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
$C=0.025 \mathrm{~mol} \mathrm{~L}^{-1}$
$\Lambda_{m}=46.1 \mathrm{Scm}^{2} \mathrm{~mol}^{-1}$
$\lambda^{0}\left(\mathrm{H}^{+}\right)=349.6 \mathrm{Scm}^{2} \mathrm{~mol}^{-1}$
$\lambda^{0}\left(\mathrm{HCOO}^{-}\right)=54.6 \mathrm{Scm}^{2} \mathrm{~mol}^{-1}$
$\Lambda_{m}^{0}(\mathrm{HCOOH})=\lambda^{0}\left(\mathrm{H}^{+}\right)+\lambda^{0}\left(\mathrm{HCOO}^{-}\right)$
$=349.6+54.6$
$=404.2 \mathrm{Scm}^{2} \mathrm{~mol}^{-1}$
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:
$K=\frac{c \propto^{2}}{(1-\propto)}$
$=\frac{\left(0.025 \mathrm{~mol} \mathrm{~L}^{-1}\right)(0.114)^{2}}{(1-0.114)}$
$=3.67 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$