The molar conductivity of `0.25 mol L^(-1)` methanoic acid is `46.1 S cm^(2) mol^(-1)`. Calculate the degree of dissociation constant. Given `: lambda_((H^(o+)))^(@)=349.6S cm^(2)mol^(-1) ` and `lambda_((CHM_(3)COO^(c-)))^(@)=54.6Scm^(2)mol^(-1)`

The molar conductivity of `0.25 mol L^(-1)` methanoic acid is `46.1 S

1.	The molar conductivity of `0.25 mol L^(-1)` methanoic acid is `46.1 S cm^(2) mol^(-1)`. Calculate the degree of dissociation constant. Given `: lambda_((H^(o+)))^(@)=349.6S cm^(2)mol^(-1) ` and `lambda_((CHM_(3)COO^(c-)))^(@)=54.6Scm^(2)mol^(-1)`
Answer» We know, `HCOOHhArrH^(+)+HCOO^(-)` `Lambda_(m)^(0)HCOOH=lambda_(H+)^(0)+lamda_(HCOO^(-))` `=349.6+54.6` `="404.2 S cm"^(2)mol^(-1)` Degree of dissociation ` alpha ` may be calculated as `alpha=(Lambda_(m)HCOOH)/(Lambda_(m)^(0)HCOOH)=(46.1)/(404.2)=0.114` Let us consider the ionisation of HCOOH. `HCOOHhArrHCOO^(-)+H^(+)` `{:(t=0,C,0,0),(teq.,C-Calpha,Calpha,Calpha):}` Dissociation constant of formic acid may be calculated as, `K=([HCOO^(-)][H^(+)])/([HCOOH])`

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The molar conductivity of `0.25 mol L^(-1)` methanoic acid is `46.1 S cm^(2) mol^(-1)`. Calculate the degree of dissociation constant. Given `: lambda_((H^(o+)))^(@)=349.6S cm^(2)mol^(-1) ` and `lambda_((CHM_(3)COO^(c-)))^(@)=54.6Scm^(2)mol^(-1)`

The molar conductivity of `0.25 mol L^(-1)` methanoic acid is `46.1 S cm^(2) mol^(-1)`. Calculate the degree of dissociation constant. Given `: lambda_((H^(o+)))^(@)=349.6S cm^(2)mol^(-1) ` and `lambda_((CHM_(3)COO^(c-)))^(@)=54.6Scm^(2)mol^(-1)`

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