A mixture containing As₂S₃ and As₂S₅ requires 20 ml of 0.05 N iodine for titration. The resulting solution is then acidified and excess of KI was added. The liberated iodine required 1.24 g hypo, Na₂S

Question

A mixture containing As₂S₃ and As₂S₅ requires 20 ml of 0.05 N iodine for titration. The resulting solution is then acidified and excess of KI was added. The liberated iodine required 1.24 g hypo, Na₂S₂O₃·5H₂O, for complete reaction. The reactions are As₂S₃ + 2I₂ + 2H₂S → As₂S₅ + 4H⁺ + 4I⁻; As₂S₅ + 4H⁺ + 2I⁻ → As₂S₃ + 2I₂ + 2H₂S. The mole per cent of As₂S₃ in the original mixture is (As = 75)

TestHub · Accepted Answer

I₂ in first titration = 20×0.05 = 1 meq = 0.5 mmol; from As₂S₃ + 2I₂, As₂S₃ = 0.25 mmol. Hypo = 1.24/248 = 5 mmol, so liberated I₂ = 2.5 mmol; from As₂S₅ giving 2I₂, total As₂S₅ present = 1.25 mmol. Original As₂S₅ = 1.25-0.25 = 1.0 mmol. Mole % As₂S₃ = 0.25/1.25×100 = 20%. Ans 0020.

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