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Exercise · Q22

Q.For a 0.01 M0.01\ \text{M} solution of acetic acid, the molar conductivity Λm\Lambda_m is 16.18 S cm2 mol−116.18\ \text{S cm}^2\ \text{mol}^{-1}. Taking Λm0(CH3COOH)=390.5 S cm2 mol−1\Lambda_m^{0}(\text{CH}_3\text{COOH}) = 390.5\ \text{S cm}^2\ \text{mol}^{-1} (as obtained by Kohlrausch's law), calculate the degree of dissociation, α\alpha, of acetic acid at this concentration.

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For a weak electrolyte, the degree of dissociation at a given concentration can be estimated as the ratio of its molar conductivity at that concentration to its limiting (infinite-dilution) molar conductivity: α=Λm(c)/Λm0\alpha = \Lambda_m(c)/\Lambda_m^{0}. The reasoning is that Λm0\Lambda_m^{0} represents the conducting power of one mole of electrolyte if it were 100% dissociated, while Λm(c)\Lambda_m(c) reflects only the fraction that is actually dissociated at concentration cc (undissociated molecules contribute essentially nothing to conduction). Substituting the given values, α=16.18390.5=0.04144\alpha = \dfrac{16.18}{390.5} = 0.04144, i.e. about 4.14%4.14\% of the acetic acid molecules are dissociated into ions a …

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