Common-Ion Effect - AP Chemistry

1. A buffer is prepared from acetic acid and acetate: $\text{HC}_2\text{H}_3\text{O}_2(aq) \rightleftharpoons \text{H}^+(aq)+\text{C}_2\text{H}_3\text{O}_2^-(aq)$. A small amount of sodium acetate is added. Qualitatively, what happens to the position of this equilibrium?​
   Answer: The equilibrium shifts to the left, forming more $\text{HC}_2\text{H}_3\text{O}_2$. 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.
2. Ammonia establishes the equilibrium $\text{NH}_3(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_4^+(aq) + \text{OH}^-(aq)$. A student adds a soluble ammonium salt (such as $\text{NH}_4\text{Cl}$) to the solution at constant temperature. How does the addition affect the equilibrium position and $[\text{OH}^-]$?
   Answer: The equilibrium shifts left and $[\text{OH}^-]$ decreases. This question tests understanding of the common ion effect on weak base equilibria. When NH₄Cl is added to the ammonia solution, it dissociates to provide NH₄⁺ ions (the common ion) to the equilibrium NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq). Le Chatelier's principle predicts that the increased NH₄⁺ concentration will cause the equilibrium to shift to the left, consuming some OH⁻ ions and forming more NH₃ molecules. This leftward shift decreases [OH⁻], making the solution less basic. Students might incorrectly choose option A thinking that adding more reactants always shifts equilibrium to the right, but this misconception fails to recognize that NH₄⁺ is a product, not a reactant. When analyzing common ion effects, always identify whether the common ion appears on the reactant or product side of the equilibrium.
3. A saturated solution of $\text{PbI}_2(s)$ is established: $\text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq)+2\text{I}^-(aq)$. A small amount of $\text{Pb(NO}_3)_2(aq)$ is added. At constant temperature, what happens to the solubility of $\text{PbI}_2$?​
   Answer: The solubility decreases because the equilibrium shifts toward $\text{PbI}_2(s)$. This question tests the common ion effect on the solubility of sparingly soluble salts. Adding Pb(NO₃)₂ increases the concentration of Pb²⁺ ions, which are already present from the PbI₂ equilibrium. According to Le Chatelier's principle, the increased [Pb²⁺] shifts the equilibrium left toward solid PbI₂, decreasing the solubility of PbI₂. Students who choose C incorrectly think that constant Ksp means constant solubility, but solubility depends on the concentrations of all ions in solution, not just the Ksp value. When analyzing solubility changes, remember that adding a common ion always decreases the solubility of the sparingly soluble salt through Le Chatelier's principle.
4. A saturated solution of $\text{BaSO}_4(s)$ is in equilibrium: $\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq)+\text{SO}_4^{2-}(aq)$. A small amount of $\text{Na}_2\text{SO}_4(aq)$ is added. At constant temperature, what happens to the solubility of $\text{BaSO}_4$?
   Answer: The solubility decreases because the equilibrium shifts toward $\text{BaSO}_4(s)$. 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.
5. A saturated solution of barium sulfate is at equilibrium: $\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq)$. Excess solid is present. A student adds a soluble sulfate salt (such as $\text{Na}_2\text{SO}_4$) at constant temperature. Which statement best describes the effect on solubility?
   Answer: The solubility decreases because the equilibrium shifts left. This question tests understanding of the common ion effect on the solubility of sparingly soluble salts. When Na₂SO₄ is added to the saturated BaSO₄ solution, it provides additional SO₄²⁻ ions (the common ion) to the equilibrium BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq). According to Le Chatelier's principle, the increased SO₄²⁻ concentration causes the equilibrium to shift to the left, forming more solid BaSO₄. This leftward shift decreases the solubility of BaSO₄ because less solid dissolves to maintain the constant Ksp value. Students might incorrectly choose option D thinking that adding ions always increases solubility, but this misconception ignores how the common ion suppresses dissolution. To solve common ion problems, remember that the solubility of a sparingly soluble salt always decreases when a common ion is added.
