Question

Molar conductivity of a weak acid HA at infinite dilution is 345.8 S cm2/mol. Calculate molar conductivity of 0.05 M HA solution. Given that alpha = 5.8 ?? 10-6.

Answer 1

Molar conductivity of a weak acid (monoprotic) is given by Ostwald's dilution law at low concentrations.

         1/Λ  =  1/Λ°   +  Λ * c / [Ka * (Λ° ) ]

Ka = activity constant (dissociation constant) = α² c
c = concentration  (Moles/L)
Λ = molar conductivity at  concentration c
Λ° = molar concentration at infinite dilution
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There is another formula for the molar conductivity of an acid:
     Λ = Λ° - K √c
  K = Kohransch constant and depends on substance and temperature.

Further, we also have that :

     α = degree of dissociation of HA into ions
        = Λ / Λ°
This is because, at infinite dilution, we assume that all the acid is 100% completely dissociated ie., α=1.  So for a dissociation of α :

   Λ = α * Λ°
      = 5.8 * 10⁻⁶ * 345.8 S cm² /mol
       = 2.00564 * 10⁻³  S cm² / mol

Actual Conductivity of 0.05 M HA solution :
    = 2.00564 * 10⁻³ S cm² /mol * 0.05 mol/1000 cm³ 
    = 1.00282 * 10⁻⁷ S / cm