The molar conductivity of $0.025 \ mol \ L^{-1}$ methanoic acid is $46.1 \ S \ cm^2 \ mol^{-1}$. Calculate its degree of dissociation and dissociation constant. Given $\lambda^o(H^{+}) = 349.6 \ S \ cm^2 \ mol^{-1}$ and $\lambda^o(HCOO^{-}) = 54.6 \ S \ cm^2 \ mol^{-1}$.

(N/A) Given:
$C = 0.025 \ mol \ L^{-1}$
$\Lambda_m = 46.1 \ S \ cm^2 \ mol^{-1}$
$\lambda^o(H^{+}) = 349.6 \ S \ cm^2 \ mol^{-1}$
$\lambda^o(HCOO^{-}) = 54.6 \ S \ cm^2 \ mol^{-1}$
Step $1$: Calculate molar conductivity at infinite dilution $\Lambda_m^o(HCOOH)$:
$\Lambda_m^o(HCOOH) = \lambda^o(H^{+}) + \lambda^o(HCOO^{-}) = 349.6 + 54.6 = 404.2 \ S \ cm^2 \ mol^{-1}$
Step $2$: Calculate degree of dissociation $(\alpha)$:
$\alpha = \frac{\Lambda_m}{\Lambda_m^o} = \frac{46.1}{404.2} \approx 0.114$
Step $3$: Calculate dissociation constant $(K_a)$:
$K_a = \frac{C \alpha^2}{1 - \alpha} = \frac{0.025 \times (0.114)^2}{1 - 0.114} = \frac{0.025 \times 0.012996}{0.886} \approx 3.67 \times 10^{-4} \ mol \ L^{-1}$