The molar conductivity of 0.007 M acetic acid is 20 S cm2 mol⁻¹. What is the dissociation constant of acetic acid ? Choose the correct option.

Correct Answer: 1.75×10⁻⁵ mol L⁻¹

Step-by-Step Explanation:

Step 1: Calculate the limiting molar conductivity (${\Lambda }_m^{\circ }$) of acetic acid (${\text{CH}}_3\text{COOH}$)
According to Kohlrausch's law of independent migration of ions, the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its individual ions.
From the provided image, we can identify the following limiting molar conductivities for the ions at infinite dilution:
${\Lambda }_{{\text{H}}^{+}}^{\circ }=350{\text{S cm}}^2{\text{mol}}^{-1}$
${\Lambda }_{{\text{CH}}_3{\text{COO}}^{-}}^{\circ }=50{\text{S cm}}^2{\text{mol}}^{-1}$
Using these values, we calculate:
${\Lambda }_m^{\circ }\left({\text{CH}}_3\text{COOH}\right)={\Lambda }_{{\text{H}}^{+}}^{\circ }+{\Lambda }_{{\text{CH}}_3{\text{COO}}^{-}}^{\circ }$
${\Lambda }_m^{\circ }\left({\text{CH}}_3\text{COOH}\right)=350+50=400{\text{S cm}}^2{\text{mol}}^{-1}$

Step 2: Find the degree of dissociation ($\alpha$)
The degree of dissociation is defined as the ratio of molar conductivity at a given concentration (${\Lambda }_m^c$) to the limiting molar conductivity (${\Lambda }_m^{\circ }$):
Given parameters in the question:
Concentration ($c$) = 0.007 M
Molar conductivity (${\Lambda }_m^c$) = 20 S cm² mol⁻¹
Substituting these values:
$\alpha =\frac{{\Lambda }_m^c}{{\Lambda }_m^{\circ }}=\frac{20}{400}=0.05$

Step 3: Calculate the dissociation constant ($K_a$)
For a weak acid like acetic acid, the dissociation constant is given by the formula:
$K_a=\frac{c{\alpha }^2}{1-\alpha }$
Since $\alpha$ is small ($\alpha =0.05$), we can use the approximation $1-\alpha \approx 1$:
$K_a\approx c{\alpha }^2$
Substituting the values:
$K_a\approx 0.007\times {\left(0.05\right)}^2$
$K_a\approx 0.007\times 0.0025$
$K_a\approx 1.75\times {10}^{-5}{\text{mol L}}^{-1}$