Question

The specific conductance of a $0.01M$ solution of acetic acid at $298K$ is $1.65\times {10}^{-4}$ ${ohm}^{-1}$ ${cm}^{-1}$. The molar conductance at infinite dilution for $H^{+}$ ion and ${CH}_{3}CO{O}^{-}$ ion are $349.1{ohm}^{-1}$ ${cm}^2{mol}^{-1}$ and $40.9{ohm}^{-1}$ ${cm}^2{mol}^{-1}$ respectively.
Calculate:

(i) Molar conductance of the solution.
(ii) Degree of dissociation of ${CH}_{3}COOH$.
(iii) Dissociation constant for acetic acid.

Answer 1

(i) Molar conductance ${\Lambda }_m=\frac{K\times 1000}{C}=\frac{1.65\times {10}^{-4}\times 1000}{0.01}=16.5{ohm}^{-1}{cm}^2{mol}^{-1}$

(ii) Degree of dissociation $\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\infty }}$

${\Lambda }_m=16.5{ohm}^{-1}{cm}^2{mol}^{-1}$

${\Lambda }_{m\left({CH}_{3}COOH\right)}^{\infty }={\Lambda }^{\infty }\left[H^{+}\right]+{\Lambda }^{\infty }\left[{CH}_{3}COOH\right]$

$=349.1+40.9=390{ohm}^{-1}{cm}^2{mol}^{-1}$

$\alpha =\frac{16.5}{390}=0.0423$

(iii) $\underset{0.01M0.01(1-\alpha )}{{CH}_{3}COOH}\rightleftharpoons \underset{00.01\alpha }{{CH}_{3}CO{O}^{-}}+\underset{00.01\alpha }{H^{+}}$

Dissociation constant $K_d=\frac{\left[{CH}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[{CH}_{3}CO{O}^{-}\right]}=\frac{0.01\alpha \times 0.01\alpha }{0.01(1-\alpha )}=\frac{0.01{\alpha }^2}{1-\alpha }=0.01{\alpha }^2$

[Taking $1-\alpha =1$, as $\alpha$ is negligible as compared to $1$]
$=0.01\times {\left(0.0423\right)}^2$

$K_d=1.86\times {10}^{-5}$

Hence, the dissociation constant ($K_d$) for acetic acid is $1.86\times {10}^{-5}$.