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For NaCl, HCl, and NaA, the molar conductivities are 126.4, 425.9, and 100.5 S cm² mol⁻¹, respectively. If the conductivity of a 0.001 M HA solution is [tex]5 \times 10^{-5} \, \text{S cm}^{-1}[/tex], what is the degree of dissociation of HA?

Answer :

Final answer:

The degree of dissociation of HA can be calculated using the formula α = √(k * C), where α is the degree of dissociation, k is the specific conductivity, and C is the concentration of the electrolyte. Given the specific conductivity of HA as 5 x 10^-5 S cm–1 and the concentration as 0.001 M, the degree of dissociation is 0.5. The correct option is d.

Explanation:

The degree of dissociation of a weak electrolyte can be calculated using the formula:

α = √(k* C)

Where α is the degree of dissociation, k is the specific conductivity, and C is the concentration of the electrolyte.

Given the specific conductivity of HA as 5 x 10^-5 S cm–1 and the concentration as 0.001 M, we can substitute the values into the formula to solve for α:

α = √((5 x 10^-5) * 0.001)

= 0.005

Therefore, the degree of dissociation of HA is 0.005, which is equal to 0.5.

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