A 0.5412 g sample of pewter, containing tin, lead, copper, and zinc, was dissolv

Question

A 0.5412 g sample of pewter, containing tin, lead, copper, and zinc, was dissolved in acid. Tin was precipitated as SnO_2-4H_2O and removed by filtration. The resulting filtrate and washings were diluted to a total volume of 250.0 mL. A 20.00 mL aliquot of this solution was buffered, and titration of the lead, copper, and zinc in solution required 34.94 mL of 0.001545 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 33.27 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 25.18 mL of the EDTA solution. Determine the percent composition by mass of each metal in the pewter sample.

Explanation / Answer

pewter sample = 0.5412 g (contains Sn + Pb + Cu + Zn)

Sn removed by precipitation

With 20 ml aliquot

moles of EDTA used (Pb + Cu + Zn) = 0.001545 M x 34.94 ml = 0.054 mmol

So,

25 ml sample would have = 0.0675 mmol

With 25 ml aliquot

Cu masked

moles of EDTA used (Pb + Zn) = 0.001545 M x 33.27 ml = 0.0514 mmol

So,

moles of Cu present in 25 ml sample = 0.0675 - 0.0514 = 0.0161 mmol

moles of Cu in 250 ml solution = 0.0161 x 250/25 = 0.161 mmol

mass of Cu present = 0.161 x 63.546/1000 = 0.0102 g

mass percent Cu in pewter = 0.0102 x 100/0.5412 = 1.885%

With 30 ml aliquot

Cu and Zn masked by cyanide

EDTA used (Pb) = 0.001545 M x 25.18 ml = 0.039 mmol

In 250 ml (Pb) present = 0.039 x 250/30 = 0.325 mmol

In 25 ml sample = 0.0325 mmol

Mass of Pb in pewter = 0.325 x 207.2 = 0.0673 g

Mass percent Pb in sample = 0.0673 x 100/0.5412 = 12.43%

moles of Zn in 25 ml sample = 0.0675 - (0.0161 + 0.0325) = 0.0189 mmol

moles of Zn in 250 ml sample = 0.0189 x 250/25 = 0.189 mmol

mass of Zn in pewter = 0.189 x 65.38 = 0.012 g

mass percent of Zn in pewter = 0.012 x 100/0.5412 = 2.22%

Mass percent Sn in pewter = 100 - (1.885 + 12.43  + 2.22) = 83.465%

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