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The molar conductivities (\(\Lambda_m^0\)) for NaCl, HCl, and NaA are 126.4, 425.9, and 100.5 S cm² mol⁻¹, respectively. If the conductivity of 0.001 M HA is \(5 \times 10^{-5}\) S cm⁻¹, what is the degree of dissociation of HA?

1. 0.125
2. 0.75
3. 0.50
4. 0.25

Answer :

Final Answer:

The degree of dissociation of HA is 0.50, thus the correct option is 3.

Explanation:

To find the degree of dissociation (α) of the weak electrolyte HA, we can use the formula for molar conductivity (Λ) of a weak electrolyte, which is given by:

Λ = Λ°m - α × ∆Λ

Where Λ°m is the molar conductivity at infinite dilution, and ∆Λ represents the decrease in molar conductivity due to the dissociation of the weak electrolyte. Rearranging the formula to solve for α, we have:

α = (Λ°m - Λ) / Λ°m

Substituting the given values, we get:

α = (100.5 - 5 × 10^-5) / 100.5 = 0.499

The calculated value for α is approximately 0.499, which is very close to 0.50. Therefore, the correct option is (3) 0.50.

In this calculation, we utilized the provided molar conductivities at infinite dilution (∧m⁰) for NaCl, HCl, and NaA, along with the conductivity of 0.001 M HA solution. By applying the formula for calculating the degree of dissociation, we determined the fraction of HA molecules that dissociate into ions in solution. This value represents the degree to which the weak electrolyte undergoes dissociation when dissolved in solution (option 3).

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