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Electrochemistry

Question

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The conductivity of 0.001 M acetic acid is 4 x 10^–5 s/cm. Calculate the dissociation constant of acetic acid, if λ°_m for acetic acid is 390.5 s cm² mol.

Solution

concentration =0.001M
conductivity =4 x10^-5 s/cm

\land = \frac{k \times 1000}{M} k = 4 \times 10^{- 5} s / cm M = 0.001 M \land = \frac{4 \times 10^{- 5} \times 1000}{0.001} = 4 \times 10^{- 5} \times 10^{3} \times 10^{3} = 40 S {cm}^{2} {mol}^{- 1} . \land_{0} = 390.5 S {cm}^{2} / mol α = \frac{\land}{\land_{0}} = \frac{40}{390.5} = 0.102 K = {Cα}^{2} = 0.001 \times (0.102)^{2} = 1.04 \times 10^{- 5} .

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