PbCrO4↓ + 4NaOH(Excess) -> Na2[Pb(OH)4] — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

Step 1: Identify starting materials: Lead(II) chromate (PbCrO4↓) is a highly insoluble yellow-orange precipitate. Step 2: Analyze the reaction with excess strong base (NaOH): Lead(II) is amphoteric and reacts with excess hydroxide ions to form the soluble tetrahydroxoplumbate(II) complex, [Pb(OH)4]2-. Step 3: Since the solid yellow precipitate dissolves completely to form a clear solution of a soluble complex, this is a precipitate dissolution reaction (type B).