Question

Calculate ∧°m for acetic acid and its degree of dissociation (α) if its molar conductivity is 48.1 Ω⁻¹ cm² mol⁻¹.

Given that

`∧_m^°` (HCl) = 426 Ω⁻¹ cm² mol⁻¹

`∧_m^°` (NaCl) = 126 Ω⁻¹ cm² mol⁻¹

`∧_m^°` (CH₃COONa) = 91 Ω⁻¹ cm² mol⁻¹

Answer 1

Given: Molar conductivity = 48.1 Ω⁻¹ cm² mol⁻¹

According to kohlrausch law,

\[\ce{HCl -> H+ + Cl-}\]

\[\ce{∧°_m of HCl -> ∧°_m of H+ + ∧°_m of Cl-}\]

426 = `∧_m^°` of H⁺ + `∧_m^°` of Cl⁻ 

426 = `∧_m^°` of HCl    ...(i)

Similarly,

91 = `∧_m^°` of CH₃COO⁻ + `∧_m^°` of Na⁺

91 = `∧_m^°` of CH₃COONa    ...(ii)

126 = `∧_m^°` of Na⁺ + `∧_m^°` of Cl⁻   

126 = `∧_m^°` of Na⁺  ...(ii)

We get eq (i) + eq (ii) − eq (iii)

`∧_m^° (CH_3COOH) = ∧_m^° (HCl) + ∧_m^° (CH_3COONa) - ∧_m^° (NaCl)`

= 426 + 91 − 126

= 391 Ω⁻¹ cm² mol⁻¹

α = `"molar conductivity"/"limiting molar conductivity"`

= `48.1/391`

= 0.123