Conductivity of 0.00241 M acetic acid is 7.896 x 10^–5 S cm^–1. Calculate its molar conductivity, if Λ° for acetic acid is 390.5 S cm² mol^–1, what is its dissociation constant?

Solution

We have given that
Concentration = 0.00241M
Conductivity =0.00241S Cm^-1
thus by using the formula
we get

\land = K \times \frac{1000}{C} C = 0.00241 M K = 7.896 \times 10^{- 5} S {cm}^{- 1} \land = \frac{7.896 \times 10^{- 5} \times 1000}{0.00241} = 32.76 α = \frac{\land}{\land_{0}} = \frac{32.76}{390.5} = 0.084 K = \frac{{Cα}^{2}}{1 - α} = \frac{0.00241 \times (0.084)^{2}}{(1 - 0.084)} = 1.856 \times 10^{- 5} .

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