Questions & explanations
1. In the conductometric titration of a mixture of 10 mL 0.1 M HCl and 10 mL 0.1 M CH3COOH with 0.1 M NaOH, what is the conductance (in arbitrary units) at the first equivalence point? Given: Λ₀(H⁺)=349.6, Λ₀(Na⁺)=50.1, Λ₀(Cl⁻)=76.3, Λ₀(CH3COO⁻)=40.9 S cm² mol⁻¹. Assume cell constant = 1 cm⁻¹ and complete dissociation of NaCl and CH3COONa.
- 40.9
- 76.3
- 50.1
- 126.4
Answer: 126.4
At the first equivalence point, all HCl is neutralised. Solution contains Na⁺ (50.1), Cl⁻ (76.3), and undissociated CH3COOH. The molar conductivity is due to NaCl only: Λ₀(Na⁺)+Λ₀(Cl⁻)=50.1+76.3=126.4 S cm² mol⁻¹. Concentration of NaCl is 0.05 M (dilution factor 2). Conductivity κ = 126.4 × 0.05 / 1000 = 0.00632 S cm⁻¹. With cell constant 1 cm⁻¹, conductance G = κ = 0.00632 S, but options are in same units as Λₘ; the question asks for conductance in arbitrary units proportional to Λₘ, so answer is 126.4.
2. In the conductometric titration of 20 mL of 0.1 M NH4OH with 0.1 M HCl, what is the conductance (in arbitrary units) at the equivalence point? Given: Λ₀(NH₄⁺) = 73.5, Λ₀(OH⁻) = 198.0, Λ₀(H⁺) = 349.6, Λ₀(Cl⁻) = 76.3 S cm² mol⁻¹. Assume cell constant = 1 cm⁻¹ and complete dissociation of NH4Cl.
- 76.3
- 149.8
- 73.5
- 198.0
Answer: 149.8
At equivalence, all NH4OH is converted to NH4Cl. The molar conductivity of NH4Cl is Λ₀(NH₄⁺) + Λ₀(Cl⁻) = 73.5 + 76.3 = 149.8 S cm² mol⁻¹. Since concentration is 0.05 M (dilution factor 2), conductivity κ = Λₘ × C / 1000 = 149.8 × 0.05 / 1000 = 0.00749 S cm⁻¹. With cell constant 1 cm⁻¹, conductance G = κ = 0.00749 S, but options are in same units as Λₘ; the question asks for conductance in arbitrary units proportional to Λₘ, so answer is 149.8.
3. The conductivity of a saturated AgCl solution is 3.4 × 10⁻⁶ S cm⁻¹ and that of water used is 1.6 × 10⁻⁶ S cm⁻¹. Given λ⁰(Ag⁺) = 61.9 and λ⁰(Cl⁻) = 76.3 S cm² mol⁻¹, what is the Ksp of AgCl?
- 1.7 × 10⁻¹⁰
- 3.4 × 10⁻¹⁰
- 1.7 × 10⁻⁸
- 3.4 × 10⁻⁸
Answer: 1.7 × 10⁻¹⁰
First, subtract water conductivity: κ_salt = 3.4×10⁻⁶ − 1.6×10⁻⁶ = 1.8×10⁻⁶ S cm⁻¹. Using Kohlrausch's law, Λ⁰(AgCl) = 61.9 + 76.3 = 138.2 S cm² mol⁻¹. For a dilute saturated solution, Λ ≈ Λ⁰. Solubility S = (κ_salt × 1000) / Λ⁰ = (1.8×10⁻⁶ × 1000) / 138.2 = 1.30×10⁻⁵ mol L⁻¹. For a 1:1 salt, Ksp = S² = (1.30×10⁻⁵)² = 1.69×10⁻¹⁰ ≈ 1.7×10⁻¹⁰.
4. In conductometric titration, the equivalence point is detected by:
- a colour change of the solution
- a sharp increase in pH of the solution
- a sudden change in the slope of the conductance vs volume graph
- the formation of a precipitate
Answer: a sudden change in the slope of the conductance vs volume graph
In conductometric titration, the equivalence point is detected by a sharp change in the slope of the conductance vs volume graph. This is because the titrant replaces one ion with another of different limiting molar conductivity, causing a linear change in conductance until the equivalence point, where the slope changes abruptly.
5. Iron rusts in acidic rainwater (pH 5) with O2 at 1 atm. Given E°(Fe2+/Fe) = -0.44 V and E°(O2/H2O) = +1.23 V, what is the cell EMF under these conditions?
- 1.37 V
- 1.67 V
- 1.52 V
- 1.23 V
Answer: 1.37 V
The cell reaction is 2Fe + O2 + 4H+ → 2Fe2+ + 2H2O. E°cell = 1.23 - (-0.44) = 1.67 V. Using Nernst equation: E = E° - (0.0591/4) log([Fe2+]^2/([H+]^4 P_O2)). With [Fe2+] = 1 M, P_O2 = 1 atm, [H+] = 10^-5 M, log term = log(1/(10^-20)) = 20, so (0.0591/4)*20 = 0.2955 V. Thus E = 1.67 - 0.2955 = 1.3745 V ≈ 1.37 V.
