The conductivity of 0.001028 mol l-1 acetic acid is 4.95 10-5 scm-1. Find out its dissociation constant, if for acetic acid is 390.5 scm-1 mol-1.
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Answer:
1000×4.95×10
−5
Scm
−1
=48.15S cm
2
mol
−1
α=
Λ
m
0
Λ
m
=
390.5S cm
2
mol
−1
48.15S cm
2
mol
−1
=0.123
K
a
=
[CH
3
COOH]
[CH
3
COO
−
][H
+
]
=
C(1−α)
Cα×Cα
K
a
=
1−α
Cα
2
K
a
=
1−0.123
0.001028×(0.123)
2
=1.78×10
−5
mol/L
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