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Question

The molar conductivity of 0.025 mol L1 methanoic acid is 46.1 S cm2 mol1. Calculate its degree of

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Solution

Given:

λ(H+)=349.6 S cm2mol1

λ(HCOO)=54.6 S cm2mol1

Concentration (c)=0.025 mol L1

Λ(HCOOH)=46.1 S cm2mol1

Degree of dissocation

The dissociation reaction is,
HCOOH(aq)H+(aq)+HCOO(aq)

t=0ct=teqc(1α)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 mol1

We know, degree of dissociation α=ΛHCOOHΛHCOOH

α=46.1404.2=0.114

Dissociation constant
For above reaction dissociation constant is given by:

Dissociation constant (K) =[H+][HCOO][HCOOH]=c2a2c(1α)
Putting the values of c and α, we get,

K=0.025×(0.114)210.114
K=3.67×104 mol L1

Final Answer:

The degree of dissociation and dissociation constant are 0.114 and 3.67×104 mol L1 respectively.


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