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Q2.11:
Conductivity of 0.00241 M acetic acid is 7.896 (\times) (10^{-5}) S (cm^{-1}). Calculate its molar conductivity. If (\Lambda^{0}_{m}) for acetic acid is 390.5 S (cm^{2} mol^{-1}), what is its dissociation constant?

Solution :

Step 1: Calculating Molar Conductivity ($\Lambda_m$)

The molar conductivity ($\Lambda_m$) is defined by the relationship between conductivity ($\kappa$) and the concentration of the electrolyte ($C$). The formula is given by:

$\Lambda_m = \frac{\kappa \times 1000}{C}$

Given Variables:

  • Conductivity ($\kappa$) = $7.896 \times 10^{-5} \, \text{S cm}^{-1}$
  • Concentration ($C$) = $0.00241 \, \text{M (mol L}^{-1})$

Calculation:

$\Lambda_m = \frac{7.896 \times 10^{-5} \, \text{S cm}^{-1} \times 1000 \, \text{cm}^3 \text{L}^{-1}}{0.00241 \, \text{mol L}^{-1}}$

$\Lambda_m = \frac{0.07896}{0.00241} \, \text{S cm}^2 \text{mol}^{-1}$

$\Lambda_m \approx 32.76 \, \text{S cm}^2 \text{mol}^{-1}$ [Per the definition of Molar Conductivity]

Step 2: Calculating Degree of Dissociation ($\alpha$)

The degree of dissociation ($\alpha$) relates the molar conductivity at a given concentration to the molar conductivity at infinite dilution ($\Lambda^0_m$).

$\alpha = \frac{\Lambda_m}{\Lambda^0_m}$

Given Variables:

  • $\Lambda_m = 32.76 \, \text{S cm}^2 \text{mol}^{-1}$
  • $\Lambda^0_m = 390.5 \, \text{S cm}^2 \text{mol}^{-1}$

Calculation:

$\alpha = \frac{32.76}{390.5} \approx 0.0839$ [By the Arrhenius theory of electrolytic dissociation]

Step 3: Calculating Dissociation Constant ($K_a$)

For a weak electrolyte like acetic acid ($CH_3COOH \rightleftharpoons CH_3COO^- + H^+$), the dissociation constant is determined using Ostwald’s Dilution Law:

$K_a = \frac{C \alpha^2}{1 - \alpha}$

Variables:

  • $C = 0.00241 \, \text{M}$
  • $\alpha = 0.0839$

Calculation:

$K_a = \frac{0.00241 \times (0.0839)^2}{1 - 0.0839}$

$K_a = \frac{0.00241 \times 0.007039}{0.9161}$

$K_a = \frac{0.00001696}{0.9161} \approx 1.85 \times 10^{-5}$

Visual Representation of Dissociation Equilibrium

Equilibrium: CH₃COOH ⇌ CH₃COO⁻ + H⁺ Start End

Final Solution: The molar conductivity of acetic acid is $32.76 \, \text{S cm}^2 \text{mol}^{-1}$ and its dissociation constant $K_a$ is $1.85 \times 10^{-5}$.


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