Question

(a) Calculate the degree of dissociation of 0.0024 M acetic acid if conductivity of this
solution is 8.0 x 10-5 S cm!
Given 2 =349.6S cm’ mol-'; %ch.coo = 40.98 cm’ mol'​

Answer 1

Answer:

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