How does the electrical conductivity of 20 ml of 0.2 M - change when 0.5 M - Ba(OH)2 solution is gra — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Initial State of the Solution We begin with $20\text{ mL}$ of $0.2\text{ M}$ magnesium sulphate ($\ce{MgSO4}$) solution. Magnesium sulphate is a soluble, strong electrolyte that dissociates completely in water to yield free, highly mobile ions: $$\ce{MgSO4(aq) -> Mg^2+(aq) + SO4^2-(aq)}$$ These free $\ce{Mg^2+}$ and $\ce{SO4^2-}$ ions are responsible for carrying the electric current. Therefore, the initial solution has a high electrical conductivity. Step 2 - Analyze the Chemical Reaction upon Addition of Barium Hydroxide When barium hydroxide ($\ce{Ba(OH)2}$) is gradually added to the magnesium sulphate solution, a double displacement precipitation reaction occurs: $$\ce{MgSO4(aq) + Ba(OH)2(aq) -> Mg(OH)2(s) v + BaSO4(s) v}$$ Let us examine the nature of both products formed: 1. **Barium sulphate ($\ce{BaSO4}$):** This is an extremely insoluble white crystalline salt with a very low solubility product ($K_{\text{sp}} \approx 1.1 \times 10^{-10}$). 2. **Magnesium hydroxide ($\ce{Mg(OH)2}$):** This is also an insoluble white precipitate with a very low solubility product ($K_{\text{sp}} \approx 1.8 \times 10^{-11}$). Step 3 - Analyze Conductivity Changes Up to the Equivalence Point As the titration progresses and $\ce{Ba(OH)2}$ is added, the added $\ce{Ba^2+}$ and $\ce{OH^-}$ ions react quantitatively with the existing $\ce{Mg^2+}$ and $\ce{SO4^2-}$ ions in the solution to precipitate out completely as solid $\ce{BaSO4}$ and $\ce{Mg(OH)2}$. Since these solid precipitates do not dissociate into mobile ions, the concentration of free charge-carrying ions in the solution decreases steadily. Consequently, the electrical conductivity of the solution **decreases continuously** and reaches a minimum near-zero value at the equivalence point. Step 4 - Analyze Conductivity Changes Beyond the Equivalence Point (In Excess) Once the equivalence point is reached, all the $\ce{Mg^2+}$ and $\ce{SO4^2-}$ ions have been completely precipitated out of the solution. When more $\ce{Ba(OH)2}$ is added in excess, no further precipitation reaction can occur. Barium hydroxide is a soluble strong base that dissociates completely in water: $$\ce{Ba(OH)2(aq) -> Ba^2+(aq) + 2OH^-(aq)}$$ The addition of excess $\ce{Ba(OH)2}$ continuously introduces free, highly mobile $\ce{Ba^2+}$ and $\ce{OH^-}$ ions into the solution. Since the concentration of mobile charge carriers in the solution starts rising, the electrical conductivity of the solution **increases continuously**. Step 5 - Evaluate and Explain the Options * **Option (A) is incorrect:** The conductivity does not decrease continuously because addition of excess $\ce{Ba(OH)2}$ after the equivalence point introduces new free ions, which raises the conductivity. * **Option (B) is incorrect:** The conductivity does not increase continuously; it must decrease first as the conducting ions are removed from the solution during precipitation. * **Option (C) is incorrect:** The conductivity first decreases due to precipitation and then increases, which is the exact opposite of this option. * **Option (D) is correct:** As shown by the precipitation of all ions to form insoluble $\ce{BaSO4}$ and $\ce{Mg(OH)2}$, followed by the addition of excess free $\ce{Ba^2+}$ and $\ce{OH^-}$ ions, the electrical conductivity first decreases and then increases. $$\text{Correct Option: } \boxed{\text{D}}$$