specific conductance of 0.1M CH3COOH is 4.7 ×10-2 Scm-1 at 298k calculate its molar conductance
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Class 12
>>Chemistry
>>Electrochemistry
>>Conductance of Electrolytic Solutions
>>The specific conductance of a 0.01M solu
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The specific conductance of a 0.01M solution of acetic acid at 298K is 1.65×10
−4
ohm
−1
cm
−1
. The molar conductance at infinite dilution for H
+
ion and CH
3
COO
−
ion are 349.1ohm
−1
cm
2
mol
−1
and 40.9ohm
−1
cm
2
mol
−1
respectively.
Calculate:
(i) Molar conductance of the solution.
(ii) Degree of dissociation of CH
3
COOH.
(iii) Dissociation constant for acetic acid.
Medium
Solution
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(i) Molar conductance Λ
m
=
C
K×1000
=
0.01
1.65×10
−4
×1000
=16.5 ohm
−1
cm
2
mol
−1
(ii) Degree of dissociation α=
Λ
m
∞
Λ
m
Λ
m
=16.5ohm
−1
cm
2
mol
−1
Λ
m(CH
3
COOH)
∞
=Λ
∞
[H
+
]+Λ
∞
[CH
3
COOH]
=349.1+40.9=390ohm
−1
cm
2
mol
−1
α=
390
16.5
=0.0423
(iii)
0.01M0.01(1−α)
CH
3
COOH
⇌
00.01α
CH
3
COO
−
+
00.01α
H
+
Dissociation constant K
d
=
[CH
3
COO
−
]
[CH
3
COO
−
][H
+
]
=
0.01(1−α)
0.01α×0.01α
=
1−α
0.01α
2
=0.01α
2
[Taking 1−α=1, as α is negligible as compared to 1]
=0.01×(0.0423)
2
K
d
=1.86×10
−5
Hence, the dissociation constant (K
d
) for acetic acid is 1.86×10
−5
.
Explanation:
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