Question

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.

Answer 1

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 S^2- 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 S^2- to form CuS

= 1.083 mmol -  0.47556 mmol = 0.60744 mmol Cu²⁺(aq)   

Since 1 mol Cu²⁺(aq) reacts with 1 mol S^2-(aq), mmol of S^2-(aq) in the solution =  0.60744 mmol S^2-(aq)

Volume of the solution = 25.0 mL

Hence concentration of S^2-(aq) in the solution = 0.60744 mmol / 25.0 mL = 0.02430 M (answer)