The reaction of white phosphorus on boiling with alkali in inert atmosphere resulted in the formatio — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

**Step 1: Identify product 'A' from white phosphorus reaction with alkali** White phosphorus (P₄) reacts with boiling alkali in an inert atmosphere via disproportionation: P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂ Product 'A' is sodium hypophosphite (NaH₂PO₂). **Step 2: Determine the structure of NaH₂PO₂** Hypophosphite has the structure: Na⁺[H₂PO₂]⁻ The anion contains one P-H bond and one P-O bond (or P-OH group depending on ionization). **Step 3: Write the reaction with AgNO₃** In aqueous medium, hypophosphite acts as a reducing agent and gets oxidized: H₂PO₂⁻ + 2Ag⁺ + 2H₂O → H₂PO₃⁻ + 2Ag↓ + 2H⁺ **Step 4: Count silver atoms produced** From the balanced equation, 1 mole of H₂PO₂⁻ produces **2 moles of Ag**. **Step 5: Account for complete reaction** Since 1 mol of NaH₂PO₂ contains 1 mol of H₂PO₂⁻, and considering that hypophosphite can be oxidized further in the presence of excess AgNO₃: H₂PO₂⁻ + 6Ag⁺ + 2H₂O → PO₄³⁻ + 6Ag↓ + 4H⁺ This gives **6 moles of Ag** per mole of hypophosphite. Therefore, the answer is **6.00**.