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 a weak aid HY (0.10M). If `lamda_(x^(-))^(0)=lamda_(y-)^(0),` the difference in their `pK_(a)` values, `pK_(a)(HX)-pK_(a)(HY)`, is (consider degree of ionization of both acids to be ltlt1). |
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Answer» Correct Answer - C `lamda_(x-)^(0)=lamda_(y-)^(0)` `implieslamda_(H^(+))^(o)+lamda_(X^(-))^(o)=lamda_(H^(+))^(o)+lamda_(Y^(-))^(o)implieslamda_(HX)^(o)=lamda_(HY)^(o)` . . . (i) Also `(lamda_(m))/(lamda_(m)^(o)=alpha,` so `lamda_(m)(HX)=lamda_(m)^(o)alpha_(1) and lamda_(m)(HY)=lamda_(m)^(o)alpha_(2)` (where `alpha_(1)` and `alpha_(2)` are degrees of dissociation of HX and HY respectively). Now, given that `lamda_(m)(HY)=10lamda_(m)(HZ)implieslamda_(m)^(o)alpha_(2)=10xxlamda_(m)^(o)alpha_(1)` `alpha_(2)=10alpha_(1)` . . . (ii) `K_(a)=(Calpha^(2))/(1-alpha)`, but `alpha ltlt 1`, therefore, `K_(a)=Calpha^(2)` `implies(K_(a)(HX))/(K_(a)(HY))=(0.01alpha_(1)^(2))/(0.1alpha_(2)^(2))=(0.01)/(0.1)xx((1)/(10))^(2)=(1)/(1000)` `implieslog(K_(a)(HX)-pK_(a)(HY))=-3` `impliespK_(a)(HX)-pK_(a)(HY)=3` |
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