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Question

At 298 K, the limiting molar conductivity of a weak monobasic acid is 4×102S cm2 mol1. At 298 K, for an aqueous solution of the acid the degree of dissociation is α and the molar conductivity is y×102S cm2 mol1 . At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y×102S cm2 mol1.

The value of α is .

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Solution

α1=cm0m=y×1024×102=y4=α

On dilution conductivity increases three times
α1=3y×1024×102=3α1=3α
HAH++ACOOC(1α)CαCα
Ka=Cα21α
Since temperature is constant Kawill be constant
C1α211α1=C2α221α2

C×α21α=(C20(3α)2)(13α)
11α=9201(13α)
2060α=99α;
α=1151=0.2156
α=0.22

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