(a) State Kohlrausch law of independent migration of ions. Write an expression for the molar conductivity of acetic acid at infinite dilution according to Kohlrausch law.
(b) Calculate `Lambda_(m)^(@)` , for acetic acid.
Given that `Lambda_(m)^(@) (HCl) = 426 S cm^(2) mol^(-1)`
`Lambda_(m)^(@) (NaCl) = 126 S cm^(2) mol^(-1)`
`Lambda_(m)^(@) (CH_(3)COONa) = 91 S cm^(2) mol^(-1)`

Explanation

Kohlrausch law states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and the cation of the electrolyte. In expression for the molar conductivity at infinite dilution.
`Lambda_(m(NaCl))^(@) = lambda_(Na^(+))^(@) + lambda_(Cl^(-))^(@)`
In general, if an electrolyte or dissociation gives n+ cations and n- anions, than its limiting molar conductivity is given by
`Lambda_(m)^(@) =n_(+)lambda_(+)^(@) + n_(-)lambda_(-)^(@)`
`Lambda_(m)^(@)(CH_(3)COOH) =Lambda_(m)^(@) (CH_(3)COONa) + Lambda_(m)^(@)(HCl) - Lambda_(m)^(@) (NaCl)`
Given that:
`Lambda_(m)^(@) (CH_(3)COONa) = 91 S cm^(2) mol^(-1)`
`Lambda_(m)^(@) (HCl) = 426 S cm^(2) mol^(-1)`
`Lambda_(m)^(@) (NaCl) = 126 S cm^(2) mol^(-1)`
Substituting the values in the equation above, we have:
`Lambda_(m)^(@) (CH_(3)COOH) = 91 + 426 - 126 = 391S cm^(2) mol^(-1)`