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For NaCl, HCl, and NaA, the conductivities are 126.4, 425.9, and 100.5 S cm² mol⁻¹, respectively. If the conductivity of 0.001 M HA 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 is calculated using the formula α = κ / κ₀, where κ is the measured conductivity and κ₀ is the conductivity of the completely dissociated acid. The value is found to be 4.98 x 10⁻⁷, indicating that HA is a weak electrolyte.

Explanation:

The degree of dissociation of HA can be determined by comparing the conductivity of the weak acid with the conductivity of its fully dissociated ions. First, we must calculate the molar conductivity of the dissociated ions of HA using its salt form, NaA. Given the conductivity of NaA is 100.5 S cm² mol⁻¹, since the concentration of HA is 0.001 M, the molar conductivity of NaA at this concentration would also be 100.5 S cm² mol⁻¹ (assuming complete dissociation).

Next, the conductivity of 0.001 M HA is given as 5 x 10⁻⁵ S cm⁻¹. Therefore, we can calculate the degree of dissociation (α) of HA via the equation α = κ / κ₀, where κ is the measured conductivity (5 x 10⁻⁵ S cm⁻¹) and κ₀ is the conductivity of the completely dissociated acid (100.5 S cm² mol⁻¹).

The degree of dissociation (α) is then:

α = (5 x 10⁻⁵ S cm⁻¹) / (100.5 S cm² mol⁻¹) = 4.98 x 10⁻⁷

This result indicates that HA is a weak electrolyte and does not dissociate completely in solution.

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