An impure sample of PCl ₅ weighing 10.425 g is completely hydrolyzed in water. The resulting solution requires 200 mL of 0.5 M Ba(OH) ₂ solution for complete neutralization. The percentage purity of PCl ₅ in the sample is (Given: Molar mass of PCl ₅ = 208.5 g mol ⁻¹ )

Step-by-step solution

The hydrolysis of PCl ₅ proceeds as follows: PCl ₅ + 4 H ₂ O H ₃ PO ₄ + 5 HCl Let the number of moles of pure PCl ₅ be n . Moles of H ₃ PO ₄ formed = n Moles of HCl formed = 5n H ₃ PO ₄ is a tribasic acid, furnishing 3 H ^+ ions. HCl is a monobasic acid, furnishing 1 H ^+ ion. Total moles of H ^+ furnished = 3n + 5n = 8n Moles of Ba(OH) ₂ used = Molarity Volume (in L) = 0.5 0.2 = 0.1 mol Since Ba(OH) ₂ furnishes 2 OH ^- ions per formula unit, moles of OH ^- = 0.1 2 = 0.2 mol For complete neutralization: Moles of H