Question

A conductivity cell when standardized with 0.01 M KCl was found to have a resistance of 189 Ω. With 0.01 M ammonia solution the resistance was 2460 Ω. Calculate the bas dissociation constant of ammonia, given the following molar cconductivities at these concentrations: λ(K⁺) = 7305 Ω^-1cm²mol^-1; λ(Cl^-) = 76.4 Ω^-1cm²mol^-1; λ(NH₄⁺) = 73.4 Ω^-1cm²mol^-1; λ(OH^-) = 198.6 Ω^-1cm²mol^-1.

Answer 1

A conductivity cell when standardized with 0.01 M KCl was found to have a resistance of 189 Ω. With 0.01 M ammonia solution the resistance was 2460 Ω. Calculate the bas dissociation constant of ammonia, given the following molar cconductivities at these concentrations:

λ(K⁺) = 7305 Ω^-1cm²mol^-1;

λ(Cl^-) = 76.4 Ω^-1cm²mol^-1;

λ(NH₄⁺) = 73.4 Ω^-1cm²mol^-1;

λ(OH^-) = 198.6 Ω^-1cm²mol^-1.

NH3 + H2O →NH4+ + OH-

(1 - α)C            αc           αc

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Kb = [NH4+][OH-]/[NH3]≈  

  = (αc)2/ (1 – α)c     = α2c/(1 – α)

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ℓim          α = 1

c → 0

NH3 + H2O →NH4+ + OH-(1 - α)cαcαcnKb [NH4+][OH-]/[NH3]≈    = (αc)2/ (1 – α)c     = α2c/(1 – α)nℓim α = 1 c → 0nα= NH3/ oNH3  (Arrhenius)

α = NH3/ oNH3  (Arrhenius)

Kb = α2c/(1 – α)

c is given,

KCl= (73.5+76.4) (0.01 mol/1000 )

=1.50 x10^3 /cm

(U/A) =(1.50 X10^10-5 CM^-1) 1 X1.89

=0.283CM^-1