6. A saturated solution of $\text{Mg(OH)}_2(s)$ is at equilibrium: $\text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+}(aq)+2\text{OH}^-(aq)$. A small amount of $\text{NaOH}(aq)$ is added. How does the solubility of $\text{Mg(OH)}_2$ change?
   Answer: It decreases because the equilibrium shifts toward $\text{Mg(OH)}_2(s)$. 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.
7. A saturated solution of $\text{AgCl}(s)$ is at equilibrium with undissolved solid: $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq)+\text{Cl}^-(aq)$. A small amount of $\text{NaCl}(s)$ is added and dissolves. Assuming temperature is constant and solid $\text{AgCl}$ remains present, how does the solubility of $\text{AgCl}$ change?
   Answer: The solubility decreases because the equilibrium shifts toward $\text{AgCl}(s)$. This question tests the common ion effect on the solubility of sparingly soluble salts. When NaCl dissolves, it increases the concentration of Cl⁻ ions in solution, which are already present from the AgCl equilibrium. According to Le Chatelier's principle, the increased [Cl⁻] causes the equilibrium to shift left toward solid AgCl, reducing the amount of AgCl that can dissolve. Students who choose C incorrectly think that because Ksp is constant, solubility must also be constant, but fail to recognize that solubility depends on the concentrations of all ions present. When solving equilibrium problems involving the common ion effect, always identify which ion is common to both compounds and apply Le Chatelier's principle to predict the shift.
8. In water, the equilibrium $\text{H}_2\text{CO}_3(aq) \rightleftharpoons \text{H}^+(aq)+\text{HCO}_3^-(aq)$ is established. A small amount of $\text{NaHCO}_3(aq)$ is added. What happens to the position of the equilibrium?
   Answer: It shifts left, producing more $\text{H}_2\text{CO}_3$. 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.
9. Carbonate ion hydrolyzes in water according to $\text{CO}_3^{2-}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{HCO}_3^-(aq) + \text{OH}^-(aq)$. A student adds sodium bicarbonate, $\text{NaHCO}_3$, which dissolves completely. What is the effect on the position of this equilibrium?
   Answer: The equilibrium shifts left because $\text{HCO}_3^-$ is a common ion (product). 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.
10. A saturated solution of lead(II) iodide is at equilibrium: $\text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq)$. A student adds a soluble iodide salt (such as KI) to the mixture. Which statement best describes the change in the amount of dissolved $\text{PbI}_2$ at equilibrium?
    Answer: Less $\text{PbI}_2$ dissolves because the equilibrium shifts toward solid $\text{PbI}_2$. 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.
11. A saturated solution of magnesium hydroxide is at equilibrium: $\text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\text{OH}^-(aq)$. A student adds a strong base such as NaOH to the solution. Which statement best describes the effect on the solubility of $\text{Mg(OH)}_2$?
    Answer: The solubility decreases because the equilibrium shifts toward solid $\text{Mg(OH)}_2$. This question tests the common-ion effect on the solubility of metal hydroxides like Mg(OH)2. Adding a strong base like NaOH introduces OH- ions, a common ion in the equilibrium Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq). Le Chatelier's principle states that the excess OH- shifts the equilibrium to the left, forming more solid Mg(OH)2 to reduce the OH- concentration. As a result, the solubility of Mg(OH)2 decreases. A tempting distractor is choice D, which incorrectly asserts solubility increases because OH- drives dissolution forward, misunderstanding that products in equilibrium suppress rather than promote dissociation. To approach solubility in basic solutions, identify common ions and use Le Chatelier's principle for shift predictions.
12. Hydrofluoric acid establishes the equilibrium $\text{HF}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{F}^-(aq)$. A student adds a soluble fluoride salt (such as NaF) to the solution. What happens to the extent of HF dissociation?
    Answer: HF dissociates less because the equilibrium shifts toward reactants. 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.
13. In water, the equilibrium $\text{H}_2\text{CO}_3(aq) \rightleftharpoons \text{H}^+(aq)+\text{HCO}_3^-(aq)$ is established. A small amount of $\text{NaHCO}_3(aq)$ is added. What happens to the position of the equilibrium?​
    Answer: It shifts left, producing more $\text{H}_2\text{CO}_3$. 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.
