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

Step 1: turn conductivity into molar conductivity.
Conductivity tells us how well a fixed volume conducts, but molar conductivity tells us how well the amount of acid in one mole conducts. The bridge between them uses the concentration:
\[ \Lambda_m = \frac{\kappa \times 1000}{c} \]
Putting $\kappa = 3.905\times10^{-5}$ and $c = 0.001$:
\[ \Lambda_m = \frac{3.905\times10^{-5}\times1000}{0.001} = 39.05\ \mathrm{S\,cm^2\,mol^{-1}} \]

Step 2: find the limiting molar conductivity.
By Kohlrausch's idea, the fully dissociated value is just the sum of the two ion conductivities:
\[ \Lambda^\circ_m = \lambda^\circ_{H^+} + \lambda^\circ_{CH_3COO^-} = 349.6 + 40.9 = 390.5\ \mathrm{S\,cm^2\,mol^{-1}} \]

Step 3: take the ratio for the degree of dissociation.
The fraction that has actually split into ions is the measured value divided by the fully dissociated value:
\[ \alpha = \frac{\Lambda_m}{\Lambda^\circ_m} = \frac{39.05}{390.5} = 0.1 \]
So only about one tenth of the acid is ionised, which fits acetic acid being weak.
\[ \boxed{\Lambda_m = 39.05\ \mathrm{S\,cm^2\,mol^{-1}},\quad \alpha = 0.1} \]