A 0.4809 g sample of pewter, containing tin, lead, copper, and zinc, was dissolv
ID: 954392 • Letter: A
Question
A 0.4809 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 34.79 mL of 0.001492 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 34.57 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 26.76 mL of the EDTA solution. Determine the percent composition by mass of each metal in the pewter sample.
a) %Cu?
b)%Zn?
c)%Pb?
d)%Sn?
Explanation / Answer
Total moles of EDTA used for Pb+Cu+Zn = 0.001492 M x 34.79 ml = 0.052 mmol in 20 ml aliquot
moles of Pb+Cu+Zn in 200 ml solution = 0.52 mmol
Total EDTA used for Pb+Zn = 0.001492 M x 34.57 ml = 0.051 mmol in 25 ml aliquot
moles of Pb+Zn in 200 ml solution = 0.41 mmol
moles of Cu present = 0.52 - 0.41 = 0.11 mmol
mass of Cu present = 0.11 mmol x 63.546 g/mol/1000 = 0.007 g
mass% Cu in sample = (0.007/0.4809) x 100 = 1.45%
Total EDTA used for Pb = 0.001492 M x 26.76 ml = 0.04 mmol
moles of Pb present in 200 ml solution = 0.04 x 6.67 = 0.27 mmol
mass of Pb present in sample = 0.27 mmol x 207.2 g/mol/1000 = 0.056 g
mass% of Pb in sample = (0.056/0.4809) x 100 = 11.64%
moles of Zn present in sample = 0.41 - 0.27 = 0.14 mmol
mass of Zn present in sample = 0.14 mmol x 65.38 g/mol/1000 = 0.0091 g
mass% of Zn in sample = (0.0091/0.4809) x 100 = 1.90%
Mass of Sn in sample = 0.4809 - (0.0091 + 0.056 + 0.007) = 0.4088 g
mass% of Sn in sample = (0.4088/0.4809) x 100 = 85.01%
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