The conductivity of 0.001028 mol l^{-1} acetic acid is 4.94*10^{-5} S cm^{-1} .can you calculate its dissociation constant if Λ ^{0} _{m} foe acetic acid is 390.5 S cm^{2} mol^{-1}.

The degree of dissociation is calculated by the following relation

α = Λ/ Λ^{0}

Where Λ is the molar conductivity and Λ^{0} is the limiting molar conductivity of the solution.

Now Λ = к/c = [к (Scm^{-1}) X 1000cm^{3}L^{-1}]/ molarity (molL^{-1})

where к = conductivity of solution and c is the concentration of solution

Given к = 4.94*10^{-5} S cm^{-1} and c = 0.001028 molL^{-1}

Therefore

Λ = [4.94 X 10^{-5} X 1000] / 0.001028 Scm^{-2}mol^{-1}

= 48.054 Scm^{-2}mol^{-1}

Given Λ^{0} = 390.5 Scm^{-2}mol^{-1}

therefore α = Λ/ Λ^{0} = 48.054/390.5 = 0.123

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