Q5(c). Equivalent conductance of NaCl from ionic mobilities

Given: u₊=5.19×10⁻⁸, u₋=7.91×10⁻⁸ m² V⁻¹ s⁻¹ (“v” is the volt); F=96 500 C mol⁻¹.

1. Relation: an ion of charge z and mobility u in a field E moves at s=uE and carries zF coulombs per mole, so its molar conductivity is λ=zuF. For a 1:1 salt one equivalent is one mole, and the law of independent migration gives Λ=F(u₊+u₋).
2. λ(Na⁺)=96 500×5.19×10⁻⁸=5.008×10⁻³ S m² mol⁻¹ = 50.1 S cm² mol⁻¹
3. λ(Cl⁻)=96 500×7.91×10⁻⁸=7.633×10⁻³ S m² mol⁻¹ = 76.3 S cm² mol⁻¹
4. Λ=96 500×(5.19+7.91)×10⁻⁸, Λ_(eq)=1.264×10⁻²S m² eq⁻¹=126.4S cm² eq⁻¹
5. Units: C mol⁻¹ × m² V⁻¹ s⁻¹ = (A s)(m²)/(V s mol) = S m² mol⁻¹.
6. Transport number: t₊=u₊/(u₊+u₋)=0.396: chloride carries about 60 % of the current.
7. Concentration: these mobilities are the limiting values in water at 25°C, so the result is really Λ°. In 0.1 M NaCl ion–ion interactions lower the mobilities, and the measured value is smaller (Kohlrausch: Λ=Λ°-K√c). With the data given, the value taken for 0.1 M is 126.4 S cm² eq⁻¹, and the conductivity is κ=Λ c=1.264×10⁻²×100=1.26 S m⁻¹.

Application. the same additivity gives Λ° of weak electrolytes (acetic acid from HCl, NaAc and NaCl), and hence the degree of dissociation α=Λ/Λ° in conductometric work.