Question

Why molar conductivity of acetic acid cannot
be obtained by plotting lambda M versus
under root c at infinite dilution ? Limiting
molar conductivity of NaCl, HCl and
CH,COONa are are 126.4, 425.9 and 91.05 cm?
mol- respectively. Calculate limiting molar
conductivity for HAC.
[4 marks)​

Answer 1

The limiting molar  conductivity for acetic acid (HAc) is  390.55 Scm²mol⁻¹

- at infinite dilution , electrolyte dissociates completely but at such low concentration the conductivity of the solution is so low that it cannot be measures accurately. Hence, the molar conductivity of acetic acid cannot  be obtained by plotting lambda M versus  $\sqrt{c}$ at infinite dilution.

- Using Kohlrausch's law, we have

λ°(NaCl) = λ°(Na⁺) +  λ°(Cl⁻)       ----1

  λ°(HCl) = λ°(H⁺) +  λ°(Cl⁻)         ----2

λ°(AcNa) = λ°(Na⁺) +  λ°(Ac⁻)      ----3

 λ°(HAc) = λ°(H⁺) +  λ°(Ac⁻)        ----4

- on adding 2 and 3 equation and subtracting 1 from it, we get equation 4

- so we have

λ°(HAc) = λ°(HCl) + λ°(AcNa) - λ°(NaCl)

λ°(HAc) =  425.9 + 91.05 - 126.4

λ°(HAc) =  390.55 Scm²mol⁻¹