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# 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 - Chemistry

ConceptConductance of Electrolytic Solutions Measurement of the Conductivity of Ionic Solutions

#### 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 λ°(H+)=349.6 S cm2 mol-1 andλ°(HCOO-) = 54.6 S cm2 mol-1

#### Solution

C = 0.025 mol L−1

Lambda_m= 46.1 S cm^2 mol^(-1)

lambda^0(H^+) = 349.6 S cm^2 mol^(-1)

lambda^0(HCOO^(-)) = 54.6 S cm^2 mol^(-1)

Lambda_m^0(HCOOH) = lambda^0(H^(+))+lambda^(0)(HCOO^(-))

= 349.6 + 54.4

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

Now, degree of dissociation

alpha = (Lambda_m(HCOOH))/(Lambda_m^0(HCOOH))

= 46.1/404.2

= 0.114 (approximately)

Thus, dissociation constant:

K = (c prop^2)/(1-prop)

=((0.025 "mol L"^(-1))(0.114)^2)/(1-0.114)

= 3.67xx 10^(-4) mol L^(-1)

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Solution 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 Concept: Conductance of Electrolytic Solutions - Measurement of the Conductivity of Ionic Solutions.
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