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JEE Advanced · Chemistry · 17. Electrochemistry

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 λX-0λY-0 , the difference in their pKa values, pKaHX-pKa(HY) , is (Consider degree of ionization of both acids to << 1)

  1. A 1
  2. B 2
  3. C 3
  4. D 4
Verified Solution

Answer & Solution

Correct Answer

(C) 3

Step-by-step Solution

Detailed explanation

λX-0λY-0
λH+0+λX-0 λH+0+λY-0
λHX0λHY0 ...... (i)
Also, λmλm0=α ,
So λmHX= λm0α1 and λmHY= λm0α2
(Where α1 and α2 are degree of dissociation of HX and HY respectively)
Now, given that
λmHY=10λmHX
λm0α2=10×λm0α1
α2=10α1 ...... (ii)
Ka=Cα21-α , But α1 , therefore Ka=Cα2
Ka(HX)Ka(HY)=0.01α120.1α22=0.010.1×1102=11000
[log(KaHX-logKaHY=-3
pKaHX-pKaHY=3
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