Sulfide ion was determined by indirect titration with EDTA. To a solution containing 25.00mL of 0.04332 M Cu(ClO4)2 plus 15mL of 1M acetate buffer (pH 4.5) was added 25.00ml of unknown sulfide solution with vigorous stirring. The CuS precipitate was filtered and washed with hot water. Ammonia was added to the filtrate (which contains excess Cu2+) until the blue color of Cu(NH3)42+ was observed. Titration of the filtrate with 0.03927M EDTA required 12.11mL of EDTA to reach the end point with the indicator murexide. Find the molarity of sulfide in the unknown.
mmoles of Cu(ClO4)2 taken = MxV(mL) = 0.04332 mol/L x 25 mL = 1.083 mmol
Since Cu(ClO4)2 is completely dissociated in water, moles of Cu2+(aq) taken = 1.083 mmol Cu2+(aq)
Since there is excess Cu2+(aq), not all of the S2- is precipitated as CuS(s).
All of the excess Cu2+(aq) react NH3 to from Cu(NH3)4(aq).
Now Cu(NH3)4(aq) react with EDTA to form complex.
Hence mmoles of excess Cu2+(aq) = mmoles of EDTA taken
= MxV(mL) = 0.03927M x 12.11 mL = 0.47556 mmol Cu2+(aq).
Hence mmoles of Cu2+(aq) that forms precipitation with S2- to form CuS
= 1.083 mmol - 0.47556 mmol = 0.60744 mmol Cu2+(aq)
Since 1 mol Cu2+(aq) reacts with 1 mol S2-(aq), mmol of S2-(aq) in the solution = 0.60744 mmol S2-(aq)
Volume of the solution = 25.0 mL
Hence concentration of S2-(aq) in the solution = 0.60744 mmol / 25.0 mL = 0.02430 M (answer)
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