1.

Hydrofluoric acid is a weak acid. Molar conductivity of 0.02 M HF solution at 25^(@)C is 176.2Omega^(-1)cm^(2)mol^(-1). If its Lamda^(@)m=405Omega^(-1)cm^(2)mol^(-1), then find out its equilibrium constant at given concentration.

Answer»

`6.7xx10^(-4)M`
`3.2xx10^(-4)M`
`6.4xx10^(-5)M`
`3.2xx10^(-5)M`

Solution :`alpha=(LamdaC)/(Lamda^(@)m)`
`=(176.2Omega^(-1)cm^(2)"mole"^(-1))/(405.2Omega^(-1)cm^(2)"mole"^(-1))=0.435`
`K_(a)=([H^(+)][F^(-1)])/([HF])=(Calpha^(2))/(1-alpha)`
`=((0.002M)(0.435)^(2))/(1-0.435)`
`=6.70xx10^(-4)M`.


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