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 
, the difference in their pK a values, pK a (HX) – pK a (HY), is (consider degree of ionization of both acids to be <<1)
Text Solution
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(3)
Sol. 
⇒ 
⇒ 
Also 
= α , So λ m (HX) = 
α 1 and λ m (HY) = 
α 2
(Where α 1 and α 2 are degrees of dissociation of HX and HY respectively.)
Now, Given that
λ m (HY) = 10 λ m (HX).
⇒ 
α 2 = 10 × 
α 1
α 2 = 10 α 1
K a = 
, but α << 1, therefore K a = C α 2 .
⇒ 
= 
= 
× 
= 
.
⇒ log (K a (HX)) – log (K a (HY)) = –3. ⇒ pK a (HX) – pK a (HY) = 3.
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