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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.

3.8The molar conductivity of 0.025 mol L^{-1} methanoic acid is 46.0 \, S cm^2\, S\, cm^2 mol ^{-1}
Calculate its degree of dissociation and dissociation constant. Given \lambda ^{0}(H+)=349.6 \, S cm^{2} mol ^{-1} and \lambda ^{0}(HCOO^{-})=54.6\,\: S\, cm^{2}\, \: mol^{-1}
 

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We know that :- 

                         \Lambda _m = \lambda^{\circ}(H^+) + \lambda^{\circ}(HCOO^-)

                                 = 349.6 +54.6

                                 = 404.2\ Scm^2\ mol^{-1}

For degree of dissociation, we have :- 

                                                          \alpha = \frac{\Lambda _m(HCOOH)}{\Lambda ^{\circ}(HCOOH)}

or                                                            \alpha = \frac{46.1}{404.2} = 0.114

For dissociation constant, we have :- 

                                            K_a = \frac{c\alpha ^2}{1-\alpha }

or                                                K_a = \frac{0.025\times(0.114)^2}{1-0.114 }

or                                                       = 3.67\times 10^{-4}\ mol\ L^{-1}

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