chemistry : class11 Common Ion Concentration : Common ion effect : Equilibrium MCQ Questions and Answers
Practice important
Common Ion Concentration
MCQ questions and answers.
Page 1
of
1
|
10
MCQ Questions
1. The solubility product of AgCl is 1.8 × 10^-10 at 25 °C. If AgCl is equilibrated with 0.100 mol L^-1 NaCl, what is the approximate molar solubility of AgCl? Assume the contribution of AgCl to [Cl^-] is negligible.
A.
1.8 × 10^-11 mol L^-1
B.
1.8 × 10^-9 mol L^-1
C.
1.8 × 10^-8 mol L^-1
D.
9.0 × 10^-10 mol L^-1
Correct Answer:
1.8 × 10^-9 mol L^-1
Explanation:
For AgCl(s) ⇌ Ag+ + Cl−, Ksp = [Ag+][Cl−]. Since [Cl−] is approximately 0.100 mol L−1, [Ag+] = 1.8 × 10−10/0.100 = 1.8 × 10−9 mol L−1. The common Cl− ion strongly suppresses AgCl dissolution.
2. For HF, Ka = 6.8 × 10^-4. What is the approximate hydrogen-ion concentration in a solution containing 0.100 mol L^-1 HF and 0.200 mol L^-1 NaF? Assume the change in the concentrations of HF and F^- is negligible.
A.
3.4 × 10^-4 mol L^-1
B.
6.8 × 10^-4 mol L^-1
C.
1.36 × 10^-3 mol L^-1
D.
6.8 × 10^-5 mol L^-1
Correct Answer:
3.4 × 10^-4 mol L^-1
Explanation:
For HF ⇌ H+ + F−, Ka = [H+][F−]/[HF]. Thus [H+] = Ka[HF]/[F−] = (6.8 × 10−4)(0.100/0.200) = 3.4 × 10−4 mol L−1. The F− supplied by NaF is the common ion.
3. Ammonia has Kb = 1.8 × 10^-5. In a solution containing 0.200 mol L^-1 NH3 and 0.100 mol L^-1 NH4Cl, 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:
3.6 × 10^-5 mol L^-1
Explanation:
For NH3 + H2O ⇌ NH4+ + OH−, Kb = [NH4+][OH−]/[NH3]. Therefore [OH−] = Kb[NH3]/[NH4+] = (1.8 × 10−5)(0.200/0.100) = 3.6 × 10−5 mol L−1. NH4+ is the common ion.
4. The solubility product of CaF2 is 4.0 × 10^-11. What is its approximate molar solubility in 0.100 mol L^-1 NaF? Assume the fluoride concentration remains approximately 0.100 mol L^-1.
A.
4.0 × 10^-11 mol L^-1
B.
4.0 × 10^-9 mol L^-1
C.
2.0 × 10^-8 mol L^-1
D.
4.0 × 10^-7 mol L^-1
Correct Answer:
4.0 × 10^-9 mol L^-1
Explanation:
For CaF2(s) ⇌ Ca2+ + 2F−, Ksp = [Ca2+][F−]^2. Hence [Ca2+] = 4.0 × 10−11/(0.100)^2 = 4.0 × 10−9 mol L−1. The added F− is the common ion and suppresses CaF2 dissolution.
5. A weak acid HA has Ka = 1.0 × 10^-5. Compared with a 0.100 mol L^-1 solution of HA alone, what is the approximate ionization percentage of HA in a solution containing 0.100 mol L^-1 HA and 0.100 mol L^-1 NaA?
A.
0.001%
B.
0.010%
C.
0.100%
D.
1.00%
Correct Answer:
0.010%
Explanation:
With A− as a common ion, [H+] ≈ Ka[HA]/[A−] = 1.0 × 10−5 mol L−1. The ionization percentage is ([H+]/[HA]) × 100 = (1.0 × 10−5/0.100) × 100 = 0.010%. The common A− ion greatly reduces ionization.
6. The solubility product of PbI2 is 8.0 × 10^-9. If PbI2 is placed in 0.020 mol L^-1 KI, what is the approximate molar solubility of PbI2? Assume [I^-] remains 0.020 mol L^-1.
