The molar conductivity of `0.025mol L^(-1)` methanoic acid is `46.1 S cm^(2)mol ^(-1).` Calculate its degree of dissociation and dissociation constant.
Given, `lamda^(0)(H^(+))=349.6S cm^(2) mol ^(-1)`
`and lamda^(0)(HCOO^(-))=54.6S cm^(2)mol^(-1)`

Explanation

Step I: Calculatio of degree of dissociatiation (a) `HCOOH`
`^^_(m)^(c)=46.1S cm^(2) mol ^(-1)`
`^^_(m(HCOOH))^(0)=^^_(m(HCOO^(-)))^(0) +lamda_(m(H^(+)))^(0)`
`=(54.6+349.6)S cm^(2) mol ^(-1)`
`=404.2 S cm ^(2)mol ^(-1)`
`alpha =(^^_(m)^(c))/(^^_(m)^(0))=((46.1)S cm^(2)mol ^(-1))/((404.2)S cm^(2) mol ^(-1))=0.1140`
Step II : Calculation of dissociation constant
`HCOOH_((aq))overset("water")hArrHCOO_((aq))^(-)+H_((aq))^(+)`
`{:("Initial conc.", C, 0, 0),("Equlibrium conc", C(1-alpha), C alpha, C alpha):}`
Dissociation constant, `K_(a)=([HCOO^(-)][H^(+)])/([HCOOH])=(CalphaxxCalpha)/(C(1-alpha))=(Calpha^(2))/((1-alpha))`
Placing values `K_(a) =((0.025 mol L^(-1))xx (0.114)^(2))/((1-0.114))`
`=((3.249xx10^(-4)mol L^(-1)))/((0.886))`
`=3.67 xx10^(-4) mol L^(-1)`
`alpha= 0.114`
`K_(a) =3.67 xx10^(-4)mol L^(-1)`