Common-Ion Effect
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AP Chemistry › Common-Ion Effect
A saturated solution of $\text{CaF}_2(s)$ is at equilibrium: $\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq)+2\text{F}^-(aq)$. A soluble fluoride salt is added, increasing $\text{F}^-$. At constant temperature, what happens to the amount of $\text{CaF}_2(s)$ that can dissolve?
It stays the same because the solution is already saturated.
It increases because extra ions help pull solid into solution.
It increases because the added salt acts as a buffer for $\text{F}^-$.
It decreases because the equilibrium shifts toward $\text{CaF}_2(s)$.
It stays the same because adding a common ion does not affect $K_{sp}$.
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.
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?
It does not shift because $\text{NH}_4^+$ makes the solution a buffer.
It shifts right because adding a salt always increases ion formation.
It does not shift because water is a pure liquid.
It shifts right, increasing $[\text{OH}^-]$.
It shifts left, decreasing $[\text{OH}^-]$.
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.
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?
The equilibrium shifts right because adding a salt always increases $\text{OH}^-$ production.
The equilibrium shifts right because $\text{HCO}_3^-$ is a strong acid that neutralizes $\text{OH}^-$.
The equilibrium position does not change because $K_b$ is constant.
The equilibrium is unchanged because adding $\text{HCO}_3^-$ only creates a buffer, not a shift.
The equilibrium shifts left because $\text{HCO}_3^-$ is a common ion (product).
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.
The weak acid equilibrium is established in water: $\text{HF}(aq) \rightleftharpoons \text{H}^+(aq)+\text{F}^-(aq)$. A small amount of $\text{NaF}(s)$ is added and dissolves. Qualitatively, what happens to the ionization of HF?
HF ionizes more because $\text{NaF}$ acts as a buffer that forces dissociation.
HF ionizes less because the equilibrium shifts toward HF.
HF ionization is unchanged because the added ion is a spectator.
HF ionizes more because adding $\text{F}^-$ increases conductivity.
HF ionization is unchanged because $K_a$ changes when salt is added.
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.
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?
It shifts right because adding a salt always increases ion formation.
It shifts left, decreasing $[\text{OH}^-]$.
It shifts right, increasing $[\text{OH}^-]$.
It does not shift because $\text{NH}_4^+$ makes the solution a buffer.
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.
Hydrofluoric acid establishes the equilibrium $\text{HF}(aq) \rightleftharpoons \text{H}^+(aq) + \text{F}^-(aq)$. A student adds a soluble fluoride salt (such as NaF) to the solution at constant temperature. Which statement best describes the effect on the equilibrium position and the extent of HF dissociation?
The solution becomes buffered, so HF must dissociate more.
The equilibrium shifts right and HF dissociates more.
The equilibrium position does not change because $K_a$ is constant.
The equilibrium shifts left and HF dissociates less.
The equilibrium shifts right because adding ions increases conductivity.
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.
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 $\text{AgCl}$?
The equilibrium shifts left and the solubility of $\text{AgCl}$ increases.
The equilibrium shifts right and the solubility of $\text{AgCl}$ increases.
The equilibrium shifts left and the solubility of $\text{AgCl}$ decreases.
The equilibrium position does not change and the solubility is unchanged.
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.
A saturated solution of lead(II) iodide is at equilibrium: $\text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq)$. Excess solid remains. A student adds a soluble iodide salt (such as KI) at constant temperature. What happens to the solubility of $\text{PbI}_2$?
The solubility is unchanged because $K_{sp}$ does not change.
The solubility increases because $\text{I}^-$ is a spectator ion.
The solubility is unchanged because the added salt creates a buffer.
The solubility increases because more ions are present to dissolve the solid.
The solubility decreases because the equilibrium shifts left.
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.
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 $\text{BaSO}_4$?
The solubility does not change because $\text{Na}^+$ is a spectator ion.
The solubility does not change because $K_{sp}$ prevents any shift.
The solubility increases because adding a salt always increases solubility.
The solubility decreases because the equilibrium shifts toward $\text{BaSO}_4(s)$.
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.
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?
$[\text{H}^+]$ decreases because the equilibrium shifts toward $\text{HNO}_2$.
$[\text{H}^+]$ increases because $\text{NO}_2^-$ consumes $\text{HNO}_2$.
$[\text{H}^+]$ is unchanged because common ions do not affect weak-acid equilibria.
$[\text{H}^+]$ is unchanged because the solution is buffered so equilibrium cannot shift.
$[\text{H}^+]$ increases because the added salt makes the acid dissociate more.
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⁺].