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Conductivity of 2.5 xx 10^(-4) M methono...

Conductivity of `2.5 xx 10^(-4)` M methonoic acid is `5.25 xx 10^(-5)` S ` cm^(-1)` . Calculate its molar Condcutivity and degree of dissociation. Given `: lambda^(@) ( H^(+) = 349.5` S `cm^(2) "mol"^(-1)` and `lambda^(@) ( HCOO^(-)) ` = 50.5 S `cm^(2) "mol"^(-1)`.

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Conductivity of methoic acid `= 5.25 xx 10^(-5) S cm^(-1)`
Concentration of methanoic acid `= 2.5 xx 10^(-4) M `
`= 2.5 xx 10^(-4) xx 1000` mol `cm^(-3) = 0.25 ` mol `cm^(-3)`
Molar conductivity,
`Lambda_(m) = ( k )/( C ) = (5.25 xx 10^(-5) S cm^(-1))/(0.25 " mol " cm^(-3))`
` = 21 xx 10^(-5) S cm^(2) " mol"^(-1)`
Given, `lambda_((H^(+)))^(@) = 349 .5 S cm^(2) "mol"^(-1)`
`lambda_((HCOO^(-)))^(@) = 50.5 S cm^(2) "mol"^(-1)`
Degree of dissociation , `alpha = ( Lambda_(m))/( Lambda_(m))` ....(i)
`lambda_(m (HCOOH))^(@) = lambda_(H^(+))^(@)+ lambda_(HCOO^(-))^(@) `
`= 349.5 + 50.5 = 400 S cm^(2) "mol"^(-1)`
Substituting the value of `Lambda_(m)` and `Lambda_(m(HCOOH))^(@)`
In Equ. (i),
we get `alpha = ( 21 xx 10^(-5) S cm^(2) "mol"^(-1))/( 400 S cm^(2) "mol"^(-1)) = 5.25 xx 10^(-7) `
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