Question

A 0.4423 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 35.87 mL of 0.001481 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 33.95 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 25.03 mL of the EDTA solution. Determine the percent composition by mass of each metal in the pewter sample

Answer 1

EDTA form 1:1 complex with copper, lead and Zinc

so the total amount of copper + lead + zince in 200 mL of the solution in which the 0.4423 g of sample after removing tin was dissolved was

(200/15)x(35.87 mL x 0.001481 mol/L)/1000 = 7.08 x 10^-4 moles

After masking copper, Lead and zinc was estimated to be

(200/20) x (33.95 mL x 0.001481 mol/L)/1000 = 5.02 x 10^-4 moles

After masking copper and zinc the lead was found to be

(200/25) x (25.03 mL x 0.001481 mol/L)/1000 = 2.96 x 10^-4 moles

2.96 x 10^-4 moles of Lead will be  2.96 x 10^-4 moles x 207.2 g/mol = 0.0614 g

The amount of zinc in this sample is 5.02 x 10^-4 moles - 2.96 x 10^-4 moles = 2.06 x 10^-4 moles

2.06 x 10^-4 moles of zinc will be 2.06 x 10^-4 moles x 65.38 g/mol = 0.0134 g

The amount of copper in the sample will be 7.08 x 10^-4 moles - 5.02 x 10^-4 moles = 2.06 x 10^-4 moles

2.06 x 10^-4 moles of copper will be 2.06 x 10^-4 moles x 63.54 g/mol = 0.0131 g

copper will be (0.0131/0.4423)x100 = 2.96%

zinc will be (0.0134/0.4423)x100 =3.03 %

Lead will be (0.0614/0.4423)x100 = 13.88%

Tin will be 100 - (2.96 + 3.03 + 13.88) = 80.12%