Question

The equivalent conductivity of \[\dfrac{N}{{10}}\] solution of acetic acid at $25^\circ $ is $14.3{(ohm)^{ - 1}}{(cm)^2}{(equiv)^{ - 1}}$. What will be the degree of dissociation of acetic acid?
$({ \wedge _{\infty C{H_3}COOH}} = 390.71{(ohm)^{ - 1}}{(cm)^2}{(equiv)^{ - 1}})$.
A. $3.66\% $
B. $3.9\% $
C. $2.12\% $
D. $0.008\% $

Answer 1

Hint: Degree of dissociation is the phenomenon of generating the free ions which are able to carry the current and they are dissociated from the fraction of the solute at a given concentration. So, it is the fraction of molecules of substance dissociating at a given time. We also have a formula for the degree of dissociation. Which tells us that the degree of dissociation is dependent on the equivalent conductivity of the compound.

Complete step by step answer:
With the help of the Kohlrausch’s law the degree of dissociation of a weak electrolyte like acetic acid can be determined.
So according to the Kohlrausch’s law for the above problem,
Degree of dissociation, ${\alpha _c} = \dfrac{{{ \wedge _{C{H_3}COOH}}}}{{{ \wedge _{\infty C{H_3}COOH}}}}$, Where ${ \wedge _{C{H_3}COOH}}$= Equivalent molar conductivity of acetic acid solution at any concentration.
And $ \wedge {\infty _{C{H_3}COOH}}$= Equivalent molar conductivity of acetic acid solution at infinite dilution.
And here we have Given with following quantities in the problem itself,
Equivalent molar conductivity of acetic acid solution at any concentration (here it is at $25^\circ C$) is,${ \wedge _{\infty C{H_3}COOH}} = 390.71{(ohm)^{ - 1}}{(cm)^2}{(equiv)^{ - 1}}$
And equivalent molar conductivity of acetic acid solution at infinite dilution is, ${ \wedge _{C{H_3}COOH}} = 14.3{(ohm)^{ - 1}}{(cm)^2}{(equiv)^{ - 1}}$
So now the degree of dissociation of acetic acid in the given problem,
${\alpha _c} = \dfrac{{{ \wedge _{C{H_3}COOH}}}}{{{ \wedge _{\infty C{H_3}COOH}}}}$
Now putting the values from the given,
${\alpha _c} = \dfrac{{14.3{{(ohm)}^{ - 1}}{{(cm)}^2}{{(equiv)}^{ - 1}}}}{{390.71{{(ohm)}^{ - 1}}{{(cm)}^2}{{(equiv)}^{ - 1}}}} = 0.0366$
So ${\alpha _c} = 0.0366$
This is the degree of dissociation, but the answers in the option are given in the percentage so the percentage degree of dissociation will be,
${\alpha _{{c_\% }}} = 0.0366 \times 100 = 3.66\% $
So, the degree of dissociation of acetic acid in the given problem is $3.66\% $.
Hence option (A) is the correct answer.

Note:
The ionic dissociation theory is used to explain electrical conductivity and many other features of the electrolytic solutions. In a modern form the theory of the ionic dissociation assumes that the even solid electrolytes are made up of some ions which are held together by electrostatic force of attraction. When these electrolytes are dissolved in a solvent then forces which are keeping the ions held together, they get weaker so the electrolyte will get dissociated into the ions. The principle of ionic dissociation is used to describe electrical conductivity and many other characteristics of electrolytic solutions. Dissociation degree refers to the amount of solvent dissociated into ions or radicals per mole. In the event of very heavy acids and bases, the degree of dissociation is about 1. Less active acids and bases can dissociate to a lesser degree.