InterviewSolution
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InterviewSolution
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The molar conductivity of a solution of a weak acid HX (0.01M) is 10 times smaller than the molar conductivity of a solution of weak acid HY (0.10M). If `lamda_(X^(-))^(@)=lamda_(Y^(-))^(@)`, the difference in their `pK_(a)` values, `pK_(a)(HX)-pK_(a)(HY)` is (consider degree of ionization of both acids to be lt lt1). |
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Answer» Correct Answer - C 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` |
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