The molar conductivity of a solution of a weak acid `HX(0.01 M)` is 10 times smalller than the molar conductivity of a solution of a weak acid `HY (0.10 M)`. If `lamda_(X^(-))^(@) =lamda_(Y^(-))^(@)`, the difference in their `pK_(a)` values, `pK_(a)(HX) - pK_(a)(HY)`, is (consider degree of ionisation of both acids to be `ltlt 1`):

Answer

C

Explanation

Suppose `wedge_(m)(HX)` is represented by `wedge_(m_(1))` and `wedge_(m)(HY)` by `wedge_(m_(2))`
Then given that `wedge_(m_(1))=(1)/(10)wedge_(m_(2))`
`HX hArr H^(+)+X^(-),K_(a)=([H^(+)][X^(-)])/([HX])`
Representing `K_a` of HX by `K_(a_(1))`, then as
`K_(a)=Calpha^(2)` and `alpha=(wedge_(m))/(wedge_(m)^(@))`, we have
`K_(a_(1))=C_(1)((wedge_(m_(1)))/(wedge_(m_(1))^(@)))^(2)` . . . (i)
Similarly, `HYhArrH^(+)+Y^(-),K_(a)=([H^(+)][H^(-)])/([HY])`
Representing `K_(a)` of HY by `K_(a_(2))`, we have
`K_(a_(2))=C_(2)((wedge_(m_(2)))/(wedge_(m_(2))^(@)))^(@)` . . . (ii)
As `lamda_(X^(-))^(@)=lamda_(Y^(-))^(@),wedge_(HX)^(@)=wedge_(HY)^(@)` or `wedge_(m_(1))^(@)=wedge_(m_(2))^(@)`
From eqns. (i) and (ii)
`(K_(a_(1)))/(K_(a_(2)))=(C_(1))/(C_(2))(wedge_(m_(1)))/(wedge_(m_(2)))^(2)=(0.01)/(0.1)((1)/(10))^(2)` ltBrgt `=0.001=10^(-3)`
or `logK_(a_(1))=logK_(a_(2))=log^(10^(-3))=-3`
or `-logK_(a_(1))-(-logK_(a_(2)))=3`
or `pK_(a_(1))-pK_(a_(2))=3`