1.

The molar conductivity of 0.025 M methanoic acid (HCOOH) is 46.15" S "cm^(2)mol^(-1). Calculate its degree of dissociation and dissociation constant. Given lambda_((H^(+)))^(@)=349.6" S "cm^(2)mol^(-1) and lambda_((HCOO^(-)))^(@)=54.6" S " cm^(2)mol^(-1).

Answer»

Solution :`Lambda_(m)^(@)(HCOOH)=Lambda_(m)^(@)+Lambda_(m)^(@)(HCOO^(-))`
`=349.6+54.6" S cm"^(2)" mol"^(-1)="404.2 S cm"^(2)" mol"^(-1)`
`Lambda_(m)^(@)="46.1 S cm"^(2)" mol"^(-1)` `alpha=(Lambda_(m)^(C ))/(Lambda_(m)^(@))=(46.1)/(404.2)=0.114`
`{:("INITIAL CONC.","C mol L"^(+),0,0),("At equil.",C(1-alpha),Calpha,Calpha):}`
`K_(4)=(Calpha^(2))/(1-alpha)=(0.025xx(0.114)^(2))/(1-0.114)`
`=3.67xx10^(-4)`


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