Question

The conductivity of a 0.2 M solution of KCl is \( 2.48 \times 10^{-2} \) S cm\(^{-1}\). Calculate its molar conductivity and degree of dissociation (\( \alpha \)).
Given: \( \lambda^0_{\text{K}^+} = 73.5 \, \text{S cm}^2 \text{mol}^{-1} \), \( \lambda^0_{\text{Cl}^-} = 76.5 \, \text{S cm}^2 \text{mol}^{-1} \)

Hint:

Molar conductivity can be calculated using the solution's conductivity and concentration. The degree of dissociation can be determined by comparing the observed molar conductivity to the molar conductivity at infinite dilution.

Answer 1

Step 1: Molar conductivity \( \Lambda_m \)
The molar conductivity \( \Lambda_m \) is given by: \[ \Lambda_m = \frac{k}{C} \] Where: - \( k = 2.48 \times 10^{-2} \, \text{S cm}^{-1} \) is the conductivity of the solution, - \( C = 0.2 \, \text{mol/L} \) is the concentration of the solution. First, we multiply the conductivity by 1000 to convert the units from \( \text{S cm}^{-1} \) to \( \text{S L mol}^{-1} \): \[ \Lambda_m = \frac{k \times 1000}{C} = \frac{1000 \times 2.48 \times 10^{-2}}{0.2} \] This gives: \[ \Lambda_m = 124 \, \text{S cm}^2 \text{mol}^{-1} \] Step 2: Calculate the molar conductivity at infinite dilution \( \Lambda^0 \)
The molar conductivity at infinite dilution is the sum of the molar conductivities of the individual ions: \[ \Lambda^0_m = \lambda^0_{\text{K}^+} + \lambda^0_{\text{Cl}^-} \] Substitute the given values: \[ \Lambda^0_m = 73.5 + 76.5 = 150 \, \text{S cm}^2 \text{mol}^{-1} \] Step 3: Calculate the degree of dissociation \( \alpha \)
The degree of dissociation \( \alpha \) is given by the ratio of the observed molar conductivity to the molar conductivity at infinite dilution: \[ \alpha = \frac{\Lambda_m}{\Lambda^0_m} = \frac{124}{150} = 0.826 \] Thus, the degree of dissociation is approximately 0.826.