#### Question

The molar conductivity of 0.025 mol L^{-1} methanoic acid is 46.1 S cm^{2} mol^{-1}. Calculate its degree of dissociation and dissociation constant Given λ^{°}(H^{+})=349.6 S cm^{2} mol^{-1} andλ^{°}(HCOO-) = 54.6 S cm^{2} 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.