chemistry : class11 Common Ion Principle : Common ion effect : Equilibrium MCQ Questions and Answers
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Common Ion Principle
MCQ questions and answers.
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MCQ Questions
1. A solution contains 0.100 mol L⁻¹ acetic acid and 0.100 mol L⁻¹ sodium acetate. If Ka for acetic acid is 1.8 × 10⁻⁵, what is the approximate hydrogen-ion concentration?
A.
1.8 × 10⁻⁵ mol L⁻¹
B.
1.8 × 10⁻³ mol L⁻¹
C.
1.0 × 10⁻⁷ mol L⁻¹
D.
5.6 × 10⁻¹⁰ mol L⁻¹
Correct Answer:
1.8 × 10⁻⁵ mol L⁻¹
Explanation:
For CH₃COOH ⇌ H⁺ + CH₃COO⁻, Ka = [H⁺][CH₃COO⁻]/[CH₃COOH]. Since equal concentrations of acid and its common ion are present, [H⁺] ≈ Ka = 1.8 × 10⁻⁵ mol L⁻¹.
2. The solubility product of AgCl is 1.8 × 10⁻¹⁰. What is the approximate molar solubility of AgCl in 0.010 mol L⁻¹ NaCl, assuming the chloride concentration remains essentially constant?
A.
1.8 × 10⁻⁸ mol L⁻¹
B.
1.8 × 10⁻⁶ mol L⁻¹
C.
1.8 × 10⁻¹⁰ mol L⁻¹
D.
1.0 × 10⁻⁵ mol L⁻¹
Correct Answer:
1.8 × 10⁻⁸ mol L⁻¹
Explanation:
For AgCl(s) ⇌ Ag⁺ + Cl⁻, Ksp = [Ag⁺][Cl⁻]. In the presence of the common ion Cl⁻ at 0.010 mol L⁻¹, solubility s satisfies 1.8 × 10⁻¹⁰ = s(0.010), giving s = 1.8 × 10⁻⁸ mol L⁻¹.
3. CaF₂ has Ksp = 3.9 × 10⁻¹¹. What is its approximate molar solubility in 0.100 mol L⁻¹ NaF?
A.
3.9 × 10⁻⁹ mol L⁻¹
B.
3.9 × 10⁻⁶ mol L⁻¹
C.
1.95 × 10⁻⁹ mol L⁻¹
D.
3.9 × 10⁻¹¹ mol L⁻¹
Correct Answer:
3.9 × 10⁻⁹ mol L⁻¹
Explanation:
For CaF₂(s) ⇌ Ca²⁺ + 2F⁻, Ksp = [Ca²⁺][F⁻]². With the common ion F⁻ maintained at approximately 0.100 mol L⁻¹, s = 3.9 × 10⁻¹¹/(0.100)² = 3.9 × 10⁻⁹ mol L⁻¹.
4. A solution contains 0.100 mol L⁻¹ NH₃ and 0.200 mol L⁻¹ NH₄Cl. If Kb for NH₃ is 1.8 × 10⁻⁵, what is the approximate hydroxide-ion concentration?
A.
9.0 × 10⁻⁶ mol L⁻¹
B.
1.8 × 10⁻⁵ mol L⁻¹
C.
3.6 × 10⁻⁵ mol L⁻¹
D.
9.0 × 10⁻⁵ mol L⁻¹
Correct Answer:
9.0 × 10⁻⁶ mol L⁻¹
Explanation:
For NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, Kb = [NH₄⁺][OH⁻]/[NH₃]. The common ion NH₄⁺ suppresses NH₃ ionization, so [OH⁻] ≈ Kb[NH₃]/[NH₄⁺] = (1.8 × 10⁻⁵)(0.100/0.200) = 9.0 × 10⁻⁶ mol L⁻¹.
5. Mg(OH)₂ has Ksp = 1.8 × 10⁻¹¹. What is its approximate molar solubility in 0.010 mol L⁻¹ NaOH?
A.
1.8 × 10⁻⁷ mol L⁻¹
B.
1.8 × 10⁻⁹ mol L⁻¹
C.
9.0 × 10⁻⁷ mol L⁻¹
D.
1.8 × 10⁻¹¹ mol L⁻¹
Correct Answer:
1.8 × 10⁻⁷ mol L⁻¹
Explanation:
For Mg(OH)₂(s) ⇌ Mg²⁺ + 2OH⁻, Ksp = [Mg²⁺][OH⁻]². The common ion OH⁻ is supplied by NaOH, so s = 1.8 × 10⁻¹¹/(0.010)² = 1.8 × 10⁻⁷ mol L⁻¹.
