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NCERT Class XII Chemistry
Chapter - Electrochemistry
Questions with Solutions

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Question : 11
Total: 32
Conductivity of 0.00241 M acetic acid is 7.896 × 10‌–5 S cm‌–1. Calculate its molar conductivity. If Λ0m m for acetic acid is 390.5 S cm‌2mol‌–1, what is its dissociation constant?
Solution:  
Λcm=
κ×1000
M
=
(7.896×10−5 )×1000
0.00241
=32.76 S cm2 mol−1
α=
Λcm
Λ0m
=
32.76
390.5
=8.4×10−2
Ka=
cα2
1−α
=
0.00241×(8.4×10−2)2
1−0.084
 =1.86×10−5
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