The molar conductivity of acetic acid solutio | vedclass

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(B) The degree of dissociation $\alpha$ is given by the formula $\alpha = \sqrt{\frac{K_a}{C}}$.Given $K_a = 1.8 	imes 10^{-5}$ and $C = 0.01 \ M = 1

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$16.54$

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(B) The degree of dissociation $\alpha$ is given by the formula $\alpha = \sqrt{\frac{K_a}{C}}$.Given $K_a = 1.8 	imes 10^{-5}$ and $C = 0.01 \ M = 10^{-2} \ M$.$\alpha = \sqrt{\frac{1.8 	imes 10^{-5}}{10^{-2}}} = \sqrt{1.8 	imes 10^{-3}} = \sqrt{18 	imes 10^{-4}} = 4.24 	imes 10^{-2} = 0.0424$.The molar conductivity $\Lambda_m$ is related to the molar conductivity at infinite dilution $\Lambda_m^\circ$ by $\Lambda_m = \alpha 	imes \Lambda_m^\circ$.$\Lambda_m = 0.0424 	imes 390 \ S \ cm^2 \ mol^{-1} = 16.536 \ S \ cm^2 \ mol^{-1} \approx 16.54 \ S \ cm^2 \ mol^{-1}$.

The molar conductivity of acetic acid solution at infinite dilution is $390 \ S \ cm^2 \ mol^{-1}$. What is the molar conductivity of $0.01 \ M$ acetic acid solution (in $S \ cm^2 \ mol^{-1}$)? (Given: $K_{a}(CH_3COOH) = 1.8 \times 10^{-5}$,assume $1-\alpha \approx 1$)

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