Question

A 0.4505 g sample of pewter, containing tin, lead, copper, and zinc, was dissolved in acid. Tin was precipitated as SnO2⋅4H2Oand removed by filtration. The resulting filtrate and washings were diluted to a total volume of 250.0 mL. A 15.00 mL aliquot of this solution was buffered, and titration of the lead, copper, and zinc in solution required 35.10 mL of 0.001497 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 33.38 mL of the 0.001497 M 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 26.02 mL of the 0.001497 M EDTA solution. Determine the percent composition by mass of each metal in the pewter sample.

%Cu:

%Zn:

%Pb:

%Sn:

Answer 1

Each of the metal ions i.e. Zn²⁺, Pb²⁺ and Cu²⁺ react with EDTA in 1:1 stoichiometric amounts.

Titration of 25.00 ml aliquot (last aliquot)

Moles of lead = 0.001497 M * 26.02 ml/1000 ml = 0.00003895 mol

Titration of 20.00 ml aliquot (second aliquot)

Moles of lead + zinc = 0.001497 M* 33.38 ml/1000 ml

= 0.00004997 mol

Moles of lead in 20.00 ml aliquot = (0.00003895 mol/25 ml)*20 ml

= 0.00003116

Moles of zinc = 0.00004997 - 0.00003116 = 0.00001881 mol

Titration of 15.00 ml aliquot (first aliquot)

Moles of lead+copper+zinc = 0.00001497 M*35.10 ml/1000 ml

= 0.00005255 mol

Moles of lead in 15 ml = 0.00003895 mol*15 ml/25 ml = 0.00002337 mol

Moles of zinc in 15 ml = 0.00001881 mol *20 ml/15 ml= 0.00001411 mol

Moles of Copper in 15 ml = 0.00005255 - 0.00001411 - 0.00002337 = 0.00001507 mol

Molar mass of lead = 207.2 g/mol

Mass of lead in 250 ml.solution = (0.00003895 mol/25 ml)*250 ml*206 g/mol

= 0.08070 g

Molar mass of zinc = 65.55 g/mol

Mass of Zinc in 250 ml solution = (0.00001881 mol/20 ml)*250 ml*65.38 g/mol

= 0.01537 g

Molar mass of copper = 63.55 g/mol

Mass of copper in 250 ml.solution = (0.00001507 mol/15 ml)*250 ml*63.55 g/mol

= 0.01596 g

% of lead = (0.08070 g/0.4505 g)*100 = 17.91 %

% of Zinc = (0.01537 g/0.4505 g)*100 = 3.412 %

% of copper = (0.01596 g/0.4505 g)*100 = 3.543 %

% of Tin = 100 - 17.91 - 3.412 - 3.543 = 75.13%