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.

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

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

Step 1: Understand the given data
Concentration of acetic acid, c = 0.007 M = 7 × 10⁻³ mol L⁻¹
Molar conductivity, ∧_m = 20 S cm² mol⁻¹
Limiting molar conductivity of H⁺, ∧°_H⁺ = 350 S cm² mol⁻¹
Limiting molar conductivity of CH₃COO⁻, ∧°_CH3COO⁻ = 50 S cm² mol⁻¹

Step 2: Calculate limiting molar conductivity of acetic acid (∧°_m)
According to Kohlrausch's Law of independent migration of ions:

${\wedge }_m^{\circ }={\wedge }_{H^{+}}^{\circ }+{\wedge }_{C{H}_{3}CO{O}^{-}}^{\circ }$

Substituting the given values:

${\wedge }_m^{\circ }=350+50=400Sc{m}^{2}mo{l}^{-1}$

Step 3: Calculate the degree of dissociation (α)
The degree of dissociation is given by the ratio of molar conductivity to limiting molar conductivity:

$\alpha =\frac{{\wedge }_m}{{\wedge }_m^{\circ }}$

Substituting the values:

$\alpha =\frac{20}{400}=0.05$

Step 4: Calculate the dissociation constant (K_a)
The dissociation constant for a weak monobasic acid is given by Ostwald's dilution law:

$K_a=c{\alpha }^2$

Substituting c = 0.007 M and α = 0.05:

$K_a=0.007\times {\left(0.05\right)}^2$

$K_a=7\times {10}^{-3}\times 2.5\times {10}^{-3}=1.75\times {10}^{-5}mol{L}^{-1}$

Therefore, the dissociation constant of acetic acid is 1.75 × 10⁻⁵ mol L⁻¹.