Which of the following graph truly represents the titration of CH3COOH solution against solution? — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Principles of Conductometric Titrations The electrical conductance ($G$) of an aqueous electrolyte solution depends directly on: 1. The total concentration (number) of ions present in the solution. 2. The charge of the ions. 3. The individual ionic mobilities (speeds) of the ions under an electric field. Among all aqueous ions, the hydrogen ion ($\ce{H^+}$) and the hydroxide ion ($\ce{OH^-}$) possess exceptionally high ionic mobilities due to the Grotthuss proton-hopping mechanism: $$\lambda^\circ(\ce{H^+}) \approx 349.6\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$$ $$\lambda^\circ(\ce{OH^-}) \approx 199.1\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$$ Other typical ions like sodium ($\ce{Na^+}$) and acetate ($\ce{CH3COO^-}$) have much lower ionic mobilities: $$\lambda^\circ(\ce{Na^+}) \approx 50.1\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$$ $$\lambda^\circ(\ce{CH3COO^-}) \approx 40.9\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$$ During a conductometric titration, as the titrant is added, chemical reactions replace ions of one mobility with ions of another mobility, changing the overall conductance. Step 2 - Analyze the Titration of \ce{CH3COOH} with \ce{NaOH} Step-by-Step We analyze the behavior of the conductance graph in four distinct regions during the titration of a weak acid ($\ce{CH3COOH}$) with a strong base ($\ce{NaOH}$): * **Region 1: Initial State (No base added yet)** Acetic acid ($\ce{CH3COOH}$) is a weak acid and dissociates only partially in water: $$\ce{CH3COOH(aq) <=> CH3COO^-(aq) + H^+(aq)}$$ Because of this extremely low degree of ionization ($\alpha \ll 1$), the concentration of free $\ce{H^+}$ and $\ce{CH3COO^-}$ ions is very low. Consequently, the initial conductance of the solution is **low**. * **Region 2: Very Beginning of Base Addition (The slight dip)** When the first few drops of the strong base $\ce{NaOH}$ are added, the highly dissociated salt sodium acetate ($\ce{CH3COONa}$) is formed. The sudden increase in the concentration of acetate ions ($\ce{CH3COO^-}$) strongly suppresses the dissociation of the remaining unreacted acetic acid due to the **common ion effect**. This initial suppression causes a very small, temporary decrease in conductance, appearing as a **slight dip** at the start of the curve. * **Region 3: Up to the Equivalence Point (Gradual increase)** As the addition of $\ce{NaOH}$ continues, the neutralization reaction proceeds: $$\ce{CH3COOH(aq) + Na^+(aq) + OH^-(aq) -> CH3COO^-(aq) + Na^+(aq) + H2O(l)}$$ In this reaction, the poorly dissociated, neutral $\ce{CH3COOH}$ molecules are converted into a highly dissociated strong electrolyte, sodium acetate ($\ce{CH3COONa}$), which exists completely as free ions ($\ce{Na^+}$ and $\ce{CH3COO^-}$). Although these ions have lower mobilities than $\ce{H^+}$, their total concentration increases continuously. This continuous increase in the overall ion population leads to a **steady and gradual increase** in conductance up to the equivalence point. * **Region 4: Beyond the Equivalence Point (Steep rise)** Once all the acetic acid has been completely neutralized, any further addition of $\ce{NaOH}$ introduces excess free sodium ions ($\ce{Na^+}$) and highly mobile hydroxide ions ($\ce{OH^-}$) into the solution: $$\ce{NaOH(aq) -> Na^+(aq) + OH^-(aq)}$$ Since the ionic mobility of $\ce{OH^-}$ is extremely high, the concentration of these highly conducting ions increases rapidly, resulting in a **sharp, steep upward rise** in the conductance curve after the equivalence point. Step 3 - Evaluate and Explain Each Option * **Option (A) is incorrect:** This graph shows a high initial conductance that drops sharply to a minimum and then rises steeply. This represents the titration of a **strong acid** (such as $\ce{HCl}$) against a **strong base** (such as $\ce{NaOH}$), where highly mobile $\ce{H^+}$ ions are initially consumed and replaced by less mobile $\ce{Na^+}$ ions, followed by excess $\ce{OH^-}$ causing a steep rise. * **Option (B) is incorrect:** This graph shows a steady increase in conductance followed by a flat plateau. This represents the titration of a **weak acid** (such as $\ce{CH3COOH}$) against a **weak base** (such as $\ce{NH4OH}$), where the excess weak base added after the equivalence point does not dissociate significantly and therefore fails to increase the conductance further. * **Option (C) is correct:** This graph perfectly captures all four stages of the weak acid–strong base titration: low initial conductance, a tiny initial dip due to suppressed ionization, a steady linear increase up to the equivalence point due to salt formation, and a sharp, steep upward break after the equivalence point due to excess highly mobile hydroxide ions. * **Option (D) is incorrect:** This curve shows a high initial conductance decreasing to a plateau, which does not match the electrochemical characteristics of a weak acid titration. $$\text{Correct Option: } \boxed{C}$$