chemistry : class11 Salt Solubility Suppression : Common ion effect : Equilibrium MCQ Questions and Answers
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Salt Solubility Suppression
MCQ questions and answers.
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MCQ Questions
1. The solubility product of AgCl is 1.8 × 10⁻¹⁰. What is the molar solubility of AgCl in 0.100 M NaCl, assuming the contribution of AgCl to [Cl⁻] is negligible?
A.
1.8 × 10⁻¹¹ M
B.
1.8 × 10⁻⁹ M
C.
1.34 × 10⁻⁵ M
D.
1.8 × 10⁻⁸ M
Correct Answer:
1.8 × 10⁻⁹ M
Explanation:
For AgCl(s) ⇌ Ag⁺ + Cl⁻, Ksp = [Ag⁺][Cl⁻] = s(0.100). Therefore, s = 1.8 × 10⁻¹⁰ ÷ 0.100 = 1.8 × 10⁻⁹ M. The high chloride concentration suppresses AgCl solubility through the common ion effect.
2. Ag₂CrO₄ has Ksp = 1.1 × 10⁻¹². Its molar solubility in 0.100 M AgNO₃ is approximately:
A.
1.1 × 10⁻¹² M
B.
1.1 × 10⁻¹⁰ M
C.
5.5 × 10⁻¹¹ M
D.
1.0 × 10⁻⁶ M
Correct Answer:
1.1 × 10⁻¹⁰ M
Explanation:
For Ag₂CrO₄(s) ⇌ 2Ag⁺ + CrO₄²⁻, Ksp = [Ag⁺]²[CrO₄²⁻]. In 0.100 M AgNO₃, [Ag⁺] ≈ 0.100 M, so Ksp = (0.100)²s. Hence s = 1.1 × 10⁻¹² ÷ 10⁻² = 1.1 × 10⁻¹⁰ M.
3. CaF₂ has Ksp = 3.9 × 10⁻¹¹. What is its molar solubility in 0.010 M NaF, assuming that the fluoride ion produced by CaF₂ dissolution is negligible compared with the initial fluoride concentration?
A.
3.9 × 10⁻⁷ M
B.
3.9 × 10⁻⁹ M
C.
1.95 × 10⁻⁷ M
D.
1.56 × 10⁻⁵ M
Correct Answer:
3.9 × 10⁻⁷ M
Explanation:
For CaF₂(s) ⇌ Ca²⁺ + 2F⁻, Ksp = [Ca²⁺][F⁻]². With [F⁻] ≈ 0.010 M and [Ca²⁺] = s, 3.9 × 10⁻¹¹ = s(0.010)². Thus s = 3.9 × 10⁻⁷ M. The common fluoride ion strongly suppresses dissolution.
4. AgCl has Ksp = 1.8 × 10⁻¹⁰. Its solubility in pure water is approximately 1.34 × 10⁻⁵ M. What is the approximate factor by which its solubility decreases in 0.010 M NaCl?
A.
7.4
B.
74
C.
744
D.
7.4 × 10⁴
Correct Answer:
744
Explanation:
In 0.010 M NaCl, the solubility is s = Ksp/[Cl⁻] = 1.8 × 10⁻¹⁰ ÷ 0.010 = 1.8 × 10⁻⁸ M. The suppression factor is (1.34 × 10⁻⁵)/(1.8 × 10⁻⁸) ≈ 744. The decrease results from the common chloride ion.
5. A sparingly soluble salt AB₂ has Ksp = 4.0 × 10⁻¹². What is its molar solubility in a solution containing 0.100 M B⁻, assuming the added B⁻ concentration remains effectively constant?
A.
4.0 × 10⁻¹⁰ M
B.
2.0 × 10⁻⁹ M
C.
4.0 × 10⁻⁸ M
D.
1.0 × 10⁻⁴ M
Correct Answer:
4.0 × 10⁻¹⁰ M
Explanation:
For AB₂(s) ⇌ A²⁺ + 2B⁻, Ksp = [A²⁺][B⁻]². With [A²⁺] = s and [B⁻] ≈ 0.100 M, 4.0 × 10⁻¹² = s(0.100)². Therefore, s = 4.0 × 10⁻¹⁰ M. The common B⁻ ion suppresses the salt's solubility.
