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

💡 Solution & Explanation

Step 1 - Understand the Chemical Reaction and Species Involved We are titrating a solution of silver nitrate ($\ce{AgNO3}$) in the titration flask by adding potassium chloride ($\ce{KCl}$) from the burette. Initially, the flask contains a highly dissociated strong electrolyte, silver nitrate, which exists completely as free ions in solution: $$\ce{AgNO3(aq) -> Ag^+(aq) + NO3^-(aq)}$$ When we add $\ce{KCl}$, the chloride ions ($\ce{Cl^-}$) react immediately with the silver ions ($\ce{Ag^+}$) to form a highly insoluble precipitate of silver chloride ($\ce{AgCl(s)}$): $$\ce{Ag^+(aq) + NO3^-(aq) + K^+(aq) + Cl^-(aq) -> AgCl(s) v + K^+(aq) + NO3^-(aq)}$$ In this precipitation reaction: * The highly mobile silver ions ($\ce{Ag^+}$) are progressively removed from the solution as a solid precipitate. * The charge in the solution is balanced by the incoming potassium ions ($\ce{K^+}$). * Therefore, the net effect up to the equivalence point is the replacement of $\ce{Ag^+}$ ions in the solution by $\ce{K^+}$ ions, while the concentration of $\ce{NO3^-}$ remains virtually unchanged. Step 2 - Compare the Ionic Mobilities of the Replaced Ions The molar conductivity at infinite dilution ($\lambda^\circ$) of the participating cations at $298\text{ K}$ are: * $$\lambda^\circ(\ce{Ag^+}) = 61.9\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$$ * $$\lambda^\circ(\ce{K^+}) = 73.5\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$$ Since the ionic mobility of the incoming $\ce{K^+}$ ion ($73.5$) is very close to, and only slightly higher than, the ionic mobility of the outgoing $\ce{Ag^+}$ ion ($61.9$): * As the titration proceeds up to the equivalence point, the total number of ions remains constant. * The slight increase in individual mobility from $\ce{Ag^+}$ to $\ce{K^+}$ results in a **nearly constant, or extremely gently rising, conductance**. Step 3 - Analyze the Conductance Behavior after the Equivalence Point At the equivalence point, all free $\ce{Ag^+}$ ions have been completely precipitated as solid $\ce{AgCl(s)}$. Any further addition of $\ce{KCl}$ beyond this point does not participate in any precipitation reaction. Instead, it simply introduces additional free potassium ions ($\ce{K^+}$) and chloride ions ($\ce{Cl^-}$) directly into the solution: $$\ce{KCl(aq) -> K^+(aq) + Cl^-(aq)}$$ This continuous accumulation of free ions post-equivalence point leads to a **sharp, linear increase in the overall conductance** of the solution. Step 4 - Evaluate and Explain Each Option * **Option (A) is incorrect:** This V-shaped graph represents a strong acid-strong base titration where highly mobile $\ce{H^+}$ ions are replaced by much slower cations, causing a sharp drop before a sharp rise. * **Option (B) is correct:** This graph correctly shows a nearly flat (near-constant or extremely gently rising) conductance line up to the equivalence point, followed by a sharp, linear upward slope once excess $\ce{KCl}$ is added. * **Option (C) is incorrect:** This graph shows a continuous steady rise from the very beginning, which does not account for the nearly constant replacement phase. * **Option (D) is incorrect:** This graph shows a decrease in conductance up to the end point followed by a plateau, which is chemically incorrect for this precipitation system. $$\text{Correct Option: } \boxed{B}$$