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.

The correct option is 1.75 x 10₋₅ mol L₋₁.

Step-by-step Explanation:
To find the dissociation constant ($K_a$) of acetic acid, we use its molar conductivity, concentration, and limiting molar conductivity.

1. Identify the given values:
Concentration of acetic acid ($C$) = $0.007\text{M}=7\times {10}^{-3}{\text{mol L}}^{-1}$
Molar conductivity (${\Lambda }_m$) = $20{\text{S cm}}^2{\text{mol}}^{-1}$

2. Determine the degree of dissociation ($\alpha$):
The limiting molar conductivity of acetic acid (${\Lambda }_m^{\circ }$) is standardly taken as $400{\text{S cm}}^2{\text{mol}}^{-1}$ in simplified textbook problems.
The formula for the degree of dissociation is:
$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}$
Substituting the values:
$\alpha =\frac{20}{400}=0.05$

3. Calculate the dissociation constant ($K_a$):
The dissociation constant is given by the formula:
$K_a=\frac{C{\alpha }^2}{1-\alpha }$
Since $\alpha =0.05$ is relatively small, we can approximate $1-\alpha \approx 1$:
$K_a\approx C{\alpha }^2$
Substituting the values:
$K_a=0.007\times {\left(0.05\right)}^2$
$K_a=0.007\times 0.0025$
$K_a=1.75\times {10}^{-5}{\text{mol L}}^{-1}$