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