Question 15 of 16 Map Sapling Learning A 0.4523 g sample of pewter, containing t

Question

Question 15 of 16 Map Sapling Learning A 0.4523 g sample of pewter, containing tin, lead, copper, and zinc, was dissolved in acid. Tin was precipitated as SnO2 4H20 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 34.05 mL of 0.001518 M EDTA. Thiosulfate was used to mask the c second 25.00 mL aliquot. Titration of the lead and zinc in this aliquot required 34.11 mL of the EDTÀ 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 24.50 mL of the EDTA solution. Determine the percent composition by mass of each metal in the pewter sample. in a Number Number %Zn Number % Pb % Sn

Explanation / Answer

Sn removed as SnO2.4H2O

Total volume of diluted filtrate solution = 250 ml

20 ml aliquot took = 0.001518 M x 34.05 ml = 0.0517 mmol EDTA

So, moles of (Cu+Zn+Pb) in 250 ml = 0.0517 mmol x 250 ml/20 ml = 0.6461 mmol

thiosulfate addition, masked Cu

25 ml aliquot took = 0.001518 M x 34.11 ml = 0.0518 mmol

So, moles of (Zn+Pb) in 250 ml = 0.0518 mmol x 250 ml/25 ml = 0.518 mmol  

moles of Cu = 0.6461 - 0.518 = 0.1281 mmol

mass of Cu in pewter sample = 0.1281 mmol x 63.546 g/mol/1000 = 0.00814 mg

mass% Cu in pewter sample = 0.00814 g x 100/0.4523 g = 1.80%

CN- added to mask Zn,

30 ml aliquot took = 0.001518 M x 24.50 ml = 0.0372 mmol

So, moles of Pb in 250 ml = 0.0372 mmol x 250 ml/25 ml = 0.310 mmol  

mass of Pb in sample = 0.310 mmol x 207.2 g/mol/1000 = 0.0642 g

mass% Pb in pewter sample = 0.0642 g x 100/0.4523 g = 14.20%

moles of Zn in sample = 0.518 - 0.310 = 0.208 mmol

mass of Zn in sample = 0.208 mmol x 65.38 g/mol/1000 = 0.0136 g

mass% Zn in pewter sample = 0.0136 g x 100/0.4523 g = 3.01%

So,

mass of Sn in sample = 0.4523 - (0.0136 + 0.0642 + 0.00814) = 0.36636 g

mass% Sn in pewter sample = 0.36636 g x 100/0.4523 g = 80.99%

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