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Question 11

Conductivity of 0.00241 M acetic acid is 7.896 × 10 - 5 S cm - 1. Calculate its molar conductivity and if Amº for acetic acid is 390.5 S cm2 mol - 1, what is its dissociation constant?

 

Answer

Given,K = 7.896 × 10 - 5 S m - 1

c = 0.00241 mol L - 1

Then, molar conductivity, Am = K/c

= 32.76S cm2 mol - 1

Again, Amº = 390.5 S cm2 mol - 1

Now,

= 0.084

Dissociation constant,

= (0.00241 mol L-1)(0.084)2  / (1-0.084)

= 1.86 × 10 - 5 mol L - 1

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