Calculate Ka of acetic acid if its 0.05 M solution has molar conductivity of 7.814 × 10^-4 Ω^-1 m^2 — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Dissociation of Weak Electrolytes and Kohlrausch's Relation Weak acids, such as acetic acid ($\ce{CH3COOH}$), do not dissociate completely in aqueous solution. Instead, they establish a dynamic chemical equilibrium: $$\ce{CH3COOH(aq) <=> CH3COO^-(aq) + H^+(aq)}$$ For any weak monobasic acid, the degree of dissociation ($\alpha$) represents the fraction of total acid molecules that have ionized in the solution. It is directly related to the molar conductivity of the acid at a given concentration $C$ ($\Lambda_m$) and its limiting molar conductivity at infinite dilution ($\Lambda_m^\circ$): $$\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$$ Step 2 - Calculate the Degree of Dissociation ($\alpha$) We are given the following experimental parameters at $25^\circ\text{C}$: * Concentration of the solution ($C$) = $0.05\text{ M}$ * Molar conductivity ($\Lambda_m$) = $7.814 \times 10^{-4}\ \Omega^{-1}\text{ m}^2\text{ mol}^{-1}$ * Limiting molar conductivity ($\Lambda_m^\circ$) = $3.907 \times 10^{-2}\ \Omega^{-1}\text{ m}^2\text{ mol}^{-1}$ Substitute these values with their respective units into Kohlrausch's formula to find $\alpha$: $$\alpha = \frac{7.814 \times 10^{-4}\ \Omega^{-1}\text{ m}^2\text{ mol}^{-1}}{3.907 \times 10^{-2}\ \Omega^{-1}\text{ m}^2\text{ mol}^{-1}}$$ Performing the division: $$\alpha = \left( \frac{7.814}{3.907} \right) \times 10^{-4 - (-2)}$$ $$\alpha = 2 \times 10^{-2} = 0.02$$ Since $\alpha = 0.02$ (or $2.0\%$), the degree of dissociation is extremely small ($\alpha \ll 1$). Step 3 - Apply Ostwald's Dilution Law to Calculate the Dissociation Constant ($K_a$) The acid dissociation constant ($K_a$) for a weak monobasic acid is defined by the equilibrium relation: $$K_a = \frac{C \cdot \alpha^2}{1 - \alpha}$$ Since $\alpha \ll 1$, the term $1 - \alpha \approx 1$. Thus, we can safely apply the standard simplified approximation of Ostwald's Dilution Law: $$K_a \approx C \cdot \alpha^2$$ Now, substitute the initial concentration ($C = 0.05\text{ M}$) and the calculated degree of dissociation ($\alpha = 0.02$) into this expression: $$K_a = 0.05\text{ mol/L} \times (0.02)^2$$ $$K_a = 0.05 \times 0.0004$$ $$K_a = 5 \times 10^{-2} \times 4 \times 10^{-4} = 20 \times 10^{-6}$$ $$K_a = \mathbf{2 \times 10^{-5}}$$ *(Note: If we calculate the exact value without using the $1 - \alpha \approx 1$ approximation: $K_a = \frac{2 \times 10^{-5}}{1 - 0.02} = \frac{2 \times 10^{-5}}{0.98} \approx 2.04 \times 10^{-5}$, which is extremely close to our approximated value and is rounded to $2 \times 10^{-5}$ in competitive examinations).* Step 4 - Evaluate and Explain the Options * **Option (A) is correct:** As calculated using Ostwald's dilution law and the experimental molar conductivity ratio, the dissociation constant ($K_a$) of acetic acid is $2 \times 10^{-5}$. * **Option (B) is incorrect:** $1.8 \times 10^{-5}$ is the standard literature value of $K_a$ for acetic acid at $25^\circ\text{C}$ under pure standard conditions, but under the specific experimental measurements provided in this problem, the calculated value is $2 \times 10^{-5}$. * **Option (C) is incorrect:** This value ($4 \times 10^{-4}$) is obtained by mathematical errors, such as squaring the molar conductivity values or using incorrect units. * **Option (D) is incorrect:** This value ($0.02$) represents the degree of dissociation ($\alpha$), not the acid dissociation constant ($K_a$). $$\text{Correct Option: } \boxed{\text{A}}$$