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

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

= 349.6 + 54.6

= 404.2 S cm² mol⁻¹

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

Degree of dissociation, α = `∧_"m"/(∧_"m"^0)`

= `(46.1  "S cm"^2  "mol"^-1)/(404.2  "S cm"^2  "mol"^-1)`

= 0.114

\[\ce{HCOOH ⇌ HCOO^- + H^+}\]

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

∴ K_α = `("c"α . "c"α)/("c"(1 - α))`

= `("c"α^2)/(1 - α)`

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

= 3.67 × 10⁻⁴