A.
1.0 × 10^-5 mol L^-1
B.
2.0 × 10^-5 mol L^-1
C.
4.0 × 10^-5 mol L^-1
D.
8.0 × 10^-6 mol L^-1
Correct Answer:
2.0 × 10^-5 mol L^-1
Explanation:
For PbI2(s) ⇌ Pb2+ + 2I−, Ksp = [Pb2+][I−]^2. Thus the solubility s = [Pb2+] = 8.0 × 10−9/(0.020)^2 = 2.0 × 10−5 mol L−1. I− from KI is the common ion.
7. A saturated solution of Ag2CrO4 is prepared in 0.010 mol L^-1 K2CrO4. If Ksp(Ag2CrO4) = 1.0 × 10^-12, what is the approximate concentration of Ag+ at equilibrium?
A.
1.0 × 10^-8 mol L^-1
B.
1.0 × 10^-7 mol L^-1
C.
2.0 × 10^-6 mol L^-1
D.
1.0 × 10^-5 mol L^-1
Correct Answer:
1.0 × 10^-5 mol L^-1
Explanation:
For Ag2CrO4(s) ⇌ 2Ag+ + CrO4^2−, Ksp = [Ag+]^2[CrO4^2−]. Therefore [Ag+] = √(1.0 × 10−12/0.010) = √(1.0 × 10−10) = 1.0 × 10−5 mol L−1. Chromate is the common ion.
8. For the equilibrium NH4+ ⇌ NH3 + H+, Ka(NH4+) = 5.6 × 10^-10. In a solution containing 0.050 mol L^-1 NH4Cl and 0.200 mol L^-1 NH3, what is the approximate pH? Use log 2 = 0.301.
A.
8.85
B.
9.15
C.
9.55
D.
10.00
Correct Answer:
9.15
Explanation:
For NH4+ ⇌ NH3 + H+, [H+] = Ka[NH4+]/[NH3] = (5.6 × 10−10)(0.050/0.200) = 1.4 × 10−10 mol L−1. Therefore pH = 10 − log 1.4 ≈ 10 − 0.146 = 9.85. However, using the corresponding base-equilibrium relation gives pOH = pKb + log([NH4+]/[NH3]) = 4.25 + log(0.25) = 3.65, so pH = 10.35. Thus none of the listed values is consistent; the correct calculated pH is 9.85.
9. The solubility product of Zn(OH)2 is 4.9 × 10^-17. What is the approximate solubility of Zn(OH)2 in a solution buffered at [OH^-] = 1.0 × 10^-3 mol L^-1?
A.
4.9 × 10^-11 mol L^-1
B.
4.9 × 10^-8 mol L^-1
C.
2.2 × 10^-8 mol L^-1
D.
4.9 × 10^-14 mol L^-1
Correct Answer:
4.9 × 10^-11 mol L^-1
Explanation:
For Zn(OH)2(s) ⇌ Zn2+ + 2OH−, Ksp = [Zn2+][OH−]^2. Thus the solubility is [Zn2+] = 4.9 × 10−17/(1.0 × 10−3)^2 = 4.9 × 10−11 mol L−1. The externally maintained OH− concentration acts as the common ion.
10. A weak acid HA has Ka = 1.0 × 10^-4. In a solution containing equal concentrations of HA and its sodium salt NaA, the concentration of HA is increased from 0.010 mol L^-1 to 0.100 mol L^-1 while the concentration of A^- remains 0.010 mol L^-1. By what factor does [H+] approximately increase?
A.
1
B.
5
C.
10
D.
100
Correct Answer:
10
Explanation:
With A− acting as the common ion, [H+] ≈ Ka[HA]/[A−]. Since [A−] remains fixed while [HA] increases by a factor of 10, [H+] also increases by a factor of 10. The common-ion concentration controls the suppression of HA ionization.
Short Questions and Answers
Practice short questions and answers related to
Common Ion Concentration
.
Short questions are currently unavailable for this page.