The conductivity of 0.001 M acetic acid at a | vedclass

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(A) $1$. Calculate the molar conductivity $(\wedge_m)$ using the formula: $\wedge_m = \frac{\kappa 	imes 1000}{C} = \frac{5.07 	imes 10^{-5} 	imes

Vedclass · Accepted Answer

$1.94 	imes 10^{-5}$

Vedclass · Answer

(A) $1$. Calculate the molar conductivity $(\wedge_m)$ using the formula: $\wedge_m = \frac{\kappa 	imes 1000}{C} = \frac{5.07 	imes 10^{-5} 	imes 1000}{0.001} = 50.7 \ S \ cm^2 \ mol^{-1}$.$2$. Calculate the degree of dissociation $(\alpha)$: $\alpha = \frac{\wedge_m}{\wedge_m^0} = \frac{50.7}{390} = 0.13$.$3$. Calculate the dissociation constant $(K_a)$ using the formula: $K_a = \frac{C \alpha^2}{1 - \alpha}$.$4$. Substituting the values: $K_a = \frac{0.001 	imes (0.13)^2}{1 - 0.13} = \frac{0.001 	imes 0.0169}{0.87} \approx 1.94 	imes 10^{-5}$.

The conductivity of $0.001 \ M$ acetic acid at a certain temperature is $5.07 \times 10^{-5} \ S \ cm^{-1}$. If $\wedge_m^0$ of acetic acid at the same temperature is $390 \ S \ cm^2 \ mol^{-1}$,the dissociation constant of acetic acid at that temperature is:

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