Hydrofluoric acid is a weak acid. At `25^(@)C`, the molar conductivity of 0.002 M HF is `176.2 Omega^(-1)cm^(2) mol^(-1)`. If its `Lambda_(m)^(@)=405.2Omega^(-1)cm^(2)mol^(-1)`. Equilibrium constant at the given concentration is

Options

A. `6.7xx10^(-4)M`
B. `3.2xx10^(-4)M`
C. `6.7xx10^(-5)M`
D. `3.2xx10^(-5)M`

Answer

A

Explanation

`alpha=(Lambda_(c))/(Lambda_(m)^(@))=(176.2Omega^(-1)"cm"^(2)"mol"^(-1))/(405.2Omega^(-1)"cm"^(2)"mol"^(-1))=0.435`
`K=([H^(+)][F^(-)])/([HF])=(Calpha^(2))/(1-alpha)=((0.002M)(0.435)^(2))/(1-0.435)`