chemistry : class12 Conductometric Titrations : Conductance in electrolytic solutions : Electrochemistry MCQ Questions and Answers
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Conductometric Titrations
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MCQ Questions
1. In the conductometric titration of hydrochloric acid with sodium hydroxide, the conductance initially decreases and then increases. Which change most directly explains the minimum at the equivalence point?
A.
Hydroxide ions replace chloride ions before equivalence
B.
Highly mobile hydrogen ions are replaced by less mobile sodium ions, followed by an excess of hydroxide ions
C.
Chloride ions are completely precipitated at equivalence
D.
Sodium ions become nonconducting after equivalence
Correct Answer:
Highly mobile hydrogen ions are replaced by less mobile sodium ions, followed by an excess of hydroxide ions
Explanation:
Before equivalence, H+ ions are neutralized and replaced by Na+ ions, whose ionic mobility is much lower, so conductance falls. Beyond equivalence, added OH− ions remain in excess and cause conductance to rise sharply.
2. A weak monoprotic acid is titrated conductometrically with a strong base. Which feature is expected in the conductance-versus-volume curve?
A.
A steep decrease followed by a less steep decrease
B.
Nearly constant conductance before equivalence followed by a decrease after equivalence
C.
A gradual increase before equivalence followed by a sharper increase after equivalence
D.
A gradual decrease before equivalence followed by a gradual increase of identical slope
Correct Answer:
A gradual increase before equivalence followed by a sharper increase after equivalence
Explanation:
The weak acid is poorly ionized initially. Addition of strong base forms its soluble salt, increasing the concentration of ions and hence conductance gradually. After equivalence, excess OH−, having high ionic mobility, produces a sharper increase.
3. During the conductometric titration of sodium chloride with silver nitrate, the conductance generally decreases slightly before the equivalence point. What is the principal reason?
A.
Na+ ions are removed completely from solution
B.
Ag+ ions precipitate chloride ions and the remaining chloride ions are replaced by nitrate ions of slightly lower ionic conductance
C.
Nitrate ions precipitate sodium ions
D.
Silver nitrate remains completely unionized before equivalence
Correct Answer:
Ag+ ions precipitate chloride ions and the remaining chloride ions are replaced by nitrate ions of slightly lower ionic conductance
Explanation:
As AgNO3 is added, Cl− is removed as AgCl. Nitrate ions enter the solution in place of chloride ions. Since nitrate has slightly lower ionic conductance than chloride, the conductance usually decreases slightly before equivalence; excess AgNO3 then causes an increase.
4. A mixture containing hydrochloric acid and acetic acid is titrated with sodium hydroxide conductometrically. Two breaks occur at titrant volumes V1 and V2. What quantities do these volumes represent?
A.
V1 corresponds to neutralization of acetic acid only, and V2 to neutralization of hydrochloric acid only
B.
V1 corresponds to neutralization of hydrochloric acid, and V2 corresponds to total neutralization of both acids
C.
V1 corresponds to precipitation of chloride, and V2 to precipitation of acetate
D.
V1 corresponds to total acid concentration, and V2 to the concentration of hydrochloric acid only
Correct Answer:
V1 corresponds to neutralization of hydrochloric acid, and V2 corresponds to total neutralization of both acids
Explanation:
The strong acid is neutralized first, giving the first change in slope. Continued addition of NaOH neutralizes acetic acid as well. Therefore, the second break represents the combined amount of hydrochloric acid and acetic acid.
5. In a conductometric titration, the corrected conductance before equivalence is represented by K = 18 + 0.60v, while after equivalence it is represented by K = 30 − 0.30v, where v is the volume of titrant in millilitres. What is the equivalence-point volume?
A.
10.0 mL
B.
13.3 mL
C.
20.0 mL
D.
30.0 mL
Correct Answer:
13.3 mL
Explanation:
The equivalence point is obtained from the intersection of the two extrapolated straight lines: 18 + 0.60v = 30 − 0.30v. Thus, 0.90v = 12 and v = 13.3 mL approximately.
6. Why is dilution correction often applied to conductance readings during a conductometric titration?
A.
To convert a weak electrolyte into a strong electrolyte
B.
To eliminate the effect of the increasing solution volume on the measured conductance
C.
To remove the ionic mobility of the titrant completely
D.
To ensure that the equivalence point always occurs at zero conductance
Correct Answer:
To eliminate the effect of the increasing solution volume on the measured conductance
Explanation:
Adding titrant increases the total volume and can lower the measured conductance by dilution even when the ionic composition favors an increase. Corrected conductance, often proportional to G(V + v), reduces this volume effect and gives a more reliable intersection of the two linear portions.
7. When hydrochloric acid is titrated conductometrically with ammonium hydroxide, the conductance falls up to equivalence and then changes only slightly. Which explanation is most appropriate?
A.
After equivalence, excess NH4OH is a weak electrolyte and produces relatively few additional ions
B.
After equivalence, chloride ions precipitate as ammonium chloride
C.
Before equivalence, H+ ions are replaced by still more mobile NH4+ ions
D.
Ammonium hydroxide completely dissociates after equivalence
Correct Answer:
After equivalence, excess NH4OH is a weak electrolyte and produces relatively few additional ions
Explanation:
Before equivalence, highly mobile H+ ions are replaced by less mobile NH4+ ions, causing a fall in conductance. Beyond equivalence, added ammonium hydroxide is weakly ionized, so its excess produces only a small conductance increase.
8. A strong acid is titrated with a strong base in a conductometric titration. If the titrant is added in equal increments, which observation most reliably identifies the equivalence point?
A.
The first reading obtained after adding titrant
B.
The point at which the raw conductance becomes exactly zero
C.
The intersection of the straight-line portions obtained before and after the conductance minimum
D.
The point at which the volume of titrant equals the initial acid volume
Correct Answer:
The intersection of the straight-line portions obtained before and after the conductance minimum
Explanation:
The equivalence point is located graphically by extrapolating the approximately linear conductance regions before and after equivalence. Their intersection gives the endpoint more accurately than selecting the lowest experimental reading.
9. In the conductometric titration of a strong acid with a strong base, the titrant is added in a large excess after equivalence. What is the principal effect on the endpoint determination from uncorrected conductance data?
A.
The post-equivalence slope may be distorted because increasing volume lowers conductance by dilution
B.
The equivalence point shifts because chloride ions are converted into hydroxide ions
C.
The acid becomes weak due to the dilution and no endpoint exists
D.
The conductance remains independent of volume because conductivity and conductance are identical quantities
Correct Answer:
The post-equivalence slope may be distorted because increasing volume lowers conductance by dilution
Explanation:
Measured conductance depends on both ionic composition and the volume-dependent concentration of ions. Excess titrant increases dilution, which can partially offset the rise caused by excess OH− and distort extrapolation unless a dilution correction is used.
10. A 25.0 mL sample of sulfuric acid is titrated conductometrically with 0.100 mol L−1 sodium hydroxide. The first equivalence point occurs at 12.5 mL of titrant and the second at 25.0 mL. Assuming both protons are titrated successively, what is the molarity of H2SO4?
A.
0.0250 mol L−1
B.
0.0500 mol L−1
C.
0.100 mol L−1
D.
0.200 mol L−1
Correct Answer:
0.0500 mol L−1
Explanation:
At the second equivalence point, moles of NaOH added are 0.100 × 0.0250 = 0.00250 mol. Each mole of H2SO4 consumes two moles of NaOH, so moles of H2SO4 are 0.00125 mol. Dividing by 0.0250 L gives 0.0500 mol L−1.
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