PbCl2↓ + H2SO4 <-> PbSO4↓ + 2HCl — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

Step 1: Identify starting materials: Lead(II) chloride (PbCl2↓) is a white precipitate (insoluble in cold water). Step 2: Analyze the reaction with sulfuric acid: Sulfuric acid reacts with PbCl2 to form lead(II) sulfate (PbSO4↓), which has a much lower solubility product (Ksp) than PbCl2, driving the replacement of the solid phase. Step 3: Since one solid precipitate (PbCl2↓) is converted into another solid precipitate (PbSO4↓), this is classified as a precipitate exchange reaction (type C).