Conductivity of 0.00241 M acetic acid is 7.896 × 10^{−5} S cm^{−1}. Calculate its molar conductivity and if `Lambda_m^0` for acetic acid is 390.5 S cm^{2} mol^{−1}, what is its dissociation constant?

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#### Solution

Given, *κ* = 7.896 × 10^{−5} S m^{−1}

c = 0.00241 mol L^{−1}

Then, molar conductivity, `Lambda_m = k/c`

= `(7.896xx10^(-5) S cm^(-1))/(0.00241 " mol L"^(-1))xx1000 cm^(3)/L`

= 32.76S cm^{2} mol^{−1}

Again, `Lambda_m^0` = 390.5 S cm^{2} mol^{−1}

Now, alpha = `Lambda_m/Lambda_m^0` = `(32.76 " S ""cm"^2""mol^(-1))/(390.5 " S cm"^2"" mol^(-1))`

= 0.084

∴Dissociation constant ,` K_alpha = (calpha^2)/(1-alpha)`

=`((0.00241 "mol L"^(-1))(0.084)^2)/((1-0.084))`

= 1.86 × 10^{−5} mol L^{−1}

Concept: Conductance of Electrolytic Solutions - Variation of Conductivity and Molar Conductivity with Concentration

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