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 stays the same because adding a common ion does not affect $K_{sp}$.
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 $\text{CaF}_2(s)$.
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.
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?
$[\text{H}^+]$ is unchanged because common ions do not affect weak-acid equilibria.
$[\text{H}^+]$ decreases because the equilibrium shifts toward $\text{HNO}_2$.
$[\text{H}^+]$ increases because $\text{NO}_2^-$ consumes $\text{HNO}_2$.
$[\text{H}^+]$ increases because the added salt makes the acid dissociate more.
$[\text{H}^+]$ is unchanged because the solution is buffered so equilibrium cannot shift.
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⁺].
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 increases because adding a salt always increases solubility.
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 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.
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 does not shift because $\text{NH}_4^+$ makes the solution a buffer.
It shifts right, increasing $[\text{OH}^-]$.
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.
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?
The equilibrium position is unchanged because $K_a$ depends only on concentration.
The equilibrium shifts to the right because adding a salt increases ionization.
The equilibrium position is unchanged because buffers prevent any equilibrium shift.
The equilibrium shifts to the right to produce more $\text{H}^+$.
The equilibrium shifts to the left, forming more $\text{HC}_2\text{H}_3\text{O}_2$.
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.
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 decreases because the equilibrium shifts toward $\text{BaSO}_4(s)$.
The solubility increases because adding a salt always increases solubility.
The solubility does not change because $K_{sp}$ prevents any shift.
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?
$[\text{H}^+]$ increases because the added salt makes the acid dissociate more.
$[\text{H}^+]$ is unchanged because the solution is buffered so equilibrium cannot shift.
$[\text{H}^+]$ increases because $\text{NO}_2^-$ consumes $\text{HNO}_2$.
$[\text{H}^+]$ is unchanged because common ions do not affect weak-acid equilibria.
$[\text{H}^+]$ decreases because the equilibrium shifts toward $\text{HNO}_2$.
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⁺].
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?
The equilibrium shifts right, but less $\text{PbI}_2$ dissolves due to ion pairing.
Less $\text{PbI}_2$ dissolves because the equilibrium shifts toward solid $\text{PbI}_2$.
The amount dissolved is unchanged because the solid does not appear in $K_{sp}$.
More $\text{PbI}_2$ dissolves because $\text{I}^-$ acts as a buffer for $\text{Pb}^{2+}$.
More $\text{PbI}_2$ dissolves because adding $\text{I}^-$ increases the total ion concentration.
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.
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?
It shifts right, producing more $\text{H}^+$.
It does not shift because $\text{HCO}_3^-$ is part of a buffer.
It shifts right because adding ions increases dissociation of weak acids.
It does not shift because $K_a$ depends on the amount of added salt.
It shifts left, producing more $\text{H}_2\text{CO}_3$.
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.
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?
It increases because $\text{OH}^-$ increases the rate of dissolution.
It increases because $\text{NaOH}$ buffers the hydroxide concentration.
It decreases because the equilibrium shifts toward $\text{Mg(OH)}_2(s)$.
It stays the same because adding solute cannot change a saturated solution.
It stays the same because strong bases do not affect $K_{sp}$.
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.