If the degree of dissociation of formic acid | vedclass

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Question

(A) Given: $\lambda^{\circ}(H^{+}) = 349.6 \ S \ cm^2 \ mol^{-1}$ and $\lambda^{\circ}(HCOO^{-}) = 54.6 \ S \ cm^2 \ mol^{-1}$.Degree of dissociation

Vedclass · Accepted Answer

$44.46 \ S \ cm^2 \ mol^{-1}$

Vedclass · Answer

(A) Given: $\lambda^{\circ}(H^{+}) = 349.6 \ S \ cm^2 \ mol^{-1}$ and $\lambda^{\circ}(HCOO^{-}) = 54.6 \ S \ cm^2 \ mol^{-1}$.Degree of dissociation $\alpha = 11.0 \% = 0.11$.The molar conductivity at infinite dilution is calculated as: $\lambda_{m}^{\circ}(HCOOH) = \lambda^{\circ}(H^{+}) + \lambda^{\circ}(HCOO^{-}) = 349.6 + 54.6 = 404.2 \ S \ cm^2 \ mol^{-1}$.The molar conductivity $\lambda_{m}$ at a given concentration is related to the degree of dissociation by the formula: $\lambda_{m} = \alpha 	imes \lambda_{m}^{\circ}$.Substituting the values: $\lambda_{m} = 0.11 	imes 404.2 = 44.46 \ S \ cm^2 \ mol^{-1}$.

If the degree of dissociation of formic acid is $11.0 \%$,the molar conductivity of $0.02 \ M$ solution of it is (Given: $\lambda^{\circ}(H^{+}) = 349.6 \ S \ cm^2 \ mol^{-1}, \lambda^{\circ}(HCOO^{-}) = 54.6 \ S \ cm^2 \ mol^{-1}$)

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