ ions (the common ion) to the equilibrium PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq). According to Le Chatelier's principle, the increased I⁻ concentration causes the equilibrium to shift to the left, forming more solid PbI₂. This leftward shift decreases the solubility of PbI₂ because less solid dissolves to maintain the Ksp value with the higher I⁻ concentration. Students might incorrectly choose option A thinking that more ions in solution means more dissolution, but this misconception fails to consider the equilibrium constraint imposed by Ksp. To analyze common ion effects on solubility, remember that adding a common ion always decreases the solubility of sparingly soluble salts.

Question 9

1. The solubility does not change because $\text{Na}^+$ is a spectator ion.
2. The solubility decreases because the equilibrium shifts toward $\text{BaSO}_4(s)$. (correct answer)
3. The solubility does not change because $K_{sp}$ prevents any shift.
4. The solubility increases because adding a salt always increases solubility.
5. The solubility increases because sulfate stabilizes $\text{Ba}^{2+}$ in solution.

Explanation: This question tests the common ion effect on the solubility of sparingly soluble salts. Adding Na₂SO₄ increases the concentration of SO₄²⁻ ions, which are already present from the BaSO₄ equilibrium. According to Le Chatelier's principle, the increased [SO₄²⁻] shifts the equilibrium left toward solid BaSO₄, decreasing the solubility of BaSO₄. Students who choose D incorrectly believe that adding any salt increases solubility, but this is only true for salts without common ions; common ions always decrease solubility of sparingly soluble salts. When analyzing solubility equilibria, identify whether the added compound contains a common ion and apply Le Chatelier's principle accordingly.

Question 10

1. $[\text{H}^+]$ increases because the added salt makes the acid dissociate more.
2. $[\text{H}^+]$ decreases because the equilibrium shifts toward $\text{HNO}_2$. (correct answer)
3. $[\text{H}^+]$ is unchanged because common ions do not affect weak-acid equilibria.
4. $[\text{H}^+]$ is unchanged because the solution is buffered so equilibrium cannot shift.
5. $[\text{H}^+]$ increases because $\text{NO}_2^-$ consumes $\text{HNO}_2$.

Explanation: This question tests the common ion effect on weak acid equilibria and its impact on H⁺ concentration. Adding KNO₂ increases the concentration of NO₂⁻ ions, which are already present as a product in the HNO₂ equilibrium. By Le Chatelier's principle, the increased [NO₂⁻] causes the equilibrium to shift left toward HNO₂, consuming H⁺ ions and thus decreasing [H⁺]. Students who choose A incorrectly think that adding a salt of the conjugate base makes the acid dissociate more, but the opposite occurs due to the common ion effect. To analyze how common ions affect H⁺ concentration, remember that adding the conjugate base always suppresses acid ionization and reduces [H⁺].

Question 11

1. The equilibrium shifts to the right to produce more $\text{H}^+$.
2. The equilibrium shifts to the left, forming more $\text{HC}_2\text{H}_3\text{O}_2$. (correct answer)
3. The equilibrium position is unchanged because buffers prevent any equilibrium shift.
4. The equilibrium shifts to the right because adding a salt increases ionization.
5. The equilibrium position is unchanged because $K_a$ depends only on concentration.

Explanation: This question tests the common ion effect on weak acid equilibria. Adding sodium acetate increases the concentration of acetate ions (C₂H₃O₂⁻), which are already present as a product in the acetic acid equilibrium. By Le Chatelier's principle, increasing the concentration of a product causes the equilibrium to shift left, forming more undissociated HC₂H₃O₂ and consuming H⁺ ions. Students who choose C incorrectly believe that buffers prevent any equilibrium shift, but buffers work precisely because equilibria do shift to resist pH changes. To analyze common ion effects on acid-base equilibria, identify the common ion and apply Le Chatelier's principle to determine the direction of the shift.

Question 12

1. The solubility increases because sulfate stabilizes $\text{Ba}^{2+}$ in solution.
2. The solubility decreases because the equilibrium shifts toward $\text{BaSO}_4(s)$. (correct answer)
3. The solubility does not change because $\text{Na}^+$ is a spectator ion.
4. The solubility increases because adding a salt always increases solubility.
5. The solubility does not change because $K_{sp}$ prevents any shift.

Explanation: This question tests the common ion effect on the solubility of sparingly soluble salts. Adding Na₂SO₄ increases the concentration of SO₄²⁻ ions, which are already present from the BaSO₄ equilibrium. According to Le Chatelier's principle, the increased [SO₄²⁻] shifts the equilibrium left toward solid BaSO₄, decreasing the solubility of BaSO₄. Students who choose D incorrectly believe that adding any salt increases solubility, but this is only true for salts without common ions; common ions always decrease solubility of sparingly soluble salts. When analyzing solubility equilibria, identify whether the added compound contains a common ion and apply Le Chatelier's principle accordingly.

