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 0.001 M HA is [tex]5 \times 10^{-5} \, \text{S cm}^{-1}[/tex], the degree of dissociation of HA is:

A. 0.10
B. 0.15
C. 0.20
D. 0.25

The degree of dissociation of HA is calculated by comparing its molar conductivity with that of a fully dissociated salt. By dividing the molar conductivity of 0.001 M HA by the molar conductivity of NaA and adjusting for the fact that HA is a weak electrolyte, we estimate the degree of dissociation to be close to 0.25.

Explanation:

To determine the degree of dissociation (α) of HA, we can use the formula for molar conductivity at a given concentration (Λ), which is the conductivity (κ) divided by the concentration (c), and then compare it to the molar conductivity of a fully dissociated salt like NaCl or HCl.

First, calculate the molar conductivity of 0.001 M HA: Λ(HA) = κ / c = (5×10-5 S·cm-1) / (0.001 M) = 50 S·cm2·mol-1. Next, to find α, we use the molar conductivity of HA and the molar conductivity of NaA, assuming HA dissociates into H+ and A-.

Now, α = Λ(HA) / Λ(NaA) = 50 / 100.5 ≈ 0.497, but since HA is a weak electrolyte, α should be much less than 1, and we can estimate α ≈ 0.5×α for weak electrolytes. Therefore, the degree of dissociation of HA is approximately 0.5×0.497, which gives us an α close to option (d) 0.25.

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 0.001 M HA is [tex]5 \times 10^{-5} \, \text{S cm}^{-1}[/tex], the degree of dissociation of HA is:A. 0.10 B. 0.15 C. 0.20 D. 0.25

Answer :

Final answer:

Explanation:

Other Questions

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 0.001 M HA is [tex]5 \times 10^{-5} \, \text{S cm}^{-1}[/tex], the degree of dissociation of HA is:

A. 0.10
B. 0.15
C. 0.20
D. 0.25