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Tardigrade
Question
Chemistry
The conductivity of 0.001 M acetic acid is 5 × 10-5 S cm-1 and ∧° is 390.5 S cm2 mol-1. The calculated value of dissociation constant of acetic acid would be
Q. The conductivity of
0.001
M
acetic acid is
5
×
1
0
−
5
S
c
m
−
1
and
∧
∘
is
390.5
S
c
m
2
m
o
l
−
1
. The calculated value of dissociation constant of acetic acid would be
1483
165
Electrochemistry
Report Error
A
81.78
×
1
0
4
B
81.78
×
1
0
−
5
C
18.78
×
1
0
−
6
D
18.78
×
1
0
−
5
Solution:
∧
m
c
of
0.001
M
solution
=
K
=
Molarity
1000
=
5
×
1
0
−
5
×
0.001
1000
=
50
Degree of dissociation
(
α
)
=
∧
m
∘
∧
m
c
=
390.5
50
=
0.128
For weak acids dissociation constant
(
K
)
=
(
1
−
α
)
C
α
2
K
=
(
1
−
0.128
)
0.001
×
(
0.128
)
2
=
18.78
×
1
0
−
6