$\Lambda^o_m$ for $ ext{NaCl}$, $ ext{HCl}$ and $ ext{CH}_3 ext{COONa}$ are $126.4$, $425.9$, and $91.05

Question

$\Lambda^o_m$ for $	ext{NaCl}$, $	ext{HCl}$ and $	ext{CH}_3	ext{COONa}$ are $126.4$, $425.9$, and $91.05	ext{ S cm}^2	ext{ mol}^{-1}$ respectively. If the conductivity of $0.001028	ext{ mol L}^{-1}$ acetic acid solution is $4.95 	imes 10^{-5}	ext{ S cm}^{-1}$, the degree of dissociation of the acetic acid solution is:

Accepted Answer

According to Kohlrausch's law of independent migration of ions:
$\Lambda^{\circ}_m(	ext{CH}_3	ext{COOH}) = \Lambda^{\circ}_m(	ext{CH}_3	ext{COONa}) + \Lambda^{\circ}_m(	ext{HCl}) - \Lambda^{\circ}_m(	ext{NaCl})$
$= 91.05 + 425.9 - 126.4 = 390.55	ext{ S cm}^2	ext{ mol}^{-1}$

The molar conductivity at a given concentration $c$ is:
$\Lambda_m = \frac{\kappa 	imes 1000}{c}$
$\Lambda_m = \frac{4.95 	imes 10^{-5} 	imes 1000}{0.001028} = \frac{0.0495}{0.001028} \approx 48.15	ext{ S cm}^2	ext{ mol}^{-1}$

The degree of dissociation, $\alpha$ is:
$\alpha = \frac{\Lambda_m}{\Lambda^{\circ}_m} = \frac{48.15}{390.55} \approx 0.1233$

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