Question

90. Calculate degree of dissociation of 0.02M acetic acid at 298K. Give that (CH3COOH)=17.37Scm²m-¹,H+=345.8Scm²m-¹, CH3COO-=40.2Scm²m-¹

Answer 1

Degree of dissociation

1. The degree of association is classified as the fraction of the total number of molecules that are connected or combined, resulting in a larger molecule being created.

2. The degree of dissociation is the phenomenon of producing free ions carrying current, which at a given concentration is dissociated from the fraction of solute.

3. The dissociation of acetic acid is given as: $C{H}_{3}COOH\left(aq\right)⟶C{H}_{3}CO{O}^{-}+H^{+}$

4. The degree of dissociation is related to molar conductivity as:

$\alpha =\frac{\lambda }{{\lambda }_0}where,\lambda =molarconductivityatgivenconcentration$

${\lambda }_0=molarconductivityatinifinitedilution$

5. The molar conductivity of acetic acid

${\left({\lambda }_0\right)}_{C{H}_{3}COOH}={\left({\lambda }_0\right)}_{C{H}_{3}CO{O}^{-}}+{\left({\lambda }_0\right)}_{H^{+}}$

$=345.8+40.2$

$=386Sc{m}^{2}mol{e}^{-1}$

6. Hence the degree of dissociation is given as:

$\alpha =\frac{{\left(\lambda \right)}_{C{H}_{3}COOH}}{{\left({\lambda }_0\right)}_{C{H}_{3}COOH}}=\frac{17.37}{386}=0.045$