An aqueous solution of 'X' is added slowly to an aqueous solution of 'Y'. The variation in conductiv — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Core Principles of Conductometric Titrations The electrical conductivity ($\kappa$) of an aqueous electrolyte solution depends directly on: 1. The concentration (number) of free ions present per unit volume. 2. The charge of these ions. 3. The individual ionic mobilities ($\lambda^\circ$) of the ions under an electric field. Among common ions, hydrogen ions ($\ce{H^+}$, $\lambda^\circ \approx 349.6\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$) and hydroxide ions ($\ce{OH^-}$, $\lambda^\circ \approx 199.1\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$) have exceptionally high ionic mobilities due to the Grotthuss proton-hopping mechanism. Standard ions such as potassium ($\ce{K^+}$, $\lambda^\circ \approx 73.5$), silver ($\ce{Ag^+}$, $\lambda^\circ \approx 61.9$), sodium ($\ce{Na^+}$, $\lambda^\circ \approx 50.1$), and acetate ($\ce{CH3COO^-}$, $\lambda^\circ \approx 40.9$) have much lower mobilities. During a titration, as X is added to Y, ions of one mobility are consumed and replaced by other ions, altering the overall conductivity. Step 2 - Analyze Pair (A): \ce{(C2H5)3N} added to \ce{CH3COOH} * **Reactants:** Triethylamine ($\ce{(C2H5)3N}$) is a weak base, and acetic acid ($\ce{CH3COOH}$) is a weak acid. * **Initial State:** $\ce{CH3COOH}$ is a weak electrolyte and dissociates only partially in water, so the initial conductivity is low. * **Before the Equivalence Point:** Adding the weak base $\ce{(C2H5)3N}$ produces a highly soluble strong electrolyte, triethylammonium acetate: $$\ce{CH3COOH(aq) + (C2H5)3N(aq) -> CH3COO^-(aq) + (C2H5)3NH^+(aq)}$$ This reaction converts neutral, unionized molecules into free ions, causing a steady **increase** in conductivity. * **After the Equivalence Point:** Once all the acetic acid is neutralized, further addition of excess weak base $\ce{(C2H5)3N}$ does not produce any significant number of new ions because its dissociation is extremely low and is further suppressed by the common ion effect. Thus, the conductivity **does not change much**. * **Matching:** **(A) $\rightarrow$ (R)**: Conductivity increases then does not change much. Step 3 - Analyze Pair (B): \ce{KI} (0.1 M) added to \ce{AgNO3} (0.01 M) * **Reactants:** Potassium iodide ($\ce{KI}$) and silver nitrate ($\ce{AgNO3}$) are both strong electrolytes. * **Initial State:** The flask contains $\ce{AgNO3}$ which is fully dissociated into highly conducting $\ce{Ag^+}$ and $\ce{NO3^-}$ ions. * **Before the Equivalence Point:** Adding $\ce{KI}$ precipitates the highly insoluble silver iodide ($\ce{AgI}$): $$\ce{Ag^+(aq) + NO3^-(aq) + K^+(aq) + I^-(aq) -> AgI(s) \downarrow + K^+(aq) + NO3^-(aq)}$$ In this reaction, highly mobile $\ce{Ag^+}$ ions are progressively replaced by $\ce{K^+}$ ions. Because the ionic mobilities of $\ce{Ag^+}$ ($61.9$) and $\ce{K^+}$ ($73.5$) are very similar, the overall conductivity of the solution remains **nearly constant** (does not change much). * **After the Equivalence Point:** Once all $\ce{Ag^+}$ is precipitated, any further addition of excess $\ce{KI}$ directly accumulates free, mobile $\ce{K^+}$ and $\ce{I^-}$ ions in the solution, leading to a sharp **increase** in conductivity. * **Matching:** **(B) $\rightarrow$ (S)**: Conductivity does not change much then increases. Step 4 - Analyze Pair (C): \ce{CH3COOH} added to \ce{KOH} * **Reactants:** Weak acetic acid ($\ce{CH3COOH}$) is added to strong potassium hydroxide ($\ce{KOH}$). * **Initial State:** The flask contains $\ce{KOH}$ which is fully dissociated, yielding a high concentration of highly mobile $\ce{OH^-}$ ions. The initial conductivity is high. * **Before the Equivalence Point:** Adding $\ce{CH3COOH}$ neutralizes the base to form potassium acetate: $$\ce{K^+(aq) + OH^-(aq) + CH3COOH(aq) -> CH3COO^-(aq) + K^+(aq) + H2O(l)}$$ Here, highly mobile $\ce{OH^-}$ ions ($\lambda^\circ \approx 199$) are consumed and replaced by much slower acetate ions ($\ce{CH3COO^-}$, $\lambda^\circ \approx 41$). This results in a sharp **decrease** in conductivity. * **After the Equivalence Point:** Once the base is completely neutralized, further addition of excess weak acetic acid ($\ce{CH3COOH}$) adds a weak electrolyte that cannot dissociate significantly in the presence of the common acetate ions. Thus, the conductivity **does not change much**. * **Matching:** **(C) $\rightarrow$ (Q)**: Conductivity decreases then does not change much. Step 5 - Analyze Pair (D): \ce{NaOH} added to \ce{HI} * **Reactants:** Strong base sodium hydroxide ($\ce{NaOH}$) is added to strong hydroiodic acid ($\ce{HI}$). * **Initial State:** The flask contains fully dissociated $\ce{HI}$, providing a high concentration of highly mobile hydronium ions ($\ce{H^+}$). The initial conductivity is extremely high. * **Before the Equivalence Point:** Adding $\ce{NaOH}$ neutralizes the strong acid: $$\ce{H^+(aq) + I^-(aq) + Na^+(aq) + OH^-(aq) -> Na^+(aq) + I^-(aq) + H2O(l)}$$ During this phase, highly mobile $\ce{H^+}$ ions ($\lambda^\circ \approx 350$) are neutralized and replaced by much slower sodium ions ($\ce{Na^+}$, $\lambda^\circ \approx 50$). This causes a sharp **decrease** in conductivity up to the equivalence point. * **After the Equivalence Point:** Beyond neutralization, excess strong base $\ce{NaOH}$ completely dissociates into free $\ce{Na^+}$ and highly mobile $\ce{OH^-}$ ions ($\lambda^\circ \approx 199$), causing a sharp **increase** in conductivity. * **Matching:** **(D) $\rightarrow$ (P)**: Conductivity decreases then increases. Step 6 - Summary of Final Column Match Combining the matched pairs from the step-by-step physical and chemical analyses: * (A) matches with (R) * (B) matches with (S) * (C) matches with (Q) * (D) matches with (P) $$\boxed{\text{A}\rightarrow\text{R};\ \text{B}\rightarrow\text{S};\ \text{C}\rightarrow\text{Q};\ \text{D}\rightarrow\text{P}}$$