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19/01/2022 6:03 pm
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Conductivity of 0.00241 M acetic acid is 7.896 × 10-5 S cm-1. Calculate its molar conductivity and if acetic acid is 390.5 S cm2 mol-1, what is its dissociation constant?
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19/01/2022 6:26 pm
Given, ΔG° = 7.896 × 10-5 S m-1
c = 0.00241 mol L-1
Then, molar conductivity, Λm = \(\frac{K}{c}\)
= \(\frac{7.896 \times 10^{-5} S cm^{-1}}{0.00241\ mol \; L^{-1}}\) x \(\frac{1000 \ cm^3}{L}\)
= 32.76S cm2 mol-1
Again, Λ0m
= 390.5 S cm2 mol-1
α = \(\frac{Λ_m}{Λ_m^0}\)
= \(\frac{32.76 \ S cm^2 \ mol^{-1}}{390.5 \ S cm^2 \ mol^{-1}}\)
Now,
= 0.084
∴ Dissociation constant, Ka = \(\frac{c \alpha^2}{(1 - \alpha)}\)
= \(\frac{(0.00241 \ mol L^{-1})(0.084)^2}{(1 - 0.084)}\)
= 1.86 x 10-5 mol L-1