chemistry : class11 Common Ion and Precipitation : Common ion effect : Equilibrium MCQ Questions and Answers
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Common Ion and Precipitation
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 0.10 M NaCl, assuming the change in chloride concentration is negligible?
A.
1.8 × 10^-11 M
B.
1.8 × 10^-9 M
C.
1.8 × 10^-8 M
D.
1.8 × 10^-7 M
Correct Answer:
1.8 × 10^-9 M
Explanation:
For AgCl(s) ⇌ Ag+ + Cl−, Ksp = [Ag+][Cl−]. In 0.10 M NaCl, [Cl−] ≈ 0.10 M, so the solubility s = Ksp/[Cl−] = (1.8 × 10^-10)/(0.10) = 1.8 × 10^-9 M. The common ion Cl− greatly decreases the solubility.
2. For Ag2CrO4, Ksp = 1.1 × 10^-12. What is its approximate molar solubility in 0.010 M AgNO3?
A.
1.1 × 10^-4 M
B.
1.1 × 10^-6 M
C.
1.1 × 10^-8 M
D.
1.1 × 10^-10 M
Correct Answer:
1.1 × 10^-8 M
Explanation:
Ag2CrO4(s) ⇌ 2Ag+ + CrO4^2−, so Ksp = [Ag+]^2[CrO4^2−]. In 0.010 M AgNO3, [Ag+] ≈ 0.010 M. Thus, solubility s = Ksp/(0.010)^2 = 1.1 × 10^-8 M. The common Ag+ ion suppresses dissolution.
3. Equal volumes of 1.0 × 10^-3 M AgNO3 and 2.0 × 10^-4 M NaCl are mixed. Given Ksp(AgCl) = 1.8 × 10^-10, what is the approximate equilibrium concentration of Cl− after precipitation?
A.
4.5 × 10^-7 M
B.
1.8 × 10^-6 M
C.
9.0 × 10^-6 M
D.
2.0 × 10^-4 M
Correct Answer:
4.5 × 10^-7 M
Explanation:
After mixing, the initial concentrations are [Ag+] = 5.0 × 10^-4 M and [Cl−] = 1.0 × 10^-4 M. Ag+ is in excess, so nearly all Cl− precipitates and [Ag+] remains approximately 4.0 × 10^-4 M. At equilibrium, [Cl−] = Ksp/[Ag+] = (1.8 × 10^-10)/(4.0 × 10^-4) = 4.5 × 10^-7 M.
4. A solution contains 0.010 M each of Cl− and I−. AgNO3 is added gradually. Given Ksp(AgCl) = 1.8 × 10^-10 and Ksp(AgI) = 8.5 × 10^-17, which statement is correct?
A.
AgCl precipitates first because Cl− is more concentrated.
B.
AgI precipitates first because it requires a lower Ag+ concentration.
C.
Both AgCl and AgI precipitate simultaneously.
D.
Neither salt precipitates until Ag+ exceeds 0.010 M.
Correct Answer:
AgI precipitates first because it requires a lower Ag+ concentration.
Explanation:
Precipitation begins when [Ag+][anion−] equals Ksp. For AgI, the required [Ag+] is 8.5 × 10^-17/0.010 = 8.5 × 10^-15 M, whereas for AgCl it is 1.8 × 10^-10/0.010 = 1.8 × 10^-8 M. Therefore, AgI precipitates first.
5. The Ksp of CaF2 is 3.9 × 10^-11. What is its approximate molar solubility in 0.10 M NaF?
A.
3.9 × 10^-3 M
B.
3.9 × 10^-5 M
C.
3.9 × 10^-7 M
D.
3.9 × 10^-9 M
Correct Answer:
3.9 × 10^-9 M
Explanation:
CaF2(s) ⇌ Ca2+ + 2F−, so Ksp = [Ca2+][F−]^2. With 0.10 M F− supplied by NaF, the solubility s = Ksp/(0.10)^2 = 3.9 × 10^-9 M. The common F− ion suppresses CaF2 dissolution.
6. Equal volumes of 0.020 M BaCl2 and 0.010 M Na2SO4 are mixed. If Ksp(BaSO4) = 1.1 × 10^-10, what is the approximate equilibrium concentration of Ba2+ after precipitation?
A.
1.1 × 10^-2 M
B.
1.0 × 10^-3 M
C.
1.05 × 10^-5 M
D.
1.1 × 10^-10 M
Correct Answer:
1.05 × 10^-5 M
Explanation:
After mixing, both Ba2+ and SO4^2− initially have concentrations of 0.010 M and precipitate nearly completely in a 1:1 ratio. At equilibrium, their residual concentrations are approximately equal, x. Therefore, Ksp = x^2, giving x = √(1.1 × 10^-10) ≈ 1.05 × 10^-5 M.
7. The Ksp of Mg(OH)2 is 1.8 × 10^-11. What is its approximate molar solubility in a solution having [OH−] = 1.0 × 10^-3 M?
A.
1.8 × 10^-2 M
B.
1.8 × 10^-5 M
C.
1.8 × 10^-8 M
D.
1.8 × 10^-11 M
Correct Answer:
1.8 × 10^-5 M
Explanation:
Mg(OH)2(s) ⇌ Mg2+ + 2OH−, so Ksp = [Mg2+][OH−]^2. With [OH−] maintained at 1.0 × 10^-3 M, the solubility is s = 1.8 × 10^-11/(1.0 × 10^-3)^2 = 1.8 × 10^-5 M. The common OH− ion lowers the solubility.
8. A solution contains 0.020 M Ag+. What minimum chloride-ion concentration is required to initiate precipitation of AgCl if Ksp(AgCl) = 1.8 × 10^-10?
A.
9.0 × 10^-12 M
B.
9.0 × 10^-10 M
C.
9.0 × 10^-9 M
D.
3.6 × 10^-8 M
Correct Answer:
9.0 × 10^-9 M
Explanation:
Precipitation begins when the ionic product equals Ksp: [Ag+][Cl−] = Ksp. Thus, [Cl−]minimum = (1.8 × 10^-10)/(0.020) = 9.0 × 10^-9 M. Any greater chloride concentration makes the solution supersaturated with respect to AgCl.
9. The Ksp of PbI2 is 8.5 × 10^-9. What is its approximate molar solubility in 0.010 M KI?
A.
8.5 × 10^-3 M
B.
8.5 × 10^-4 M
C.
8.5 × 10^-5 M
D.
8.5 × 10^-7 M
Correct Answer:
8.5 × 10^-5 M
Explanation:
PbI2(s) ⇌ Pb2+ + 2I−, so Ksp = [Pb2+][I−]^2. In 0.010 M KI, [I−] is approximately 0.010 M. Therefore, the solubility s = 8.5 × 10^-9/(0.010)^2 = 8.5 × 10^-5 M. The common I− ion reduces the solubility of PbI2.
10. A saturated solution of Ag2CrO4 is prepared in 0.010 M AgNO3. If Ksp(Ag2CrO4) = 1.1 × 10^-12, what is the approximate concentration of CrO4^2− at equilibrium?
A.
1.1 × 10^-2 M
B.
1.1 × 10^-4 M
C.
1.1 × 10^-8 M
D.
1.1 × 10^-12 M
Correct Answer:
1.1 × 10^-8 M
Explanation:
For Ag2CrO4, Ksp = [Ag+]^2[CrO4^2−]. Since [Ag+] is approximately 0.010 M due to the common Ag+ ion, [CrO4^2−] = (1.1 × 10^-12)/(0.010)^2 = 1.1 × 10^-8 M. Thus, the common ion strongly suppresses the chromate concentration produced by dissolution.
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