Conductivity of 0.00241 M acetic acid is 7.896 × 10–5 S cm–1. Calculate its molar conductivity. If 0m for acetic acid is 390.5 S cm2mol–1, what is its dissociation constant?

Given -


Molarity, C = 0.00241 M


Conductivity, κ = 7.896 × 10–5 S cm–1


Molar conductivity, m = ?


for acetic acid = 390.5 S cm2mol–1


Molar conductivity, m = S cm2 mol-1



m = 32.76 S cm2 mol-1


To calculate the dissociation constant, Ka, we use


Ka = Equation 1


Here, we need to find the value of α (degree of dissociation), by the formula,




α = 8.4×10-2 Equation 2


Thus, substituting Equation 2 in Equation 1, we get,


Ka =


=


Ka = 1.86×10 - 5


The molar conductivity, m is 32.76 S cm2 mol-1 and the dissociation constant, Ka is 1.86×10-5


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