Question

A 0.4858 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 35.85 mL of 0.001523 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 33.79 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 24.38 mL of the EDTA solution. Determine the percent composition by mass of each metal in the pewter sample. %Cu %Kn %Pb %Sn

Answer 1

25 ml of aliquot has = 0.001523 M x 24.38 ml = 0.03713074 mmol of Pb

So, 200 ml solution would have = 0.297 mmol Pb

Mass of Pb in sample = 0.297 x 207.2/1000 = 0.06155 g

% mass of Pb in pewter sample = 0.06155 x 100/0.4858 = 12.67%

15 ml of solution (Zn + Cu + Pb) has = 0.001523 M x 35.85 ml = 0.0546 mmol

So, 200 ml solution contains = 0.0546 x 200/25 = 0.437 mmol (Zn + Cu + Pb)

20 ml of solution (Zn + Pb) needed = 0.001523 M x 33.79 ml = 0.05146 mmol

So 200 ml solution has = 0.4117 mmol (Zn + Pb)

Remainder Cu in 200 ml sample = 0.437 - 0.4117 = 0.0253 mol

mass of Cu in sample = 0.0253 x 63.546 = 0.00161 g

% mass of Cu in pewter = 0.00161 x 100/0.4858 = 0.33%

Moles of Zn in sample = 0.4117 - 0.297 = 0.1147 mmol

mass of Zn in sample = 0.1147 x 65.38/1000 = 0.0075 g

% mass of Zn in pewter = 0.0075 x 100/0.4858 = 1.54%

So,

% Mass of Sn in pewter = 100 - (1.54 + 0.33 + 12.67) = 85.46%