Question

A 0.4627 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 20.00 mL aliquot of this solution was buffered, and titration of the lead, copper, and zinc in solution required 34.11 mL of 0.001500 M EDTA. Thiosulfate was used to mask the copper in a second 25.00 mL aliquot. Titration of the lead and zinc in this aliquot required 34.45 mL of the EDTA solution. Finally, cyanide was used to mask the copper and the zinc in a third 30.00 mL aliquot. Titration of the lead in this aliquot required 24.39 mL of the EDTA solution. Determine the percent composition by mass of each metal in the pewter sample.

Answer 1

Total concentration of Pb, Cu and Zn

M1V1 = M2V2

M1 *20 = 34.11 * 0.0015

M1 = 2.56 *10^-3

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total concentration of lead and zinc

M1*25 = 34.45 *0.0015

M1 = 2.067 *10-^3

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concentration of lead

M1 * 30 = 24.39 *0.0015

M1 = 1.218 * 10^-3 M

      = 1.218 * 10^-3 * 207.2 = 0.252 g/L

amount of lead present in 200 mL = 0.05 g

mass percentage of Lead = 0.05*100/0.4627 = 10.91 %

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Concentration of zinc

= total concentration of lead + zinc - concentration of lead

= 2.067 *10-^3 - 1.218 * 10^-3

= 0.849 *10^-3 M

= 0.849 *10^-3 *65.38= 0.056 g/L

amount of Zinc in 200mL = 0.01

mass percentage of zinc = 0.01*100/0.4627 = 2.39 %

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Concentration of copper

total concentration of Cu + lead + zinc - (conc of zn + lead)

= 2.56 *10^-3 - 2.067 *10-^3

= 0.031 g/L

amount of copper in 200 mL = 0.0063 g

mass percentage of copper = 0.0063*100/0.4627 = 1.35 %

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amount of Tin = 0.4627 - (0.05 + 0.01 + 0.0063) = 0.3964 g

mass percentage = 0.3964 *100/0.4627 = 85.67 %