Given:
λ∘(H+)=349.6 S cm2mol−1
λ∘(HCOO−)=54.6 S cm2mol−1
Concentration (c)=0.025 mol L−1
Λ(HCOOH)=46.1 S cm2mol−1
Degree of dissocation
The dissociation reaction is,
HCOOH(aq)⇌H+(aq)+HCOO−(aq)
t=0c−−t=teqc(1−α)cαcα
Using Kohlrausch's law, we get,
Λ∘(HCOOH)=λ∘(H+)+λ∘(HCOO−)
By putting the values, we get,
Lambda∘(HCOOH)=349.6+54.6=404.2 S cm2 mol−1
We know, degree of dissociation α=ΛHCOOHΛ∘HCOOH
⇒α=46.1404.2=0.114
Dissociation constant
For above reaction dissociation constant is given by:
Dissociation constant (K) =[H+][HCOO−][HCOOH]=c2a2c(1−α)
Putting the values of c and α, we get,
K=0.025×(0.114)21−0.114
⇒K=3.67×10−4 mol L−1
Final Answer:
The degree of dissociation and dissociation constant are 0.114 and 3.67×10−4 mol L−1 respectively.