Conductivity of 0.00241 M acetic acid is

Question:

Conductivity of $0.00241 \mathrm{M}$ acetic acid is $7.896 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}$. Calculate its molar conductivity and if $\Lambda_{m}^{0}$ for acetic acid is $390.5 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$, what is its dissociation constant?

Solution:

Given, κ = 7.896 × 10−5 S m−1

c = 0.00241 mol L−1

Then, molar conductivity, $\Lambda_{m}=\frac{\kappa}{\mathrm{c}}$

$=\frac{7.896 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}}{0.00241 \mathrm{~mol} \mathrm{~L}^{-1}} \times \frac{1000 \mathrm{~cm}^{3}}{\mathrm{~L}}$

= 32.76S cm2 mol−1

Again, $\Lambda_{m}^{0}=390.5 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$

Now, $\alpha=\frac{\Lambda_{m}}{\Lambda_{m}^{0}}=\frac{32.76 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}}{390.5 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}}$

= 0.084

$\therefore$ Dissociation constant, $K_{a}=\frac{\mathrm{c} \alpha^{2}}{(1-\alpha)}$

$=\frac{\left(0.00241 \mathrm{~mol} \mathrm{~L}^{-1}\right)(0.084)^{2}}{(1-0.084)}$

= 1.86 × 10−5 mol L−1