PbCl2↓ + Hot water -> Pb2+(aq.) + 2Cl-(aq.) — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

Step 1: Identify starting materials: Lead(II) chloride (PbCl2↓) is sparingly soluble in cold water. Step 2: Observe the temperature effect: The dissolution of PbCl2 is highly endothermic. Heating the mixture dramatically increases its solubility product (Ksp), causing the precipitate to dissolve completely. Step 3: Since the insoluble precipitate is dissolved to form clear solvated ions, this is a precipitate dissolution reaction (type B).