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. 11.4%, `3.67xx10^(-4) "mol L"^(-1)`
B. 22.8%, `1.83xx10^(-4) "mol L"^(-1)`
C. 52.2%, `4.25xx10^(-4) "mol L"^(-1)`
D. 1.14%, `3.67xx10^(-6) "mol L"^(-1)`

Answer

A

Explanation

`lambda_"HCOOH"^@=lambda_(H^+)^@ + lambda_(HCOO^-)^@ = 349.6 + 54.6 = 404.2 S "cm"^(2) "mol"^(-1)`
`alpha=Lambda_m/Lambda_m^@=46.1/404.2 `=11.4%
`K_a=(C alpha^2)/(1-alpha)=(0.025xx(0.114)^2)/(1-0.114)`
`=(0.025xx0.114xx0.114)/0.886=3.67xx10^(-4) "mol" L^(-1)`