Question

The molar conductivity of acetic acid solution at infinite dilution is 390 S cm$^2$ mol$^{-1}$. What is the molar conductivity of 0.01 M acetic acid solution (in S cm$^2$ mol$^{-1}$)? (Given: $K_a(\text{CH}_3\text{COOH})=1.8\times10^{-5}$, assume $1-\alpha \approx 1$)

• A. 10.64
• B. 16.54
• C. 51.64
• D. 15.64

Hint:

For weak electrolytes, the degree of dissociation $\alpha$ can be found using Ostwald's dilution law: $K_a = \frac{C\alpha^2}{1-\alpha}$. For very weak electrolytes, this simplifies to $\alpha = \sqrt{K_a/C}$. The molar conductivity is then found using Kohlrausch's law in the form $\Lambda_m^C = \alpha \Lambda_m^\circ$.

Answer 1

The Correct Option is B

For a weak electrolyte like acetic acid (CH$_3$COOH), the molar conductivity at a given concentration C, denoted $\Lambda_m^C$, is related to the molar conductivity at infinite dilution, $\Lambda_m^\circ$, by the degree of dissociation, $\alpha$.
$\Lambda_m^C = \alpha \Lambda_m^\circ$.
We are given $\Lambda_m^\circ = 390$ S cm$^2$ mol$^{-1}$. To find $\Lambda_m^C$, we first need to calculate $\alpha$.
The dissociation of acetic acid is: CH$_3$COOH $\rightleftharpoons$ CH$_3$COO$^-$ + H$^+$.
The acid dissociation constant, $K_a$, is given by:
$K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}$.
We are given $K_a = 1.8 \times 10^{-5}$ and the concentration $C = 0.01$ M.
We are also told to assume $1-\alpha \approx 1$, which is valid for weak acids.
So, $K_a \approx C\alpha^2$.
$1.8 \times 10^{-5} = (0.01)\alpha^2$.
$\alpha^2 = \frac{1.8 \times 10^{-5}}{10^{-2}} = 1.8 \times 10^{-3} = 18 \times 10^{-4}$.
$\alpha = \sqrt{18 \times 10^{-4}} = \sqrt{9 \times 2} \times 10^{-2} = 3\sqrt{2} \times 10^{-2}$.
Using $\sqrt{2} \approx 1.414$:
$\alpha \approx 3 \times 1.414 \times 10^{-2} = 4.242 \times 10^{-2} = 0.04242$.
Now, we can calculate the molar conductivity at 0.01 M concentration:
$\Lambda_m^C = \alpha \Lambda_m^\circ = 0.04242 \times 390$.
$\Lambda_m^C \approx 16.5438$ S cm$^2$ mol$^{-1}$.
This is approximately 16.54 S cm$^2$ mol$^{-1}$.