Question

A 1.5000-g sample of an alloy was dissolved in acid and the tin was precipitated as H2SnO3 · xH2O.  After filtration and ignition, the resulting precipitate of SnO2 weighed 0.1430 g.  The filtrate was diluted to 1000.0 mL.  A 20.00-mL aliquot was adjusted to the appropriate pH and required 33.18 mL of 0.01047 M EDTA for titration of the Cu and Pb.  A 25.00-mL aliquot of the filtrate was treated with Na2S2O3 to mask the Cu and was then titrated with 4.72 mL of the EDTA solution.  

a. Calculate the percentage of tin in the alloy.

b. Calculate the percentage of lead in the alloy.

c. Calculate the percentage of copper in the alloy.

Answer 1

mass of SnO2 = 0.1430 g

mass of Sn in SnO2 = 0.1430 g x 118.71 g/mol/150.71 g/mol = 0.113 g

a. percentage of Sn in alloy = 0.113 g x 100/1.500 g = 7.51%

20 ml aliquot (Cu + Pb) EDTA moles = 0.01047 M x 33.18 ml = 0.3474 mmol

amount (Cu + Pb) in 1000 ml = 0.3474 mmol x 1000 ml/20 ml = 17.37 mmol

25 ml aliquot (masked Cu) EDTA used = 0.01047 M x 4.72 ml = 0.05 mol

amount Pb in 1000 ml = 0.05 mol x 1000/25 ml = 2 mmol

b. percentage of Pb in alloy = 2 mmol x 207.2 g/mol x 100/1.500 g x 1000 = 27.63%

c. mass Cu in sample = (17.37 - 2) mmol x 63.55 g/mol/1000 = 0.98 g

percentage of Cu in alloy = 0.98 g x 100/1.500 g = 65.12%