chemistry : class11 Le Chatelier Applications : Common ion effect : Equilibrium MCQ Questions and Answers
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Le Chatelier Applications
MCQ questions and answers.
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MCQ Questions
1. A saturated solution of AgCl is treated separately with equal concentrations of NaCl and KNO3. In which solution will the molar solubility of AgCl decrease, and why?
A.
It decreases in NaCl solution because Cl⁻ shifts AgCl(s) ⇌ Ag⁺ + Cl⁻ to the left
B.
It decreases in KNO3 solution because K⁺ combines with Ag⁺
C.
It decreases equally in both solutions because both salts increase ionic strength
D.
It increases in NaCl solution because Cl⁻ promotes ionization of AgCl
Correct Answer:
It decreases in NaCl solution because Cl⁻ shifts AgCl(s) ⇌ Ag⁺ + Cl⁻ to the left
Explanation:
NaCl supplies Cl⁻, which is a common ion in the dissolution equilibrium of AgCl. According to Le Chatelier’s principle, adding Cl⁻ shifts the equilibrium toward undissolved AgCl, decreasing its solubility. KNO3 provides no common ion.
2. A saturated solution of CaF₂ is at equilibrium according to CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq). If NaF is added without changing the temperature, which change is expected initially?
A.
The concentration of Ca²⁺ increases because the dissolution equilibrium shifts right
B.
The concentration of F⁻ decreases because Na⁺ removes fluoride ions
C.
The equilibrium shifts left and the solubility of CaF₂ decreases
D.
The equilibrium remains unchanged because NaF is completely soluble
Correct Answer:
The equilibrium shifts left and the solubility of CaF₂ decreases
Explanation:
NaF increases the concentration of F⁻, a common ion in the CaF₂ dissolution equilibrium. The system responds by shifting left, causing precipitation of CaF₂ and reducing its molar solubility.
3. For the equilibrium CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq), sodium acetate is added to a solution of acetic acid. Which statement correctly describes the Le Chatelier effect?
A.
Ionization of acetic acid increases and the concentration of H⁺ increases
B.
Ionization of acetic acid decreases and the concentration of H⁺ decreases relative to the original solution
C.
Ionization remains unchanged because sodium acetate is a strong electrolyte
D.
The equilibrium shifts right because Na⁺ reacts with acetate ions
Correct Answer:
Ionization of acetic acid decreases and the concentration of H⁺ decreases relative to the original solution
Explanation:
Sodium acetate dissociates to provide CH₃COO⁻, a common ion. Its addition shifts the acetic acid ionization equilibrium to the left, suppressing ionization and lowering the H⁺ concentration compared with acetic acid alone.
4. A solution contains NH₄OH in equilibrium with its ions: NH₄OH(aq) ⇌ NH₄⁺(aq) + OH⁻(aq). When NH₄Cl is added, which observation is most consistent with Le Chatelier’s principle?
A.
The concentration of OH⁻ increases because NH₄Cl is soluble
B.
The ionization of NH₄OH is suppressed and the concentration of OH⁻ decreases
C.
The concentration of NH₄⁺ decreases because it reacts completely with Cl⁻
D.
The equilibrium shifts right because NH₄Cl supplies an electrolyte
Correct Answer:
The ionization of NH₄OH is suppressed and the concentration of OH⁻ decreases
Explanation:
NH₄Cl dissociates to produce NH₄⁺, which is a common ion in the weak-base equilibrium. The increased NH₄⁺ concentration shifts the equilibrium left, decreasing NH₄OH ionization and OH⁻ concentration.
5. Hydrogen sulfide ionizes in water through the equilibrium H₂S ⇌ 2H⁺ + S²⁻. During qualitative analysis, passing H₂S through an acidic solution reduces the concentration of S²⁻. What is the direct Le Chatelier explanation?
A.
H⁺ is a common ion and shifts the ionization equilibrium toward H₂S
B.
H⁺ reacts with H₂S and forces complete dissociation
C.
The acid increases S²⁻ concentration by neutralizing sulfide ions
D.
The equilibrium shifts right because the solution contains more ions
Correct Answer:
H⁺ is a common ion and shifts the ionization equilibrium toward H₂S
Explanation:
Acid addition increases H⁺, a product of H₂S ionization. The equilibrium therefore shifts left, strongly reducing S²⁻ concentration. This is a Le Chatelier application used to control sulfide precipitation in qualitative analysis.
6. Why does adding NH₄Cl help prevent the precipitation of Mg(OH)₂ from a solution containing Mg²⁺ and NH₄OH?
A.
