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.

$\left[\begin{array}{c}{\wedge }_{H^{+}}^{\circ <!--\; \circ \; -->}=350\mathrm{ S }c{m}^2\mathrm{ m}o{l}^{-<!--\; -\; -->1}\\ {\wedge }_{C{H}_{3}CO{O}^{-<!--\; -\; -->}}^{\circ <!--\; \circ \; -->}=\mathrm{50 }\mathrm{S }c{m}^2\mathrm{ m}o{l}^{-<!--\; -\; -->1}\end{array}\right]$

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

Step-by-step Explanation:

1. Identify the given values:
Molarity (concentration) of acetic acid, $C=0.007\text{ M}=7\times {10}^{-3}{\text{ mol L}}^{-1}$
Molar conductivity of acetic acid at concentration C, ${\wedge }_m=20{\text{ S cm}}^2{\text{ mol}}^{-1}$
Limiting molar conductivity of ${\text{H}}^{+}$ ion, ${\lambda }_{{\text{H}}^{+}}^{\circ }=350{\text{ S cm}}^2{\text{ mol}}^{-1}$
Limiting molar conductivity of acetate ion $\left({\text{CH}}_3{\text{COO}}^{-}\right)$ , ${\lambda }_{{\text{CH}}_3{\text{COO}}^{-}}^{\circ }=50{\text{ S cm}}^2{\text{ mol}}^{-1}$

2. Calculate the limiting molar conductivity of acetic acid $\left({\wedge }_m^{\circ }\right)$ using Kohlrausch's Law:
According to Kohlrausch's Law:
${\wedge }_m^{\circ }\left({\text{CH}}_3\text{COOH}\right)={\lambda }_{{\text{H}}^{+}}^{\circ }+{\lambda }_{{\text{CH}}_3{\text{COO}}^{-}}^{\circ }$
Substituting the given values:
${\wedge }_m^{\circ }=350+50=400{\text{ S cm}}^2{\text{ mol}}^{-1}$

3. Calculate the degree of dissociation $(\alpha )$ :
The degree of dissociation is defined as the ratio of molar conductivity at concentration C to limiting molar conductivity:
$\alpha =\frac{{\wedge }_m}{{\wedge }_m^{\circ }}$
Substituting the values:
$\alpha =\frac{20}{400}=0.05$

4. Calculate the dissociation constant $\left(K_a\right)$ :
For a weak monobasic acid like acetic acid, the dissociation constant is given by Ostwald's dilution law:
$K_a=\frac{C{\alpha }^2}{1-\alpha }$
Since $\alpha =0.05$ is small, we can approximate $1-\alpha \approx 1$ , or we can use the exact values:
Using the exact expression:
$K_a=\frac{0.007\times {\left(0.05\right)}^2}{1-0.05}$
$K_a=\frac{0.007\times 0.0025}{0.95}$
$K_a=\frac{1.75\times {10}^{-5}}{0.95}\approx 1.84\times {10}^{-5}{\text{ mol L}}^{-1}$
If we use the standard approximation for weak electrolytes:
$K_a\approx C{\alpha }^2$
$K_a\approx 0.007\times {\left(0.05\right)}^2$
$K_a\approx 0.007\times 0.0025$
$K_a\approx 1.75\times {10}^{-5}{\text{ mol L}}^{-1}$

This matches the correct option of 1.75×10⁻⁵ mol L⁻¹.