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Question

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(0.10 M). If λXλY, the difference in their pKa values, pKa(HX)pKa(HY), is (consider degree of ionizaiton of both acids to be <<1)

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Solution

λXλY
λH++λXλH++λYλHXλHY(1)
Also λmλm= α, So λm(HX)=λmα1 and λm(HY)=λmα2
Now,
λm(HY)=10λm(HX)
λmα2=10×λmα1
α2=10α1(2)
Ka=Cα21α, but α<<1, therefore Ka=Cα2
Ka(HX)Ka(HY)=0.01α210.1α22=0.010.1×(110)2=11000.
log(Ka(HX))log(Ka(HY))=3
pKa(HX)pKa(HY)=3

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