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 λ⁰ₓ⁻ ≈ λ⁰y⁻ the difference in their pKa values,
pKa (HX) - pKa (HY), is (consider degree of ionization of both acids to be << 1)
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Answer:
Λ
m
(HX)=
10
x
Λ
m
(HY)=m
Λ
m
(HY)/Λ
m
o
(HY)
Λ
m
(HX)/Λ
m
o
(HX)
=
x/Λ
m
o
(HY)
(x/10)/Λ
m
o
(HX)
=
α
2
α
1
=
10
1
---- 1
HX⇌H
+
+X
−
0.1 - -
0.1(1-α
1
) 0.1α
1
0.1α
1
K
a
1
=0.1α
1
2
----- 2
HY⇌H
+
+Y
−
0.1 - -
0.1(1-α
2
) 0.1α
2
0.1α
2
K
a
2
=0.1α
2
2
------- 3
From 1, 2 and 3 equation
K
a
2
K
a
1
=
α
2
2
α
1
2
=
100
1
logK
a
1
−logK
a
2
=−2
pK
a
1
−pK
a
2
=2
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