Question
Download Solution PDFΛm° for NaCl, HCl, and NaA is 126.4, 425.9, and 100.5 S cm2mol-1, respectively. If the conductivity of 0.001 M HA is 5 × 10-5 S cm-1, the degree of dissociation of HA is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
Kohlrausch's law states that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the anions and cations. If a salt is dissolved in water, the conductivity of the solution is the sum of the conductances of the anions and cations.
Calculation:
According to Kohlrausch’s law, the molar conductivity of HA at infinite dilution is given as,
= 425.9 + 100.5 – 126.4
= 400 S cm2 mol-1
Also, molar conductivity (Λm°) at given concentration is given as,
Given, k = conductivity ⟹ 5 × 10-5 S cm-1
M = Molarity ⟹ 0.001 M
= 50 S cm2 mol-1
Therefore, degree of dissociation (α), of HA is,
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