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ChemistryEdu Logo Electrochemistry | Questions based on conductivity#

Questions#

Resistance of a conductivity cell filled with 0.1 molL-1 KCl solution is 100 Ω. If the resistance of the same cell filled with 0.02 molL-1 KCl solution is 520 Ω, calculate the conductivity and molar conductivity of 0.02 molL-1 KCl solution. The conductivity of 0.1 molL-1 KCL solution is 1.29 S/m.

  • Cell constant of 0.1 molL-1 KCl solution:
\[G^* = Κ \times R\]
\[G^* = 1.29 S/m \times 100 Ω\]
\[G^* = 129 m^{-1}\]
  • Conductivity of 0.02 molL-1 KCl solution:
\[Κ = {G^* \over R}\]
\[Κ = {129 \over 520} = 0.248 Sm^{-1}\]
  • Concentration, C:
\[C = 0.02\ mol\ L^{-1} = 20\ mol\ m^{-3}\]
  • Molar conductivity of 0.02 molL-1 KCl solution:
\[λ_m = {Κ \over C}\]
\[λ_m = {0.248 \over 20}\]
\[λ_m = 124 \times 10^{-4}\ Sm^2mol^{-1}\]

The electrical resistance of a column of 0.05 molL-1 NaOH solution of diameter 1 cm and length 50 cm is 5.55 x 103 ohm. Calculate its resistivity, conductivity and molar conductivity.

  • Resistivity, ρ is given by:
\[ρ = {RA \over l}\]
\[ρ = {5.55 \times 10^3 \times π \times (10^{-2})^2 \times 25 \over 50 \times 10^{-2} \times 100}\]
\[ρ = 0.87135\ Ωm = 87.135\ Ωcm\]
  • Conductivity, Κ is given by:
\[Κ = {1 \over ρ}\]
\[Κ = {1 \over 87.135}\ Scm^{-1} = 0.01148\ Scm^{-1}\]
  • Molar conductivity, λm is given by:
\[λ_m = {Κ \over C}\]
\[λ_m = {0.01148 \times 1000 \over 0.05}\]
\[λ_m = 229.6\ Scm^{2}mol^{-1}\]

Calculate λ0m for CaCl2 and MgSO4 if:
\(λ^0_{(Ca^+)} = 119.0\ Scm^2mol^{-1}\)
\(λ^0_{(Cl^-)} = 76.3\ Scm^2mol^{-1}\)
\(λ^0_{(Mg^{2+})} = 106.0\ Scm^2mol^{-1}\)
\(λ^0_{({SO_4}^{2-})} = 160.0\ Scm^2mol^{-1}\)

λ0m for CaCl2 and MgSO4 can be calculated from Kohlraush law:

\[(a)\ CaCl_2 → Ca^{2+} + 2Cl^-\]
\[ λ^0_{m(CaCl_2)} = λ^0_{m(Ca^{2+})} + 2 \times λ^0_{m(Cl^-)}\]
\[ λ^0_{m(CaCl_2)} = 119.0 + 2 \times 76.3\]
\[ λ^0_{m(CaCl_2)} = 271.6\ Scm^2mol^{-1}\]
\[(b)\ MgSO_4 → Mg^{2+} + SO_4^{2-}\]
\[ λ^0_{m(MgSO_4)} = λ^0_{m(Mg^{2+})} + λ^0_{m(SO_4^{2-})}\]
\[ λ^0_{m(MgSO_4)} = 106 + 160\]
\[ λ^0_{m(MgSO_4)} = 266\ Scm^2mol^{-1}\]

λ0m for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 Scm2mol-1 respectively. Calculate λ0m for HAc.

\[ λ^0_{m(HAc)} = λ^0_{m(H^{+})} + λ^0_{m(Ac^-)}\]
\[ λ^0_{m(HAc)} = λ^0_{m(H^{+})} + λ^0_{m(Cl^{-})} + λ^0_{m(Ac^-)} + λ^0_{m(Na^+)} - λ^0_{m(Cl^-)} - λ^0_{m(Na^+)}\]
\[ λ^0_{m(HAc)} = λ^0_{m(HCl)} + λ^0_{m(NaAc)} - λ^0_{m(NaCl)}\]
\[ λ^0_{m(HAc)} = (425.9 + 91.0 - 126.4) Scm^2mol^{-1}\]
\[ λ^0_{m(HAc)} = 390.5\ Scm^2mol^{-1}\]

The conductivity of 0.001028 molL-1 acetic acid is 4.95 x 10-5 Scm-1. Calculate its dissociation constant if λ0m for acetic acid is 390.5 Scm2mol-1.

\[ λ_m = {Κ \over C} \]
\[ λ_m = {4.95 \times 10^{-5} \over 0.001028} \]
\[ λ_m = 48.15\ Scm^2mol^{-1} \]

Degree of dissociation, α:

\[ α = {λ_m \over λ^0_m} \]
\[ α = {48.15 \over 390.5} = 0.1233 \]
\[ Κ = {Cα^2 \over (1-α)} \]
\[ Κ = {0.001028 \times (0.1233)^2 \over (1-0.1233)} \]
\[ Κ = 1.78 \times 10^{-5}\ molL^{-1}\]

The molar conductivity of 0.025 molL-1 methanoic acid is 46.1 Scm2mol-1. Calculate its degree of dissociation and dissociation constant. Given λ0m(H+) = 349.6 Scm2mol-1 and λ0m(HCOO-1) = 54.6 Scm2mol-1.

\[ λ_{m(HCOOH)} = (349.6 + 54.6)\ Scm^2mol^{-1} \]
\[ λ_{m(HCOOH)} = 404.2\ Scm^2mol^{-1} \]

Degree of dissociation:

\[ α = {λ_m \over λ^0_m}\]
\[ α = {46.1 \over 404.2} = 0.114\]

Dissociation constant:

\[ K_a = {Cα^2 \over (1-α)}\]
\[ K_a = {0.025 \times (0.114)^2 \over (1-0.114)}\]
\[ K_a = 3.67 \times 10^{-4}\ molL^{-1}\]

Suggest a way to determine the λ0m value of water.

Molar conductivity of water can be calculated using the below method:

\[ H_2O ⇋ H^+ + OH^- \]
\[ λ_{m(H_2O)} = λ_{m(H^+)} + λ_{m(OH^-)} \]
\[ λ_{m(H_2O)} = λ_{m(H^+)} + λ_{m(Cl^-)} + λ_{m(Na^+)} + λ_{m(OH^-)} - (λ_{m(Na^+)} + λ_{m(Cl^-)}) \]
\[ λ_{m(H_2O)} = λ_{m(HCl)} + λ_{m(NaOH)} - λ_{m(NaCl)} \]

Why does the conductivity of a solution decrease with dilution?

The conductivity of a solution is related with the number of ions present per unit volume of the solution. When the solution is diluted, the number of ions per unit volume of solution decreases. Hence, conductivity of the solution also decreases.