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Question
Calculate the degree of dissociation (α) of acetic acid if its molar conductivity (Λm) is 39.05 S cm2 mol−1.
(Given \[\ce{\lambda^{\circ}_{(H^+)}}\] = 349.6 S cm2 mol−1 and \[\ce{\lambda^{\circ}_{(CH_3COO^-)}}\] = 40.95 S cm2 mol−1)
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Solution
The degree of dissociation of a weak electrolyte is given by the following relation:
α = `Lambda_m/Lambda^circ` ... (1)
Where,
α = Degree of dissociation of the electrolyte
Λm = Molar conductivity
Λ° = Molar conductivity at infinite dilution
We are given that the ionic molar conductivity at infinite dilution of acetate and hydrogen ions are 349.8 and 40.9 S cm2 mol−1, respectively.
Hence, the limiting molar conductivity of acetic acid would be written as:
\[\ce{\Lambda^{\circ}_{\text{acetic acid}} = \lambda^{\circ}_{(H^+)} + \lambda^{\circ}_{(CH_3COO^-)}}\]
= 40.9 + 349.8
= 390.7 cm2 mol−1
Now, since we are given that the molar conductivity of acetic acid is:
Λm = 39.05 S cm2 mol−1
Therefore, substituting the values in equation (1), we get:
Degree of dissociation of acetic acid (α) = `Lambda_m/Lambda^circ`
= \[\ce{\frac{39.05}{390.7}}\]
= 0.0999
≅ 0.1
Hence, the degree of dissociation is 0.1.
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