Question

A 0.4640 g sample of pewter, containing tin, lead, copper, and zinc, was dissolved in acid. Tin was precipitated as SnO2·4H2O and removed by filtration. The resulting filtrate and washings were diluted to a total volume of 200.0 mL. A 15.00 mL aliquot of this solution was buffered, and titration of the lead, copper, and zinc in solution required 34.88 mL of 0.001523 M EDTA. Thiosulfate was used to mask the copper in a second 20.00 mL aliquot. Titration of the lead and zinc in this aliquot required 33.83 mL of the EDTA solution. Finally, cyanide was used to mask the copper and the zinc in a third 25.00 mL aliquot. Titration of the lead in this aliquot required 26.21 mL of the EDTA solution. Determine the percent composition by mass of each metal in the pewter sample.

Answer 1

Volume of EDTA required for the titration of lead = 26.21 mL

Hence mmoles of EDTA =  26.21 mL x 0.001523 M = 0.03992 mmol

Hence mmoles of lead(Pb) = 0.03992 mmol = 0.03992 mmol x (1 mol / 1000 mmol) = 3.992x10^-5 mol

Hence mass of lead(Pb2+) = 3.992x10^-5​ mol x 207.2 g/mol = 0.00827 g Pb2+

Hence percent composition of lead(Pb2+) = (0.00827 g / 0.4640 g) x 100 = 1.78 % (answer)

Volume of EDTA required for the titration of lead(Pb2+) and zinc(Zn2+) = 33.83 mL

Hence mmoles of EDTA = 33.80 mL x 0.001523 M = 0.0514774 mmol

Hence mmoles of lead(Pb2+) and zinc(Zn2+) = 0.03992 mmol

= 0.0514774 mmol x (1 mol / 1000 mmol) = 5.15x10^-5 mol

=> moles of lead(Pb2+) + moles of zinc(Zn2+) =  5.15x10^-5 mol

=> moles of zinc(Zn2+) =  5.15x10^-5 mol - moles of lead(Pb2+) =  5.15x10^-5 mol -  3.992x10^-5 mol

=> moles of zinc(Zn2+) = 1.158x10^-5 mol

Hence mass of Zn in the pewter =mol x molar mass of Zn = 1.158x10^-5 mol x 65.38 g/mol = 0.0007571 g

Hence percent composition of lead(Pb) = (0.0007571 g / 0.4640 g) x 100 = 0.163 % (answer)

Volume of EDTA required for the titration of lead(Pb2+), zinc(Zn2+) and copper (Cu2+)= 34.88 mL

Hence mmoles of EDTA = 34.88 mL x 0.001523 M = 0.05312224 mmol

Hence mmoles of lead(Pb2+), zinc(Zn2+) and copper (Cu2+) =  0.05312224 mmol

= 0.05312224 mmol  x (1 mol / 1000 mmol) = 5.312224x10^-5 mol

=> moles of lead(Pb2+) + moles of zinc(Zn2+) + moles of copper(Cu2+) = 5.312224x10^-5 mol

=> moles of copper(Cu2+) =  5.15x10^-5 mol - moles of lead(Pb2+) - moles of zinc(Zn2+)

=  5.312224x10^-5 mol -  3.992x10^-5 mol - 1.158x10^-5 mol

=> moles of copper(Cu2+) = 0.163x10^-5 mol

Hence mass of Cu in the pewter =mol x molar mass of Cu = 0.163x10^-5 mol x 63.55 g/mol = 0.0001036 g

Hence percent composition of lead(Pb) = (0.0001036 g / 0.4640 g) x 100 = 0.0223 % (answer)

Hence percent composition of tin(Sn) = 100 - 1.78 - 0.163 - 0.0223 = 98.0 % (answer)