AP Chemistry Practice Quiz: Common-Ion Effect
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 7 questions to check your progress.
Question 1 of 7
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A) The reduction in the solubility of a salt when it is dissolved in a solution that already contains one of the ions from the salt.
B) The increase in the solubility of a salt when it is dissolved in a solution that already contains one of the ions from the salt.
C) The phenomenon where two different salts with a shared ion precipitate simultaneously regardless of their Ksp values.
D) A change in the Ksp value of a salt due to the presence of an additional, non-reacting salt in the solution.
Correct Answer: A
The provided content explicitly states that the solubility of a salt is reduced when dissolved into a solution that already contains one of its ions, and this is known as the 'common-ion effect'.
A) 0.1 M NaNO₃
B) 0.1 M CaCl₂
C) 0.1 M K₂SO₄
D) Pure water
Correct Answer: B
The solubility of PbCl₂ is reduced by the common-ion effect. The CaCl₂ solution contains the common ion Cl⁻. According to Le Châtelier's principle, the presence of Cl⁻ from CaCl₂ will shift the dissolution equilibrium PbCl₂(s) ⇌ Pb²⁺(aq) + 2Cl⁻(aq) to the left, decreasing the solubility of PbCl₂.
A) The equilibrium will shift to the left, decreasing the concentration of Ba²⁺ ions.
B) The equilibrium will shift to the right, increasing the concentration of Ba²⁺ ions.
C) The equilibrium will not shift, but the Ksp value for BaSO₄ will decrease.
D) The equilibrium will shift to the right, causing more BaSO₄ to dissolve.
Correct Answer: A
Adding Na₂SO₄ increases the concentration of the common ion, SO₄²⁻. Le Châtelier's principle states that the system will counteract this change by shifting the equilibrium to the left, which favors the formation of solid BaSO₄. This shift consumes both Ba²⁺ and SO₄²⁻ ions, thus decreasing the concentration of Ba²⁺ and reducing the overall solubility of the salt.
A) Ksp = (s)(2s)²
B) Ksp = (2s)²(0.10 + s)
C) Ksp = (0.10 + 2s)²(s)
D) Ksp = (0.10 + s)(2s)
Correct Answer: C
The dissolution equilibrium is Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq). The initial concentration of the common ion Ag⁺ from AgNO₃ is 0.10 M. At equilibrium, the concentration of CrO₄²⁻ will be 's', and the concentration of Ag⁺ will be the initial 0.10 M plus the 2s from the dissolved Ag₂CrO₄. Therefore, [Ag⁺] = 0.10 + 2s and [CrO₄²⁻] = s. The Ksp expression is Ksp = [Ag⁺]²[CrO₄²⁻], which becomes Ksp = (0.10 + 2s)²(s).
A) The common-ion effect
B) A change in the Ksp value
C) A complex ion formation
D) A hydrolysis reaction
Correct Answer: A
The provided text defines the reduction in solubility of a salt due to the presence of one of its ions in the solution as the 'common-ion effect'. Here, the Ca²⁺ ion from CaCl₂ is the common ion that reduces the solubility of CaCO₃.
A) The concentration of M²⁺ at equilibrium will be greater than it would be in pure water.
B) The value of Ksp for MX₂ will be smaller in the NaX solution than in pure water.
C) The concentration of M²⁺ at equilibrium can be calculated based on the Ksp and the initial concentration of the common ion X⁻.
D) The molar solubility 's' will be equal to half the concentration of X⁻ at equilibrium.
Correct Answer: C
The provided content states that the solubility of a salt (and/or the Ksp) can be calculated based on the concentration of a common ion. The presence of the common ion X⁻ from NaX reduces the solubility of MX₂. The new, lower solubility ('s', which equals the equilibrium concentration of M²⁺) can be found by setting up the Ksp expression: Ksp = [M²⁺][X⁻]² = (s)(0.050 + 2s)². This demonstrates that the solubility is calculated from Ksp and the common ion concentration.
A) Hess's Law
B) The Law of Conservation of Mass
C) Le Châtelier’s Principle
D) The First Law of Thermodynamics
Correct Answer: C
The provided text explicitly states that the common-ion effect 'can be understood qualitatively using Le Châtelier’s principle'. Adding a common ion is like adding a product to a dissolution equilibrium, causing the equilibrium to shift away from the products and toward the reactants (the solid salt), thus reducing solubility.