The molar conductivity of `0.025 mol L^(-1)` methanoic acid is `46.1 S cm^(2) mol^(-1)`. Its degree of dissociation `(alpha)` and dissociation constant. Given `lambda^(@)(H^(+))=349.6 S cm^(-1)` and `lambda^(@)(HCOO^(-)) = 54.6 S cm^(2) mol^(-1)`.

Options

A. `K_(a)=3.67xx10^(4)alpha=0.214`
B. `K_(a)=3.67xx10^(-4)alpha=0.114`
C. `K_(a)=2.25xx10^(-4)alpha=0.150`
D. `K_(a)=2.25xx10^(-2)alpha=0.314`

Answer

B

Explanation

`Lambda_(m)^(@)(HCOOH)=lambda^(@)(H^(+))+lambda^(@)(HCOO^(-))=349.6+54.6S cm^(2) mol^(-)`
`=404.2 S cm^(2) mol^(-)`
`Lambda_(m)^(c )=46.1S cm^(2) mol^(-1)` (Given)
`alpha=(Lambda_(m)^(c ))/(Lambda_(m)^(@))=(46.1)/(404.2)=0.114`
`{:(HCOOH hArr HCOO^(-)+H^(+)),("Initial conc. C mol L"^(-1)),("Conc. at eqm."C(1-alpha)C alpha " " C alpha):}`
`K_(alpha)=(C alpha.C alpha)/(C(1-alpha))=(C alpha^(2))/(1-alpha)=(0.025xx(0.114)^(2))/(1-0.114)=3.67xx10^(-4)`