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 \[\ce{λ^0_{(H^+)}}\] = 349.6 S cm² mol⁻¹ and \[\ce{λ^0_{(HCOO^-)}}\] = 54.6 S cm² mol⁻¹.

Answer 1

Given: \[\ce{\Lambda_{m(HCOOH)}}\] = 46.1 S cm² mol⁻¹

\[\ce{λ^0_{(H^+)}}\] = 349.6 S cm² mol⁻¹

\[\ce{λ^0_{(HCOO^-)}}\] = 54.6 S cm² mol⁻¹

\[\ce{\Lambda^0_{m(HCOOH)} = \lambda^0_{(H^+)} + \lambda^0_{(HCOO^-)}}\]

= 349.6 + 54.6

= 404.2 S cm² mol⁻¹

Degree of dissociation (α) = \[\ce{\frac{\Lambda_m}{\Lambda{^0_m}}}\]

= `46.1/404.2`

= 0.114

⇒ α = 11.4%

\[\ce{HCOOH <=> HCOO- + H+}\]

Initial concentration	c mol L−1	0	0
Concentration at equilibrium	c(1 − α)	cα	cα

∴ K_α = `(c alpha * c alpha)/(c (1 - alpha))`

= `(c alpha^2)/(1 - alpha)`

= `(0.025 xx (0.114)^2)/(1 - 0.114)`

= 3.67 × 10⁻⁴