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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper I covers physical and inorganic chemistry.

UPSC Civil Services (Mains), Chemistry optional · 2017 · Paper I

Q6(c). Molar conductivity, drift velocity and transport numbers from mobilities

  • 15 marks
  • Answered
Question

In aqueous solution, at 298 K, the mobility of NH₄⁺ ion is 7.62×10⁻⁸ m² s⁻¹ V⁻¹. Calculate its (i) molar conductivity; (ii) velocity if 15.0 V is applied across the electrodes of 25.0 cm apart. Calculate the transport numbers of the ions in CH₃COONH₄ (aq) solution if the mobility of CH₃COO⁻ ion is 4.24×10⁻⁸ m² s⁻¹ V⁻¹ under same condition.

Answer in brief

(i) λ(NH₄⁺)=7.35×10⁻³ S m² mol⁻¹ (73.5 S cm² mol⁻¹); (ii) v=4.57×10⁻⁶ m s⁻¹; t₊=0.642, t₋=0.358.

Key points

  • (i) Molar ionic conductivity. An ion of mobility u and charge z contributes λ=zuF: λ(NH₄⁺)=(1)(7.62×10⁻⁸)(96500) λ=7.35×10⁻³S m² mol⁻¹=73.5S cm² mol⁻¹ (units: m² s⁻¹ V⁻¹× C mol⁻¹ = m² A V⁻¹ mol⁻¹ = S m² mol⁻¹.)
  • (ii) Drift velocity. Field strength E=Δφ/l=15.0/0.250=60.0 V m⁻¹; v=uE: v=(7.62×10⁻⁸)(60.0) v=4.57×10⁻⁶m s⁻¹ That is about 4.6 μm per second: the ion would need about 15 hours to cross the 25 cm.
  • Transport numbers. The fraction of current carried by each ion is proportional to its mobility (same |z|, same concentration): t₊=u₊u₊+u₋=(7.62)/(7.62+4.24)=(7.62)/(11.86) t₊(NH₄⁺)=0.642 t₋(CH₃COO⁻)=1-t₊=0.358
  • λ(CH₃COO⁻)=4.09×10⁻³ S m² mol⁻¹; Λ°(CH₃COONH₄)=λ₊+λ₋=1.144×10⁻² S m² mol⁻¹, and t₊=λ₊/Λ°=0.642 again.

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

CONDUCTIVITYλ=zuFVELOCITYv=uETRANSPORTt_±=u_±/(u₊+u₋)

Given: u₊=7.62×10⁻⁸, u₋=4.24×10⁻⁸ m² s⁻¹ V⁻¹; z=1; F=96500 C mol⁻¹; 15.0 V over 0.250 m.

  1. (i) Molar ionic conductivity. An ion of mobility u and charge z contributes λ=zuF: λ(NH₄⁺)=(1)(7.62×10⁻⁸)(96500) λ=7.35×10⁻³S m² mol⁻¹=73.5S cm² mol⁻¹ (units: m² s⁻¹ V⁻¹× C mol⁻¹ = m² A V⁻¹ mol⁻¹ = S m² mol⁻¹.)
  2. (ii) Drift velocity. Field strength E=Δφ/l=15.0/0.250=60.0 V m⁻¹; v=uE: v=(7.62×10⁻⁸)(60.0) v=4.57×10⁻⁶m s⁻¹ That is about 4.6 μm per second: the ion would need about 15 hours to cross the 25 cm.
  3. Transport numbers. The fraction of current carried by each ion is proportional to its mobility (same |z|, same concentration): t₊=u₊u₊+u₋=(7.62)/(7.62+4.24)=(7.62)/(11.86) t₊(NH₄⁺)=0.642 t₋(CH₃COO⁻)=1-t₊=0.358
  4. Check: λ(CH₃COO⁻)=4.09×10⁻³ S m² mol⁻¹; Λ°(CH₃COONH₄)=λ₊+λ₋=1.144×10⁻² S m² mol⁻¹, and t₊=λ₊/Λ°=0.642 again.
  5. The values are limiting (infinite dilution) quantities. The large organic acetate ion moves more slowly than NH₄⁺, whose mobility is almost that of K⁺ (7.62×10⁻⁸ m² s⁻¹ V⁻¹).

Application. ammonium and potassium ions have nearly equal mobilities, which is why KCl or NH₄NO₃ is used in salt bridges: the liquid-junction potential is then very small.

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Model answer to Q6(c), 2017 Paper I: solution page 27
Solution page 27 of 38
Read the whole 2017 solution2017 question paper

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