For NaCl, HCl, and NaA with conductivities of 126.4, 425.9, and 100.5 S cm²mol⁻¹ respectively, if the conductivity of 0.001 M HA is [tex]5 \times 10^{-5}[/tex] S cm⁻¹, the degree of dissociation of HA is:

A) 0.02
B) 0.04
C) 0.06
D) 0.08

Question

High School

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Answer :

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Micheal156412 Micheal156412 · Answer 1 · 2024-01-16T11:49:54-05:00

For 0.001 M HA with a conductivity of 5 x 10⁻⁵ S cm⁻¹, the degree of dissociation is 0.02. The correct option is a) 0.02.

The degree of dissociation of HA can be calculated using the formula:

α = σ/(C × λ₀)

where:

For 0.001 M HA with a conductivity of 5 x 10⁻⁵ S cm⁻¹, we can substitute the values into the formula:

α = (5 x 10⁻⁵)/(0.001 x 100.5) = 0.02

Therefore, the degree of dissociation of HA is 0.02, which corresponds to option a) in the question.