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Electrochemistry
Question
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Conductivity of 0.00241 M acetic acid solution is 7.896 x 10–5 S cm-1. Calculate its molar conductivity in this solution. If Λ°m for acetic acid be 390.5 S cm2 mol–1, what would be its dissociation constant?

Answer

Given that ,        
 κ = 7.896 × 10−5 S m−1
C=M= 0.00241 mol L−1
The formula of molar conductivity,
Λm = (k  × 1000)/M
Plug the value we get
Λm = (7.896 × 10−5  ×  1000)/ 0.00241
      = 32.76S cm2 mol−1
The formula of degree of dissociation
α    = Λm/ Λom
Plug the value we get
α    = 32.76S/390.5
                         = 0.084
The formula of dissociation constant
K = Cα/(1 – α)
Plug the values we get
K = 0.00241 × 0.084/(1– 0.084)
                      = 1.86 × 10−5 mol L−1

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