Question

Molar conductivity of acetic acid at infinite dilution if limiting molar conductivity for nacl hcl ch3coona are 126.45 426.16 91

Answer 1

HCL+CH3COONA - NACL= CH3COOH
426.16+91-126.45= 390.71

Answer 2

Answer: The molar conductivity of acetic acid will be $390.71Scm^2mol^{-1}$

Explanation:

Acetic acid is a weak electrolyte because it does not completely dissociates into its ions when dissolved in water.

Molar conductivity of a weak electrolyte at infinite dilution is calculated by using Kohlrausch's Law. This law states that equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductance of the anions and cations.

For calculating the molar conductivity of acetic acid at infinite dilution:

$\Lambda^o_{CH_3COO^-}+\Lambda^o_{H^+}=(\Lambda^o_{CH_3COO^-}+\Lambda^o_{Na^+})+(\Lambda^o_{H^+}+\Lambda^o_{Cl^-})-(\Lambda^o_{Na^+}+\Lambda^o_{Cl^-})\\\\\Lambda^o_{CH_3COOH}=\Lambda^o_{CH_3COONa}+\Lambda^o_{HCl}-\Lambda^o_{NaCl}$

We are given:

$\Lambda^o_{CH_3COONa}=91Scm^2mol^{-1}\\\Lambda^o_{HCl}=426.16Scm^2mol^{-1}\\\Lambda^o_{NaCl}=126.45Scm^2mol^{-1}$

Putting values in above equation, we get:

$\Lambda^o_{CH_3COOH}=[(91+426.16)-126.45]Scm^2mol^{-1}\\\\\Lambda^o_{CH_3COOH}=390.71Scm^2mol^{-1}$

Hence, the molar conductivity of acetic acid will be $390.71Scm^2mol^{-1}$