6. The solubility of BaSO₄ decreases when Na₂SO₄ is added to a saturated BaSO₄ solution. Which change in the equilibrium is responsible?
A.
The equilibrium shifts toward solid BaSO₄ because SO₄²⁻ is a common ion
B.
The equilibrium shifts toward dissolved ions because Na⁺ reacts with BaSO₄
C.
The value of Ksp decreases because the solution contains more ions
D.
The equilibrium is unaffected because Na₂SO₄ is a soluble salt
Correct Answer:
The equilibrium shifts toward solid BaSO₄ because SO₄²⁻ is a common ion
Explanation:
BaSO₄(s) ⇌ Ba²⁺ + SO₄²⁻. Adding Na₂SO₄ increases the concentration of the common ion SO₄²⁻, causing the equilibrium to shift left and reducing BaSO₄ solubility. The Ksp value itself remains constant at a fixed temperature.
7. A solution contains 0.100 mol L⁻¹ HF and 0.200 mol L⁻¹ NaF. If Ka for HF is 6.8 × 10⁻⁴, what is the approximate pH of the solution?
A.
3.47
B.
2.17
C.
4.17
D.
5.47
Correct Answer:
3.47
Explanation:
For HF ⇌ H⁺ + F⁻, the common ion F⁻ from NaF suppresses HF ionization. Thus [H⁺] ≈ Ka[HF]/[F⁻] = (6.8 × 10⁻⁴)(0.100/0.200) = 3.4 × 10⁻⁴ mol L⁻¹, giving pH ≈ 3.47.
8. AgI and PbI₂ are placed separately in 0.010 mol L⁻¹ KI. Given Ksp(AgI) = 8.5 × 10⁻¹⁷ and Ksp(PbI₂) = 8.5 × 10⁻⁹, what is the ratio of their approximate molar solubilities, s(AgI) : s(PbI₂)?
A.
1 : 10⁶
B.
1 : 10⁴
C.
10⁶ : 1
D.
1 : 10²
Correct Answer:
1 : 10⁶
Explanation:
For AgI, s = Ksp/[I⁻] = 8.5 × 10⁻¹⁵ mol L⁻¹. For PbI₂, Ksp = s[I⁻]², so s = 8.5 × 10⁻⁵ mol L⁻¹. Therefore s(AgI):s(PbI₂) = 10⁻¹⁰, or approximately 1:10⁶. The iodide ion acts as a common ion in both cases.
9. Equal volumes of 0.020 mol L⁻¹ AgNO₃ and 0.020 mol L⁻¹ NaCl are mixed. If Ksp(AgCl) = 1.8 × 10⁻¹⁰, what is the approximate equilibrium concentration of Ag⁺ after precipitation, assuming the precipitate forms and the remaining chloride concentration is approximately 0.010 mol L⁻¹?
A.
1.8 × 10⁻⁸ mol L⁻¹
B.
9.0 × 10⁻⁹ mol L⁻¹
C.
1.8 × 10⁻¹⁰ mol L⁻¹
D.
1.0 × 10⁻² mol L⁻¹
Correct Answer:
1.8 × 10⁻⁸ mol L⁻¹
Explanation:
After mixing, chloride initially has concentration 0.010 mol L⁻¹ and acts as the common ion. At equilibrium, Ksp = [Ag⁺][Cl⁻], so [Ag⁺] ≈ 1.8 × 10⁻¹⁰/0.010 = 1.8 × 10⁻⁸ mol L⁻¹.
10. For the equilibrium NH₄OH ⇌ NH₄⁺ + OH⁻, which change produces the greatest suppression of ionization, assuming equal final volumes and comparable concentrations?
A.
Adding NH₄Cl, which increases the concentration of NH₄⁺
B.
Adding NaCl, which increases the concentration of an unrelated ion
C.
Adding distilled water, which decreases the concentration of NH₄⁺
D.
Adding a small amount of NH₄OH, which increases the weak-electrolyte concentration
Correct Answer:
Adding NH₄Cl, which increases the concentration of NH₄⁺
Explanation:
NH₄⁺ is a product and the common ion in the ionization equilibrium of NH₄OH. Adding NH₄Cl increases [NH₄⁺], shifting the equilibrium toward undissociated NH₄OH and thereby producing the common-ion suppression most directly.
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