chemistry : class11 Common Ion in Ammonium Hydroxide : Common ion effect : Equilibrium MCQ Questions and Answers
Practice important
Common Ion in Ammonium Hydroxide
MCQ questions and answers.
Page 1
of
1
|
10
MCQ Questions
1. For 0.10 M NH4OH containing 0.10 M NH4Cl at 25 °C, given Kb(NH4OH) = 1.8 × 10^-5, what is the approximate pH of the solution?
A.
8.255
B.
9.255
C.
10.745
D.
11.255
Correct Answer:
9.255
Explanation:
NH4Cl provides the common ion NH4+, suppressing the ionization of NH4OH. Thus, [OH−] ≈ Kb[NH4OH]/[NH4+] = 1.8 × 10^-5. Therefore, pOH ≈ 4.745 and pH ≈ 9.255.
2. A 0.10 M solution of NH4OH has [OH−] ≈ 1.34 × 10^-3 M in the absence of NH4Cl. When 0.10 M NH4Cl is added and the final concentration of NH4OH remains 0.10 M, [OH−] becomes approximately 1.8 × 10^-5 M. What is the percentage decrease in [OH−]?
A.
1.34%
B.
86.6%
C.
98.66%
D.
99.86%
Correct Answer:
98.66%
Explanation:
The percentage decrease is [(1.34 × 10^-3 − 1.8 × 10^-5)/(1.34 × 10^-3)] × 100, which is approximately 98.66%. The large decrease results from the common ion NH4+ suppressing NH4OH ionization.
3. A solution contains 0.10 M NH4OH. What concentration of NH4Cl is required to produce approximately [OH−] = 1.0 × 10^-5 M? Assume Kb(NH4OH) = 1.8 × 10^-5 and neglect the change in concentrations due to ionization.
A.
0.018 M
B.
0.056 M
C.
0.18 M
D.
1.80 M
Correct Answer:
0.18 M
Explanation:
For the common-ion equilibrium, [OH−] ≈ Kb[NH4OH]/[NH4+]. Hence, [NH4+] = (1.8 × 10^-5 × 0.10)/(1.0 × 10^-5) = 0.18 M. NH4Cl supplies this NH4+ concentration.
4. Equal volumes of 0.20 M NH4OH and 0.10 M NH4Cl are mixed. Given Kb(NH4OH) = 1.8 × 10^-5, what is the approximate pH of the resulting solution?
A.
8.444
B.
9.255
C.
9.556
D.
10.556
Correct Answer:
9.556
Explanation:
After mixing, [NH4OH] = 0.10 M and [NH4Cl] = [NH4+] = 0.050 M. Therefore, [OH−] ≈ (1.8 × 10^-5 × 0.10)/0.050 = 3.6 × 10^-5 M. Thus pOH ≈ 4.444 and pH ≈ 9.556.
5. For the equilibrium NH4OH(aq) ⇌ NH4+(aq) + OH−(aq), what is the principal effect of adding NH4Cl to an NH4OH solution?
A.
It increases the ionization of NH4OH by removing NH4+.
B.
It decreases the ionization of NH4OH because NH4+ is a common ion.
C.
It increases [OH−] by completely dissociating NH4OH.
D.
It converts NH4OH into a strong base without affecting equilibrium.
Correct Answer:
It decreases the ionization of NH4OH because NH4+ is a common ion.
Explanation:
NH4Cl dissociates to provide NH4+, which is already present in the NH4OH ionization equilibrium. By Le Chatelier's principle, the equilibrium shifts left, reducing the ionization of NH4OH and lowering [OH−].
6. A solution contains 0.10 M NH4OH and enough NH4Cl to make [NH4+] = 0.18 M. Given Kb(NH4OH) = 1.8 × 10^-5, what is the approximate pH?
A.
8.000
B.
9.000
C.
10.000
D.
11.000
Correct Answer:
9.000
Explanation:
Using [OH−] ≈ Kb[NH4OH]/[NH4+], [OH−] = (1.8 × 10^-5 × 0.10)/0.18 = 1.0 × 10^-5 M. Therefore pOH = 5.00 and pH = 9.00.
7. The degree of ionization of 0.10 M NH4OH is approximately 1.34% in the absence of NH4Cl. If 0.10 M NH4Cl is added so that [OH−] becomes 1.8 × 10^-5 M, what is the new degree of ionization of NH4OH?
A.
0.0018%
B.
0.018%
C.
0.18%
D.
1.80%
Correct Answer:
0.018%
Explanation:
The degree of ionization is approximately [OH−]/[NH4OH] × 100 = (1.8 × 10^-5/0.10) × 100 = 0.018%. The common NH4+ ion greatly suppresses NH4OH ionization.
8. Two solutions contain the same concentration of NH4OH: solution X has no NH4Cl, while solution Y contains 0.20 M NH4Cl. Which comparison is correct?
A.
Solution Y has higher [OH−] and higher NH4OH ionization.
B.
Solution Y has lower [OH−] and lower NH4OH ionization.
C.
Both solutions have identical [OH−] because NH4Cl is neutral.
D.
Solution Y has lower [OH−] but higher NH4OH ionization.
Correct Answer:
Solution Y has lower [OH−] and lower NH4OH ionization.
Explanation:
NH4Cl supplies NH4+, the common ion in the NH4OH equilibrium. Increasing [NH4+] shifts the equilibrium toward undissociated NH4OH, reducing both its ionization and the resulting [OH−].
9. A solution is prepared by mixing equal volumes of 0.20 M NH4OH and 0.20 M NH4Cl. If Kb(NH4OH) = 1.8 × 10^-5, what is the approximate hydroxide-ion concentration after mixing?
A.
1.8 × 10^-6 M
B.
1.8 × 10^-5 M
C.
3.6 × 10^-5 M
D.
1.8 × 10^-4 M
Correct Answer:
1.8 × 10^-5 M
Explanation:
After equal-volume mixing, both NH4OH and NH4Cl become 0.10 M. Therefore, [OH−] ≈ Kb[NH4OH]/[NH4+] = (1.8 × 10^-5 × 0.10)/0.10 = 1.8 × 10^-5 M.
10. A solution contains 0.010 M NH4OH and 0.100 M NH4Cl. Given Kb(NH4OH) = 1.8 × 10^-5, what is the approximate pH?
A.
7.255
B.
8.255
C.
9.255
D.
10.255
Correct Answer:
8.255
Explanation:
The common-ion approximation gives [OH−] ≈ (1.8 × 10^-5 × 0.010)/0.100 = 1.8 × 10^-6 M. Hence pOH ≈ 5.745 and pH ≈ 8.255. The high NH4+ concentration strongly suppresses NH4OH ionization.
Short Questions and Answers
Practice short questions and answers related to
Common Ion in Ammonium Hydroxide
.
Short questions are currently unavailable for this page.