Question

The molar conductivity of 0.025 mol L methanoic acid is 46.1 S cm² mol.
Calculate its degree of dissociation and dissociation constant. Given 2°(H)
349.6 S cm? mol-' and 2° (HCOO7) = 54.6 S cm² mol.​

Answer 1

Answer:

Answer:Given that

λ0(H+)= 349.6 S cm2 mol–1

λ0(HCOO–) = 54.6 S cm2 mol–1

Concentration ,C = 0.025 mol L-1

λ(HCOOH) = 46.1 S cm2 mol−1

use formula

λ°(HCOOH) = λ0(H+) + λ0(HCOO–)

plug the values we get

λ°(HCOOH) = 0.349.6 + 54.6

=404.2 S cm2 mol−1

Formula of degree of dissociation:

ά = λ°(HCOOH)/ λ°(HCOOH)

ά = 46.1 / 404.2

ά = 0.114

Formula of dissociation constant:

K = (c ά2)/(1 – ά)

Plug the values we get

K = 3.67 × 10–4 mol per liter