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The molar conductivity of a solution of ...

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`):

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The correct Answer is:
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Given `(Lambda_(m)(HX))/(Lambda_(m)(HY))=1/10`
`alpha_(1)/alpha_(2)=(Lambda_(m)(HX))/(Lambda_(m)(HX))//(Lambda_(m)(HY))/(Lambda_(m) (HY))=1/10`
`(K_(a)(HX))/(K_(a) (HY))=(C_(1)alpha_(1)^(2))/(C_(2)alpha_(2)^(2))`
`=0.01/0.1xx(1/10)^(2)=1/1000`
`log K_(a)(HX)-log K_(a) (HY)=-3`
`:. pK_(a) (HX)-pK_(a)(HY)=3`
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