A 0.4467 g sample of pewter, containing tin, lead, copper, and zinc, was dissolv

Question

A 0.4467 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 35.30 mL of 0.001543 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 33.13 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 24.75 mL of the EDTA solution. Determine the percent composition by mass of each metal in the pewter sample.

% Cu=

% Pb=

% Zn=

% Sn=

Explanation / Answer

Sample weight = 0.4467 g

Sn precipitated out

concentration of Pb + Cu + Zn in 200 ml solution = 0.001543 M x 35.30 ml x 200/15 = 0.72624 mmol

concentration of Pb + Zn in 200 ml solution = 0.001543 M x 33.13 ml x 200/20 = 0.51119 mmol

So, mass of Cu in sample = (0.72624 - 0.51119) mmol x 63.546 g/mol/1000 = 0.0136 g

# Cu = 0.0136 x 100/0.4467 = 3.044%

concentration of Pb in 200 ml solution = 0.001543 M x 24.75 ml x 200/25 = 0.3055 mmol

Mass of Pb in sample = 0.3055 mmol x 207.2 g/mol/1000 = 0.0633 g

% Pb = 0.0633 x 100/0.4467 = 14.170%

Mass of Zn in sample = (0.72624 - 0.51119) mmol x 65.38 g/mol/1000 = 0.014 g

% Zn = 0.014 x 100/0.4467 = 31.34%

% Sn = 100 - (31.34 + 14.170 + 3.044) = 51.446%

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