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1)

 The molar conductivity  of a solution of a weak acid HX (0.01 M)  is 10 times smaller than the molar conductivity  of a solution  of a weak  acid HY  (0.01 M), If  λ0Xλ0Y, the  difference  in their pKa  values, PKa(HX)PKa(HY)   is (consider  degree of ionisation  of both acids to be <<1)


A) 2

B) 3

C) 4

D) 1

Answer:

Option B

Explanation:

Degree of ionisation (α)=  m

Let           m(HY)=x

                     m(HX)=X10

       m(HX)m(HY)=110=α(HX)α(HY)

                                                                                                [(HX)=(HY)]

         Also :   Ka(HX)=(0.01)[α(HX)]2

                        Ka(HX)=(0.10)[α(HX)]2

                                           =(0.10)[10α(HX)]2

                                =10[α(HX)]2    ........(ii)

                     Ka(HX)Ka(HY)=0.0110

      logKa(HX)logKa(HY)=3

      logKa(HX)[logKa(HY)]=3

    pKa(HX)pKa(HY)=3