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A solution of a weak acid HA has a concentration of 0.0400 mol/L and a conductivity of 0.0330 S/m…

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A solution of a weak acid HA has a concentration of $0.0400\ \text{mol/L}$ and a conductivity of $0.0330\ \text{S/m}$ . The limiting molar conductivity of the acid is $390.5\ \text{S cm}^2/\text{mol}$ . Find the molar conductivity, the degree of dissociation, and $K_a$ in $\text{mol/L}$ .

Need a reminder? The key ideaShow

The molar conductivity is $\Lambda_m = \dfrac{\kappa}{c}$ . With $\kappa$ in $\text{S/m}$ and $c$ in $\text{mol/L}$ , $\Lambda_m = \dfrac{10\,\kappa}{c}$ in $\text{S cm}^2/\text{mol}$ . With $\kappa$ in $\text{S/cm}$ , $\Lambda_m = \dfrac{1000\,\kappa}{c}$ . A strong electrolyte follows $\Lambda_m = \Lambda_m^\circ - A\sqrt{c}$ . For a weak electrolyte, $\alpha = \dfrac{\Lambda_m}{\Lambda_m^\circ}$ .

The worked solution

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