6. PbI₂ has Ksp = 8.0 × 10⁻⁹. Its molar solubility in 0.020 M KI is approximately:
A.
2.0 × 10⁻⁵ M
B.
4.0 × 10⁻⁵ M
C.
8.0 × 10⁻⁹ M
D.
1.6 × 10⁻⁷ M
Correct Answer:
2.0 × 10⁻⁵ M
Explanation:
For PbI₂(s) ⇌ Pb²⁺ + 2I⁻, Ksp = [Pb²⁺][I⁻]². Since [I⁻] ≈ 0.020 M from KI, s = 8.0 × 10⁻⁹ ÷ (0.020)² = 2.0 × 10⁻⁵ M. The iodide common ion reduces the solubility of PbI₂.
7. For which situation is the solubility suppression of a salt most accurately represented by Ksp = sC², where C is the concentration of a common ion supplied by a soluble electrolyte?
A.
AgCl dissolving in a solution containing C M Cl⁻
B.
CaF₂ dissolving in a solution containing C M F⁻
C.
Ag₂CrO₄ dissolving in a solution containing C M Ag⁺
D.
BaSO₄ dissolving in a solution containing C M Ba²⁺
Correct Answer:
CaF₂ dissolving in a solution containing C M F⁻
Explanation:
For CaF₂(s) ⇌ Ca²⁺ + 2F⁻, Ksp = [Ca²⁺][F⁻]². If the common fluoride concentration is C and the salt solubility is s, then Ksp ≈ sC². The other salts either have a first-power common-ion term or a squared concentration of the common cation.
8. Zn(OH)₂ has Ksp = 3.0 × 10⁻¹⁷. What is its approximate molar solubility in 0.010 M NaOH, neglecting the hydroxide produced by Zn(OH)₂?
A.
3.0 × 10⁻¹³ M
B.
3.0 × 10⁻¹¹ M
C.
1.5 × 10⁻¹³ M
D.
3.0 × 10⁻⁹ M
Correct Answer:
3.0 × 10⁻¹³ M
Explanation:
For Zn(OH)₂(s) ⇌ Zn²⁺ + 2OH⁻, Ksp = [Zn²⁺][OH⁻]². In 0.010 M NaOH, [OH⁻] ≈ 0.010 M, so s = 3.0 × 10⁻¹⁷ ÷ (0.010)² = 3.0 × 10⁻¹³ M. The common hydroxide ion causes extreme solubility suppression.
9. AgCl has Ksp = 1.8 × 10⁻¹⁰. What chloride ion concentration is required to reduce its molar solubility to 1.0 × 10⁻⁹ M, assuming [Cl⁻] added is much larger than the chloride released by AgCl?
A.
0.018 M
B.
0.180 M
C.
1.80 × 10⁻⁹ M
D.
5.56 × 10⁻⁹ M
Correct Answer:
0.180 M
Explanation:
For AgCl, Ksp = s[Cl⁻]. Therefore, [Cl⁻] = Ksp/s = (1.8 × 10⁻¹⁰)/(1.0 × 10⁻⁹) = 0.180 M. This large common-ion concentration is needed to suppress the solubility to the stated value.
10. BaSO₄ has Ksp = 1.1 × 10⁻¹⁰. Assuming sulfate remains the dominant sulfate species, what is its approximate molar solubility in 0.0010 M Na₂SO₄?
A.
1.1 × 10⁻⁷ M
B.
1.1 × 10⁻¹⁰ M
C.
3.3 × 10⁻⁷ M
D.
1.0 × 10⁻⁴ M
Correct Answer:
1.1 × 10⁻⁷ M
Explanation:
For BaSO₄(s) ⇌ Ba²⁺ + SO₄²⁻, Ksp = [Ba²⁺][SO₄²⁻]. With [SO₄²⁻] ≈ 0.0010 M and [Ba²⁺] = s, s = 1.1 × 10⁻¹⁰ ÷ 0.0010 = 1.1 × 10⁻⁷ M. The sulfate common ion suppresses BaSO₄ solubility.
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