Question

The equivalent conductivity of $N/10$ solution of acetic acid at ${25}^o$C is $14.3$ $oh{m}^{-1}$ $c{m}^2$ $equi{v}^{-1}$. What will be the degree of dissociation of acetic acid?
$({\Lambda }_{\infty C{H}_{3}COOH}=390.71$ $oh{m}^{-1}$ $c{m}^2$ $equi{v}^{-1}$).

• A. $3.66\%$
• B. $3.9\%$
• C. $2.12\%$
• D. $0.008\%$

Answer 1

Answer: (A)

$3.66\%$

The equivalent conductivity of an electrolyte is defined as the conductivity of a volume of solution containing one equivalent weight of dissolved substance when placed between two parallel electrodes 1 cm apart, and large enough to contain between them all of the solution.

Given,

${\Lambda }_{m\left(C{H}_{3}COOH\right)}^{\infty }=390.71oh{m}^{-1}c{m}^{2}e{q}^{-1}$

${\Lambda }_{m\left(C{H}_{3}COOH\right)}=14.3oh{m}^{-1}c{m}^{2}e{q}^{-1}$

Degree of dissociation is given by,

$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^0}=\frac{14.3}{390.71}=0.0366⟹\alpha =3.66\%$