14. The equilibrium $\text{HNO}_2(aq) \rightleftharpoons \text{H}^+(aq)+\text{NO}_2^-(aq)$ is established in solution. A small amount of $\text{KNO}_2(aq)$ is added. Qualitatively, how is the concentration of $\text{H}^+$ affected after the system reestablishes equilibrium?
    Answer: $[\text{H}^+]$ decreases because the equilibrium shifts toward $\text{HNO}_2$. 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⁺].
15. A saturated solution of $\text{PbI}_2(s)$ is established: $\text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq)+2\text{I}^-(aq)$. A small amount of $\text{Pb(NO}_3)_2(aq)$ is added. At constant temperature, what happens to the solubility of $\text{PbI}_2$?
    Answer: The solubility decreases because the equilibrium shifts toward $\text{PbI}_2(s)$. This question tests the common ion effect on the solubility of sparingly soluble salts. Adding $\text{Pb(NO}_3)_2$ increases the concentration of $\text{Pb}^{2+}$ ions, which are already present from the $\text{PbI}_2$ equilibrium. According to Le Chatelier's principle, the increased $[\text{Pb}^{2+}]$ shifts the equilibrium left toward solid $\text{PbI}_2$, decreasing the solubility of $\text{PbI}_2$. Students who choose C incorrectly think that constant $K_{sp}$ means constant solubility, but solubility depends on the concentrations of all ions in solution, not just the $K_{sp}$ value. When analyzing solubility changes, remember that adding a common ion always decreases the solubility of the sparingly soluble salt through Le Chatelier's principle.
16. A saturated solution of silver chloride is established: $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$. Solid $\text{AgCl}$ remains in the container. A student adds a soluble chloride salt (such as NaCl) while keeping temperature constant. What happens to the equilibrium position and the solubility of $\text{AgCl}$?
    Answer: The equilibrium shifts left and the solubility of $\text{AgCl}$ decreases. 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.
17. An aqueous ammonia system is at equilibrium: $\text{NH}_3(aq)+\text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_4^+(aq)+\text{OH}^-(aq)$. A small amount of $\text{NH}_4\text{Cl}(s)$ is added and dissolves. How does the equilibrium shift?
    Answer: It shifts left, decreasing $[\text{OH}^-]$. 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.
18. An aqueous ammonia system is at equilibrium: $\text{NH}_3(aq)+\text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_4^+(aq)+\text{OH}^-(aq)$. A small amount of $\text{NH}_4\text{Cl}(s)$ is added and dissolves. How does the equilibrium shift?​
    Answer: It shifts left, decreasing $[\text{OH}^-]$. 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.
19. Ammonia establishes the equilibrium $\text{NH}_3(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_4^+(aq) + \text{OH}^-(aq)$. A student adds solid ammonium chloride, $\text{NH}_4\text{Cl}$, which dissolves completely. What is the effect on the equilibrium position?
    Answer: The equilibrium shifts left because adding $\text{NH}_4^+$ is a common-ion addition. 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.
20. A saturated solution of silver chloride is at equilibrium: $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$. A student adds a small amount of soluble sodium chloride, NaCl, and stirs until dissolved. Which statement best describes the effect on $\text{AgCl}$ solubility?
    Answer: The solubility decreases because the equilibrium shifts toward solid $\text{AgCl}$. This question tests the common-ion effect on the solubility of sparingly soluble salts like AgCl. Adding sodium chloride introduces Cl- ions, a common ion in the equilibrium AgCl(s) ⇌ Ag+(aq) + Cl-(aq). By Le Chatelier's principle, the excess Cl- shifts the equilibrium to the left, promoting the formation of solid AgCl to consume the added Cl-. Consequently, the solubility of AgCl decreases as more precipitate forms and less remains dissolved. A tempting distractor is choice D, which incorrectly suggests solubility increases because Cl- removes Ag+, based on the misconception that common ions enhance rather than suppress solubility. When evaluating solubility changes, identify common ions and use Le Chatelier's principle to predict shifts toward undissolved species.

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