Question 13

1. $[\text{H}^+]$ increases because the added salt makes the acid dissociate more.
2. $[\text{H}^+]$ decreases because the equilibrium shifts toward $\text{HNO}_2$. (correct answer)
3. $[\text{H}^+]$ is unchanged because common ions do not affect weak-acid equilibria.
4. $[\text{H}^+]$ is unchanged because the solution is buffered so equilibrium cannot shift.
5. $[\text{H}^+]$ increases because $\text{NO}_2^-$ consumes $\text{HNO}_2$.

Explanation: This question tests the common ion effect on weak acid equilibria and its impact on H⁺ concentration. Adding KNO₂ increases the concentration of NO₂⁻ ions, which are already present as a product in the HNO₂ equilibrium. By Le Chatelier's principle, the increased [NO₂⁻] causes the equilibrium to shift left toward HNO₂, consuming H⁺ ions and thus decreasing [H⁺]. Students who choose A incorrectly think that adding a salt of the conjugate base makes the acid dissociate more, but the opposite occurs due to the common ion effect. To analyze how common ions affect H⁺ concentration, remember that adding the conjugate base always suppresses acid ionization and reduces [H⁺].

Question 14

1. More $\text{PbI}_2$ dissolves because adding $\text{I}^-$ increases the total ion concentration.
2. Less $\text{PbI}_2$ dissolves because the equilibrium shifts toward solid $\text{PbI}_2$. (correct answer)
3. The amount dissolved is unchanged because the solid does not appear in $K_{sp}$.
4. More $\text{PbI}_2$ dissolves because $\text{I}^-$ acts as a buffer for $\text{Pb}^{2+}$.
5. The equilibrium shifts right, but less $\text{PbI}_2$ dissolves due to ion pairing.

Explanation: This question tests the common-ion effect on the solubility of salts with non-1:1 stoichiometry like PbI2. Adding a soluble iodide salt like KI introduces I- ions, a common ion in the equilibrium PbI2(s) ⇌ Pb2+(aq) + 2I-(aq). Le Chatelier's principle predicts a shift to the left due to the excess I-, forming more solid PbI2 to reduce the I- concentration. Therefore, less PbI2 remains dissolved at equilibrium. A tempting distractor is choice A, which claims more PbI2 dissolves due to increased ion concentration, based on the misconception that adding ions generally increases solubility rather than decreasing it via the common-ion effect. For solubility equilibria, identify the common ion and use Le Chatelier's principle to determine if solubility decreases.

Question 15

1. It does not shift because $\text{HCO}_3^-$ is part of a buffer.
2. It shifts right because adding ions increases dissociation of weak acids.
3. It does not shift because $K_a$ depends on the amount of added salt.
4. It shifts left, producing more $\text{H}_2\text{CO}_3$. (correct answer)
5. It shifts right, producing more $\text{H}^+$.

Explanation: This question tests the common ion effect on weak acid equilibria. Adding NaHCO₃ increases the concentration of HCO₃⁻ ions, which are already present as a product in the carbonic acid equilibrium. By Le Chatelier's principle, increasing [HCO₃⁻] causes the equilibrium to shift left, forming more H₂CO₃ and consuming H⁺ ions. Students who choose C incorrectly think that buffer components prevent equilibrium shifts, but buffers resist pH changes precisely because the equilibrium does shift in response to added acids or bases. To predict common ion effects, always identify which side of the equilibrium contains the common ion and apply Le Chatelier's principle to determine the shift direction.

Question 16

1. It increases because $\text{OH}^-$ increases the rate of dissolution.
2. It decreases because the equilibrium shifts toward $\text{Mg(OH)}_2(s)$. (correct answer)
3. It stays the same because strong bases do not affect $K_{sp}$.
4. It increases because $\text{NaOH}$ buffers the hydroxide concentration.
5. It stays the same because adding solute cannot change a saturated solution.

Explanation: This question tests the common ion effect on the solubility of metal hydroxides. Adding NaOH increases the concentration of OH⁻ ions, which are already present from the Mg(OH)₂ equilibrium. According to Le Chatelier's principle, the increased [OH⁻] shifts the equilibrium left toward solid Mg(OH)₂, decreasing the solubility of Mg(OH)₂. Students who choose E incorrectly believe that saturated solutions cannot change, but adding a common ion changes the equilibrium position even though the solution remains saturated with respect to the solid. When a common ion is added to a hydroxide equilibrium, the solubility of the metal hydroxide always decreases due to the equilibrium shift.

