Understanding Molar Conductivity and Dissociation Constant of Acetic Acid This solution explains how to calculate the dissociation constant ($K_a$) of a weak electrolyte, specifically acetic acid, using its molar conductivity ($\Lambda_m$) and the limiting molar conductivities of its constituent ions. We will use key principles like Kohlrausch's Law and Ostwald's Dilution Law.

Key Concepts for Acetic Acid Dissociation Calculation

Before diving into the calculations, let's review the important concepts:

• Molar Conductivity ($\Lambda_m$): This measures the conducting ability of an electrolyte solution per mole of electrolyte at a specific concentration. It is given in units like S cm² mol^-1.
• Limiting Molar Conductivity ($\Lambda_m^\circ$): This is the molar conductivity of the electrolyte at infinite dilution (zero concentration), where the electrolyte is fully dissociated.
• Kohlrausch's Law: This law states that the limiting molar conductivity of an electrolyte is the sum of the limiting ionic conductivities of its constituent ions. For acetic acid (CH₃COOH), which dissociates into H⁺ and CH₃COO^- ions, it is expressed as: $$ \Lambda_{m(\text{CH}_3\text{COOH})}^{\circ} = \Lambda_{\text{H}^{+}}^{\circ} + \Lambda_{\text{CH}_{3}\text{COO}^{-}}^{\circ} $$
• Degree of Dissociation ($\alpha$): This is the fraction of the electrolyte molecules that dissociate into ions at a given concentration. It can be calculated using the molar conductivity ($\Lambda_m$) and the limiting molar conductivity ($\Lambda_m^\circ$): $$ \alpha = \frac{\Lambda_m}{\Lambda_m^\circ} $$
• Dissociation Constant ($K_a$): For a weak acid like acetic acid, the dissociation constant relates the concentration of the acid and the degree of its dissociation. According to Ostwald's Dilution Law: $$ K_a = \frac{C\alpha^2}{1-\alpha} $$ Where $C$ is the molar concentration of the acid. For weak electrolytes where $\alpha$ is small, the approximation $1-\alpha \approx 1$ is often used, simplifying the equation to $K_a \approx C\alpha^2$.

Step-by-Step Calculation for Acetic Acid Dissociation Constant

We are given the following data:

• Concentration of acetic acid, $C = 0.01$ M
• Molar conductivity of 0.01 M acetic acid, $\Lambda_m = 10$ S cm² mol^-1
• Limiting molar conductivity of H⁺ ion, $\Lambda_{\text{H}^{+}}^{\circ} = 345$ S cm² mol^-1
• Limiting molar conductivity of CH₃COO^- ion, $\Lambda_{\text{CH}_{3}\text{COO}^{-}}^{\circ} = 55$ S cm² mol^-1

Step 1: Calculate the Limiting Molar Conductivity of Acetic Acid ($\Lambda_{m(\text{CH}_3\text{COOH})}^\circ$) Using Kohlrausch's Law: $$ \Lambda_{m(\text{CH}_3\text{COOH})}^{\circ} = \Lambda_{\text{H}^{+}}^{\circ} + \Lambda_{\text{CH}_{3}\text{COO}^{-}}^{\circ} $$ Substitute the given values: $$ \Lambda_{m(\text{CH}_3\text{COOH})}^{\circ} = 345 ~\text{S}~\text{cm}^{2}\text{mol}^{-1} + 55~\text{S}~\text{cm}^{2}\text{mol}^{-1} $$ $$ \Lambda_{m(\text{CH}_3\text{COOH})}^{\circ} = 400 ~\text{S}~\text{cm}^{2}\text{mol}^{-1} $$ Step 2: Calculate the Degree of Dissociation ($\alpha$) Using the formula relating molar conductivity and limiting molar conductivity: $$ \alpha = \frac{\Lambda_m}{\Lambda_m^\circ} $$ Substitute the values for acetic acid: $$ \alpha = \frac{10 ~\text{S}~\text{cm}^{2}\text{mol}^{-1}}{400 ~\text{S}~\text{cm}^{2}\text{mol}^{-1}} $$ $$ \alpha = \frac{1}{40} = 0.025 $$ Step 3: Calculate the Dissociation Constant ($K_a$) Using Ostwald's Dilution Law. Since the degree of dissociation ($\alpha = 0.025$) is relatively small, we can use the approximation $1-\alpha \approx 1$. $$ K_a \approx C\alpha^2 $$ Substitute the concentration ($C = 0.01$ M) and the calculated degree of dissociation ($\alpha = 0.025$): $$ K_a \approx (0.01~\text{mol L}^{-1}) \times (0.025)^2 $$ First, calculate $\alpha^2$: $$ (0.025)^2 = 0.000625 $$ Now, calculate $K_a$: $$ K_a \approx (0.01) \times (0.000625) $$ $$ K_a \approx 0.00000625 ~\text{mol L}^{-1} $$ Expressing this in scientific notation: $$ K_a \approx 6.25 \times 10^{-6} ~\text{mol L}^{-1} $$

Conclusion

The calculated dissociation constant for 0.01 M acetic acid, based on the provided conductivity data and standard values, is approximately $6.25 \times 10^{-6}$ mol L^-1.