6. What happens to the EMF of a Daniell cell when water is added to the anode compartment (ZnSO₄ solution)?
- EMF decreases
- EMF increases
- EMF remains unchanged
- EMF becomes zero
Answer: EMF increases
In a Daniell cell, the cell reaction is Zn + Cu²⁺ → Zn²⁺ + Cu. The Nernst equation is E = E° - (0.0591/2) log([Zn²⁺]/[Cu²⁺]). Adding water to the anode dilutes Zn²⁺, decreasing the ratio [Zn²⁺]/[Cu²⁺]. Since log of a smaller number is smaller, subtracting a smaller term increases E. Hence EMF increases.
7. 100 mL mixture of HCl and CH₃COOH is titrated with 0.20 M NaOH. Conductometric plot shows first break at 12.5 mL and second at 35.0 mL. What is the molarity of CH₃COOH?
- 0.035 M
- 0.025 M
- 0.070 M
- 0.045 M
Answer: 0.045 M
First break at 12.5 mL neutralizes HCl. Second break at 35.0 mL corresponds to total neutralization of both acids. Volume for CH₃COOH = 35.0 - 12.5 = 22.5 mL. Moles NaOH = 0.20 × 0.0225 = 0.0045 mol. Since 1:1 reaction, moles CH₃COOH = 0.0045 mol in 0.100 L, so molarity = 0.0045/0.100 = 0.045 M.
8. In the conductometric titration of 20 mL of 0.1 M NaCl with 0.1 M AgNO3, what is the approximate conductance (in arbitrary units) halfway to the equivalence point? Given: Λ₀(Na⁺)=50.1, Λ₀(Cl⁻)=76.3, Λ₀(Ag⁺)=61.9, Λ₀(NO₃⁻)=71.4 S cm² mol⁻¹. Assume cell constant = 1 cm⁻¹ and complete dissociation.
- 126.4
- 121.5
- 133.3
- 138.2
Answer: 133.3
Halfway to equivalence, half of Cl⁻ is precipitated. Ions present: Na⁺ (Λ₀=50.1), remaining Cl⁻ (half of 76.3=38.15), and added NO₃⁻ (half of 71.4=35.7). Sum of molar conductivities = 124.0 S cm² mol⁻¹. Conductance in arbitrary units is proportional to this sum, so the closest option is 126.4.
9. A conductivity cell calibrated with 0.1 M KCl (κ = 1.29×10⁻² S cm⁻¹) gives R = 100 Ω. With 0.0010 M CH₃COOH, R = 2.47×10⁴ Ω. Given λ⁰(H⁺)=349.6, λ⁰(CH₃COO⁻)=40.9 S cm² mol⁻¹, what is Kₐ of CH₃COOH?
- 3.6×10⁻⁵ M
- 1.8×10⁻⁵ M
- 9.0×10⁻⁶ M
- 7.2×10⁻⁵ M
Answer: 1.8×10⁻⁵ M
Cell constant G* = κ_std / G = 1.29×10⁻² / (1/100) = 1.29 cm⁻¹. For CH₃COOH, G = 1/2.47×10⁴ = 4.05×10⁻⁵ S, κ = G × G* = 5.22×10⁻⁵ S cm⁻¹. Λₘ = κ×1000/C = 52.2 S cm² mol⁻¹. Λₘ⁰ = 349.6+40.9 = 390.5 S cm² mol⁻¹. α = Λₘ/Λₘ⁰ = 0.1337. Kₐ ≈ Cα² = 0.0010×(0.1337)² = 1.79×10⁻⁵ M ≈ 1.8×10⁻⁵ M.
10. Which of the following statements about the effect of temperature on conductivity is correct?
- Conductivity of metals decreases while that of electrolytes increases with temperature.
- Conductivity of metals increases while that of electrolytes decreases with temperature.
- Conductivity of both metals and electrolytes increases with temperature.
- Conductivity of both metals and electrolytes decreases with temperature.
Answer: Conductivity of metals decreases while that of electrolytes increases with temperature.
In metals, increasing temperature increases lattice vibrations, scattering electrons and decreasing conductivity. In electrolytes, increasing temperature decreases viscosity and increases ionic mobility, and for weak electrolytes also increases dissociation, so conductivity increases.
11. Which of the following water samples has the highest specific conductivity at 298 K?
- Deionised water
- Seawater
- Tap water
- Distilled water
Answer: Seawater
Specific conductivity increases with the concentration of dissolved ions. Seawater contains the highest amount of dissolved salts (NaCl, MgCl₂, etc.), giving it the highest conductivity among the options. Deionised water has the lowest, followed by distilled water, then tap water.
12. During discharge of a lead-acid battery, the specific gravity of H2SO4 decreases. What is the reason?
- PbSO4 dissolves in the acid, increasing volume.
- H2SO4 is produced and water is consumed.
- H2SO4 is consumed and water is produced.
- The temperature rises, expanding the electrolyte.
Answer: H2SO4 is consumed and water is produced.
The overall discharge reaction is Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O. H2SO4 is consumed and water is produced, diluting the acid and decreasing its specific gravity. A hydrometer reading of ~1.10 g/mL indicates a discharged battery, while ~1.28 g/mL indicates a fully charged one.