Question

Conductivity of 0.00241 M acetic acid is 7.896 × 10-5 S cm-1 . Calculate its molar conductivity and if $\land^0_m$ for acetic acid is 390.5 S cm2 mol-1 , what is its dissociation constant?

Answer 1

Given, $k$ = 7.896 × 10-5 S m-1 c
= 0.00241 mol L-1
Then, molar conductivity, $\land_m = \frac{\kappa}{c}$

= $\frac{7.896\times10^{-5} S cm^{-1}}{0.00241 mol L^{-1}}\times \frac{1000 cm^3}{L}$

= 32.76S cm2 mol-1
Again, $\land^0_m$= 390.5 S cm2 mol-1

$\alpha = \frac{\land_m}{\land^0_m}$ = $\frac{32.76 \text{S cm}^2 \text{mol}^{-1}}{390.5 S \text{cm}^2 \text{mol}^{-1}}$
Now,
= 0.084
Dissociation constant, $\kappa_a = \frac{c\alpha^2}{(1-\alpha)}$
= $\frac{(0.00241 mol L^{-1})(0.084)^2}{(1-0.084)}$
= 1.86 $\times$ 10-5 mol L-1