NH₄Cl supplies OH⁻, causing Mg(OH)₂ to dissolve completely
B.
NH₄⁺ suppresses NH₄OH ionization, lowering OH⁻ concentration and shifting Mg(OH)₂ dissolution toward the solid
C.
Cl⁻ removes Mg²⁺ by forming an insoluble magnesium chloride precipitate
D.
NH₄Cl increases NH₄OH ionization by removing NH₄⁺ from solution
Correct Answer:
NH₄⁺ suppresses NH₄OH ionization, lowering OH⁻ concentration and shifting Mg(OH)₂ dissolution toward the solid
Explanation:
NH₄Cl provides the common ion NH₄⁺, which suppresses the ionization of NH₄OH and lowers OH⁻ concentration. With less OH⁻ available, the ionic product for Mg(OH)₂ remains below the precipitation condition, so Mg(OH)₂ is prevented from precipitating.
7. A saturated solution of CaCO₃ is divided into two portions. Na₂CO₃ is added to one portion and NaNO₃ to the other. Which comparison is correct?
A.
CaCO₃ solubility decreases only in the Na₂CO₃-treated portion
B.
CaCO₃ solubility decreases only in the NaNO₃-treated portion because nitrate is highly soluble
C.
CaCO₃ solubility increases only in the Na₂CO₃-treated portion because carbonate promotes dissolution
D.
CaCO₃ solubility changes equally in both portions because both salts contain sodium
Correct Answer:
CaCO₃ solubility decreases only in the Na₂CO₃-treated portion
Explanation:
The dissolution equilibrium is CaCO₃(s) ⇌ Ca²⁺ + CO₃²⁻. Na₂CO₃ adds the common ion CO₃²⁻ and shifts the equilibrium left. NaNO₃ does not provide a common ion, so it does not produce this common-ion suppression.
8. A sparingly soluble salt MX dissolves according to MX(s) ⇌ M⁺(aq) + X⁻(aq). If a soluble salt NX, which provides X⁻ but does not react otherwise, is added, which combination of changes is expected after re-equilibration?
A.
The concentration of M⁺ decreases, the concentration of X⁻ increases, and some MX precipitates
B.
The concentration of M⁺ increases, the concentration of X⁻ decreases, and MX dissolves
C.
Both M⁺ and X⁻ concentrations decrease because NX removes both ions
D.
Both M⁺ and X⁻ concentrations increase because the dissolution equilibrium shifts right
Correct Answer:
The concentration of M⁺ decreases, the concentration of X⁻ increases, and some MX precipitates
Explanation:
Added X⁻ is a common ion and shifts MX(s) ⇌ M⁺ + X⁻ to the left. Consequently, some MX precipitates, lowering the dissolved M⁺ concentration. Although X⁻ is consumed partly by precipitation, its final concentration is higher because it was added in excess of the amount removed.
9. In a saturated Ag₂CrO₄ solution, the equilibrium is Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq). If AgNO₃ is added, which statement best describes the equilibrium response?
A.
The equilibrium shifts right because nitrate reacts with chromate
B.
The equilibrium shifts left because Ag⁺ is a common ion, causing Ag₂CrO₄ precipitation
C.
The equilibrium remains unchanged because AgNO₃ is a strong electrolyte
D.
The equilibrium shifts left because NO₃⁻ is a common ion in the dissolution equilibrium
Correct Answer:
The equilibrium shifts left because Ag⁺ is a common ion, causing Ag₂CrO₄ precipitation
Explanation:
AgNO₃ increases the concentration of Ag⁺, which is a product of Ag₂CrO₄ dissolution. Le Chatelier’s principle predicts a shift to the left, resulting in precipitation and a decrease in the molar solubility of Ag₂CrO₄.
10. A saturated solution of PbI₂ is treated with KI. The dissolution equilibrium is PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq). Which explanation is correct if the amount of KI is small and no complex-ion formation is considered?
A.
I⁻ acts as a common ion, shifting the equilibrium left and decreasing the solubility of PbI₂
B.
K⁺ acts as a common ion, shifting the equilibrium left and decreasing the solubility of PbI₂
C.
I⁻ shifts the equilibrium right because it increases the number of dissolved ions
D.
KI has no effect because PbI₂ is already at saturation
Correct Answer:
I⁻ acts as a common ion, shifting the equilibrium left and decreasing the solubility of PbI₂
Explanation:
KI supplies I⁻, one of the ions produced when PbI₂ dissolves. The increased iodide concentration shifts the dissolution equilibrium toward solid PbI₂, so precipitation occurs and its solubility decreases, provided complex formation is neglected.
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