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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

Which of the following best describes the common-ion effect?

All Questions (7)

Which of the following best describes the common-ion effect?

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'.

In which of the following aqueous solutions would solid lead(II) chloride, PbCl₂, be least soluble?

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.

Consider the dissolution equilibrium for barium sulfate: BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq). According to Le Châtelier’s principle, what is the effect of adding a small amount of sodium sulfate (Na₂SO₄) to a saturated solution of BaSO₄?

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.

The Ksp for silver chromate, Ag₂CrO₄, is 1.1 x 10⁻¹². If solid Ag₂CrO₄ is dissolved in a 0.10 M solution of silver nitrate (AgNO₃), which expression correctly represents the Ksp relationship at equilibrium, where 's' is the molar solubility of Ag₂CrO₄?

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).

The solubility of calcium carbonate, CaCO₃, is significantly reduced when it is dissolved in a solution of calcium chloride, CaCl₂. This phenomenon is best identified as:

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 sparingly soluble salt, MX₂, is added to a 0.050 M solution of NaX. Given the dissolution equilibrium MX₂(s) ⇌ M²⁺(aq) + 2X⁻(aq), which statement correctly describes the molar solubility, 's', of MX₂ in this solution?

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.

The common-ion effect is a practical application of which fundamental chemical principle?

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.