The conductivity of $0.001028 \, mol \, L^{-1}$ acetic acid is $4.95 \times 10^{-5} \, S \, cm^{-1}$. Calculate its dissociation constant if $\Lambda_m^\circ$ for acetic acid is $390.5 \, S \, cm^2 \, mol^{-1}$.
(N/A) First,calculate the molar conductivity $(\Lambda_m)$ using the formula $\Lambda_m = \frac{\kappa \times 1000}{c}$.
$\Lambda_m = \frac{4.95 \times 10^{-5} \, S \, cm^{-1} \times 1000 \, cm^3 \, L^{-1}}{0.001028 \, mol \, L^{-1}} = 48.15 \, S \, cm^2 \, mol^{-1}$.
Next,calculate the degree of dissociation $(\alpha)$ using $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$.
$\alpha = \frac{48.15}{390.5} = 0.1233$.
Finally,calculate the dissociation constant $(K_a)$ using $K_a = \frac{c \alpha^2}{1 - \alpha}$.
$K_a = \frac{0.001028 \times (0.1233)^2}{1 - 0.1233} = 1.78 \times 10^{-5} \, mol \, L^{-1}$.
$\Lambda_m = \frac{4.95 \times 10^{-5} \, S \, cm^{-1} \times 1000 \, cm^3 \, L^{-1}}{0.001028 \, mol \, L^{-1}} = 48.15 \, S \, cm^2 \, mol^{-1}$.
Next,calculate the degree of dissociation $(\alpha)$ using $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$.
$\alpha = \frac{48.15}{390.5} = 0.1233$.
Finally,calculate the dissociation constant $(K_a)$ using $K_a = \frac{c \alpha^2}{1 - \alpha}$.
$K_a = \frac{0.001028 \times (0.1233)^2}{1 - 0.1233} = 1.78 \times 10^{-5} \, mol \, L^{-1}$.
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