chemistry : class11 Calculation of Solubility : Common ion effect : Equilibrium MCQ Questions and Answers
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Calculation of Solubility
MCQ questions and answers.
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MCQ Questions
1. The solubility product constant of AgCl is 1.0 × 10^-10. What is the molar solubility of AgCl in 0.10 M NaCl?
A.
1.0 × 10^-10 M
B.
1.0 × 10^-9 M
C.
1.0 × 10^-5 M
D.
1.0 × 10^-4 M
Correct Answer:
1.0 × 10^-9 M
Explanation:
For AgCl, Ksp = [Ag+][Cl-]. In 0.10 M NaCl, [Cl-] is approximately 0.10 M. Therefore, 1.0 × 10^-10 = s × 0.10, giving s = 1.0 × 10^-9 M.
2. For PbI2, Ksp = 8.0 × 10^-9. What is its molar solubility in 0.10 M KI?
A.
8.0 × 10^-9 M
B.
8.0 × 10^-8 M
C.
8.0 × 10^-7 M
D.
8.0 × 10^-6 M
Correct Answer:
8.0 × 10^-7 M
Explanation:
PbI2 dissociates as PbI2 ⇌ Pb2+ + 2I-. In 0.10 M KI, [I-] is approximately 0.10 M. Thus, Ksp = s(0.10)^2 = 0.01s. Hence, s = 8.0 × 10^-9 ÷ 0.01 = 8.0 × 10^-7 M.
3. The Ksp of CaF2 is 4.0 × 10^-11. What is the molar solubility of CaF2 in 0.10 M NaF?
A.
4.0 × 10^-11 M
B.
4.0 × 10^-9 M
C.
4.0 × 10^-8 M
D.
4.0 × 10^-6 M
Correct Answer:
4.0 × 10^-9 M
Explanation:
CaF2 dissociates as CaF2 ⇌ Ca2+ + 2F-. The common ion concentration is [F-] = 0.10 M. Therefore, Ksp = s(0.10)^2, so s = 4.0 × 10^-11 ÷ 0.01 = 4.0 × 10^-9 M.
4. Ag2CrO4 has Ksp = 1.0 × 10^-12. What is its molar solubility in 0.10 M AgNO3?
A.
1.0 × 10^-12 M
B.
1.0 × 10^-10 M
C.
1.0 × 10^-8 M
D.
1.0 × 10^-6 M
Correct Answer:
1.0 × 10^-10 M
Explanation:
Ag2CrO4 dissociates as Ag2CrO4 ⇌ 2Ag+ + CrO4^2-. In 0.10 M AgNO3, [Ag+] is approximately 0.10 M. Thus, Ksp = (0.10)^2s = 0.01s, giving s = 1.0 × 10^-10 M.
5. The Ksp of BaSO4 is 1.0 × 10^-10. Calculate its molar solubility in 0.010 M Na2SO4.
A.
1.0 × 10^-10 M
B.
1.0 × 10^-9 M
C.
1.0 × 10^-8 M
D.
1.0 × 10^-6 M
Correct Answer:
1.0 × 10^-8 M
Explanation:
BaSO4 dissociates as BaSO4 ⇌ Ba2+ + SO4^2-. The common ion concentration is [SO4^2-] = 0.010 M. Therefore, Ksp = s × 0.010, so s = 1.0 × 10^-10 ÷ 0.010 = 1.0 × 10^-8 M.
6. For a salt MX, Ksp = 9.0 × 10^-8. What is its molar solubility in 0.010 M MX-containing solution, assuming the common ion concentration remains 0.010 M?
A.
9.0 × 10^-10 M
B.
9.0 × 10^-8 M
C.
9.0 × 10^-6 M
D.
9.0 × 10^-4 M
Correct Answer:
9.0 × 10^-6 M
Explanation:
For MX ⇌ M+ + X-, Ksp = [M+][X-]. With the common ion concentration equal to 0.010 M, Ksp = s × 0.010. Thus, s = 9.0 × 10^-8 ÷ 0.010 = 9.0 × 10^-6 M.
7. Mg(OH)2 has Ksp = 1.8 × 10^-11. What is its molar solubility in 0.10 M NaOH?
A.
1.8 × 10^-11 M
B.
1.8 × 10^-9 M
C.
1.8 × 10^-7 M
D.
1.8 × 10^-5 M
Correct Answer:
1.8 × 10^-9 M
Explanation:
Mg(OH)2 dissociates as Mg(OH)2 ⇌ Mg2+ + 2OH-. Since [OH-] is approximately 0.10 M, Ksp = s(0.10)^2 = 0.01s. Therefore, s = 1.8 × 10^-11 ÷ 0.01 = 1.8 × 10^-9 M.
8. The Ksp of AgBr is 5.0 × 10^-13. What is the molar solubility of AgBr in 0.050 M KBr?
A.
1.0 × 10^-13 M
B.
1.0 × 10^-11 M
C.
1.0 × 10^-10 M
D.
1.0 × 10^-8 M
Correct Answer:
1.0 × 10^-11 M
Explanation:
AgBr dissociates as AgBr ⇌ Ag+ + Br-. In 0.050 M KBr, [Br-] is approximately 0.050 M. Hence, Ksp = s × 0.050, and s = 5.0 × 10^-13 ÷ 0.050 = 1.0 × 10^-11 M.
9. For CaCO3, Ksp = 3.6 × 10^-9. What is its molar solubility in 0.020 M Na2CO3?
A.
1.8 × 10^-10 M
B.
1.8 × 10^-9 M
C.
1.8 × 10^-7 M
D.
1.8 × 10^-5 M
Correct Answer:
1.8 × 10^-7 M
Explanation:
CaCO3 dissociates as CaCO3 ⇌ Ca2+ + CO3^2-. With [CO3^2-] = 0.020 M, Ksp = s × 0.020. Therefore, s = 3.6 × 10^-9 ÷ 0.020 = 1.8 × 10^-7 M.
10. The Ksp of Zn(OH)2 is 4.0 × 10^-17. Find its molar solubility in 0.010 M NaOH.
A.
4.0 × 10^-17 M
B.
4.0 × 10^-15 M
C.
4.0 × 10^-13 M
D.
4.0 × 10^-11 M
Correct Answer:
4.0 × 10^-13 M
Explanation:
Zn(OH)2 dissociates as Zn(OH)2 ⇌ Zn2+ + 2OH-. In 0.010 M NaOH, [OH-] is approximately 0.010 M. Thus, Ksp = s(0.010)^2 = 1.0 × 10^-4s, so s = 4.0 × 10^-17 ÷ 1.0 × 10^-4 = 4.0 × 10^-13 M.
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