Question

The conductivity of 0·001 M acetic acid is $\mathrm{3.905\times10^{-5}}$ S $\mathrm{cm^{-1}}$. Calculate its molar conductivity and degree of dissociation $\mathrm{(\alpha)}$. [Given : $\mathrm{\lambda^\circ_{CH_3COO^-} = 40.9}$, $\mathrm{\lambda^\circ_{H^+} = 349.6\ S\,cm^2\,mol^{-1}}$]

Hint:

Λ_m = κ×1000/c; α = Λ_m / Λ°_m.

Answer 1

Concept: First we turn the given conductivity into molar conductivity. Then the degree of dissociation is just how big the molar conductivity is compared with its full value at infinite dilution.

Step 1: Find the molar conductivity
Use $\mathrm{\Lambda_m = \dfrac{\kappa\times 1000}{c}}$ with $\mathrm{\kappa = 3.905\times10^{-5}}$ S cm$^{-1}$ and $\mathrm{c = 0.001\ M}$:\[ \Lambda_m = \frac{3.905\times10^{-5}\times 1000}{0.001} = 39.05\ S\,cm^2\,mol^{-1} \]

Step 2: Find the limiting molar conductivity
By Kohlrausch's law we add the ion values:\[ \Lambda^\circ_m = \lambda^\circ_{H^+} + \lambda^\circ_{CH_3COO^-} = 349.6 + 40.9 = 390.5\ S\,cm^2\,mol^{-1} \]

Step 3: Find the degree of dissociation
\[ \alpha = \frac{\Lambda_m}{\Lambda^\circ_m} = \frac{39.05}{390.5} = 0.1 \]So only about one tenth of the acetic acid has split into ions.

Answer: $\mathrm{\Lambda_m = 39.05\ S\,cm^2\,mol^{-1}}$ and $\mathrm{\alpha = 0.1}$ (about 10%).