Question 17

1. HF dissociates more because added ions increase the rate of ionization.
2. HF dissociates less because the equilibrium shifts toward reactants. (correct answer)
3. HF dissociation is unchanged because $K_a$ does not change with concentration.
4. HF dissociates more because $\text{F}^-$ removes $\text{H}_3\text{O}^+$ from solution.
5. HF dissociation is unchanged because NaF creates a buffer that prevents any shift.

Explanation: This question tests the common-ion effect on the dissociation of weak acids like HF. Adding a soluble fluoride salt like NaF introduces F- ions, a common ion in the equilibrium HF(aq) + H2O(l) ⇌ H3O+(aq) + F-(aq). According to Le Chatelier's principle, the increased F- concentration shifts the equilibrium to the left, reforming more HF to consume the excess F-. This reduces the extent of HF dissociation. A tempting distractor is choice C, which says dissociation is unchanged because Ka is constant, incorrectly assuming constant Ka prevents equilibrium shifts despite concentration perturbations. In weak acid problems, look for common ions and apply Le Chatelier's principle to predict changes in dissociation extent.

Question 18

1. The equilibrium position does not change because $K_a$ is constant.
2. The equilibrium shifts right because acetate is a base and consumes $\text{H}_3\text{O}^+$.
3. The equilibrium shifts left only if the solution is already buffered; otherwise no shift occurs.
4. The equilibrium shifts left because $\text{CH}_3\text{COO}^-$ is a common ion. (correct answer)
5. The equilibrium shifts right because adding ions always increases dissociation.

Explanation: This question tests the common-ion effect on the dissociation of weak acids. Adding solid sodium acetate introduces CH3COO- ions, which are a common ion in the equilibrium CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO-(aq). Le Chatelier's principle dictates that the increased CH3COO- concentration shifts the equilibrium to the left to reduce the excess CH3COO-, decreasing the dissociation of CH3COOH. This results in fewer H3O+ and CH3COO- ions from the acid at the new equilibrium. A tempting distractor is choice D, which wrongly claims the equilibrium shifts right because acetate consumes H3O+, misunderstanding that acetate is a weak base but here acts primarily as a common ion suppressing dissociation. For such equilibria, always check for common ions and apply Le Chatelier's principle to determine the direction of the shift.

Question 19

1. The amount dissolved is unchanged because $\text{BaSO}_4$ is a solid.
2. Less $\text{BaSO}_4$ dissolves because the equilibrium shifts toward the solid. (correct answer)
3. More $\text{BaSO}_4$ dissolves because $\text{SO}_4^{2-}$ acts as a buffer for $\text{Ba}^{2+}$.
4. The equilibrium shifts right, but less dissolves because the ions cancel in $K_{sp}$.
5. More $\text{BaSO}_4$ dissolves because the added sulfate increases total solute.

Explanation: This question tests the common-ion effect on the solubility of 1:1 salts like BaSO4. Adding a soluble sulfate salt like Na2SO4 introduces SO4^2- ions, a common ion in the equilibrium BaSO4(s) ⇌ Ba2+(aq) + SO4^2-(aq). Le Chatelier's principle predicts a leftward shift to form more solid BaSO4, consuming the excess SO4^2-. Thus, less BaSO4 remains dissolved at equilibrium. A tempting distractor is choice A, which wrongly suggests more dissolves due to increased solute, confusing total concentration with the suppressive effect of common ions on solubility. When adding salts to saturated solutions, identify common ions and use Le Chatelier's principle to forecast solubility changes.

Question 20

1. The equilibrium shifts right because adding a salt always increases ion formation.
2. The equilibrium position does not change because $K_b$ is constant.
3. The equilibrium shifts right because $\text{Cl}^-$ is a strong base.
4. The equilibrium shifts left because adding $\text{NH}_4^+$ is a common-ion addition. (correct answer)
5. The equilibrium shifts left only if a buffer is already present; otherwise it is unchanged.

Explanation: This question tests the common-ion effect on the equilibrium of weak bases. Adding ammonium chloride introduces NH4+ ions, a common ion in the equilibrium NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq). Le Chatelier's principle indicates that the increased NH4+ concentration shifts the equilibrium to the left to consume the excess NH4+, reducing the production of OH-. This decreases the extent of NH3 dissociation at the new equilibrium. A tempting distractor is choice C, which states the equilibrium position does not change because Kb is constant, mistakenly confusing the constancy of Kb with no shift in position despite concentration changes. To solve these problems, recognize common ions and apply Le Chatelier's principle to anticipate equilibrium adjustments.