Calculate the degree of dissociation of 0.0024 M acetic acid if conductivity of this solution is 8.0 × 10-5 S cm-1.
Given λ⁰H₊ = 349.6 S cm² mol⁻¹; λ⁰CH₃COO⁻ = 40.9 S cm² mol⁻¹
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6
K =8.0×10^-5 s cm^-1
M = 0.0024
Molar conductance = (k× 1000)/m
= (8.0×10^-5×1000)/0.0024
= 33.33 s cm^2 mol^-1
Now degree of dissociation = molar conductance/sum of conductance of H+ and CH3COO-
= 33.33/390.5
= 0.0853
Answered by
0
The degree of dissociation of 0.0024 M acetic acid if conductivity of this solution is 8.0 × 10-5 S cm-1 is 0.085
- ∧°(CH3COOH) = λ∘(H+) + λ°(CH3COO−)
∧
∧
α=(∧m)/(∧∘m)
Therefore,
α = =0.085
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