# The molar conductivity of 0.025 mol L−1 methanoic acid is

Question:

The molar conductivity of 0.025 mol L−1 methanoic acid is 46.1 S cm2 mol−1.

Calculate its degree of dissociation and dissociation constant. Given λ °(H+)

= 349.6 S cm2 mol−1 and λ °(HCOO−) = 54.6 S cm2 mol

Solution:

$C=0.025 \mathrm{~mol} \mathrm{~L}^{-1}$

$\Lambda_{m}=46.1 \mathrm{Scm}^{2} \mathrm{~mol}^{-1}$

$\lambda^{0}\left(\mathrm{H}^{+}\right)=349.6 \mathrm{Scm}^{2} \mathrm{~mol}^{-1}$

$\lambda^{0}\left(\mathrm{HCOO}^{-}\right)=54.6 \mathrm{Scm}^{2} \mathrm{~mol}^{-1}$

$\Lambda_{m}^{0}(\mathrm{HCOOH})=\lambda^{0}\left(\mathrm{H}^{+}\right)+\lambda^{0}\left(\mathrm{HCOO}^{-}\right)$

$=349.6+54.6$

$=404.2 \mathrm{Scm}^{2} \mathrm{~mol}^{-1}$

Now, degree of dissociation:

$\alpha=\frac{\Lambda_{m}(\mathrm{HCOOH})}{\Lambda_{m}^{0}(\mathrm{HCOOH})}$

$=\frac{46.1}{404.2}$

$=0.114($ approximately $)$

Thus, dissociation constant:

$K=\frac{c \propto^{2}}{(1-\propto)}$

$=\frac{\left(0.025 \mathrm{~mol} \mathrm{~L}^{-1}\right)(0.114)^{2}}{(1-0.114)}$

$=3.67 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$