A 0.5412 g sample of pewter, containing tin, lead, copper, and zinc, was dissolv

Question

A 0.5412 g sample of pewter, containing tin, lead, copper, and zinc, was dissolved in acid. Tin was precipitated as SnO_2 4H_2O 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 34.94 mL of 0.001545 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 33.27 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 25.18 mL of the EDTA solution. Determine the percent composition by mass of each metal in the pewter sample.

Explanation / Answer

Titration of lead requires 25.18 mL 0.001545 M EDTA.

Molarity of Only Pb solution = 25.18 mL* 0.001545 M /30 mL =1.29*10^-3 M

250mL solution contain 1.29*10^-3*250ml/1000mL =0.323*10^-3 moles

mass of Pb in 250 mL solution = 0.323*10^-3 moles*207.2 gm/mol =0.067 gm

mass % = 0.067gm*100/0.5412 gm = 12.34%

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Pb and Zn requires total 33.27 mL EDTA. So molarity of this solution is :

M = 33.27mL*0.001545M/25 mL = 2.056*10^-3 M = 2.056*10^-3 moles/1000mL solution

moles of Zn present in 1000ml solution =2.056*10^-3- 1.29*10^-3 =0.766*10^-3 moles

moles of Zn in 250 ml solution = 0.766*10^-3 moles*250mL/1000mL =0.192 *10^-3 moles

mass of Zn = 0.192 *10^-3 moles*65.38 g/mol =0.0125 gm

mass % = (0.0125gm/0.5412)*100 =2.31 %

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Pb + Zn +Cu requires 34.94 mL EDTA for titration.

Molarity = 34.94*0.001545M/20mL =2.69*10^-3 M

amount of onlu Cu in 1000 mL solution = 2.69*10^-3- (1.29*10^-3 +0.766*10^-3) =0.634 *10^-3 moles

moles of Cu in 250 mL = 0.634 *10^-3 *250mL/1000mL =0.1585 *10^-3 moles

mass of Cu = 0.1585 *10^-3 moles*63.5gm/moles = 0.010 gm

mass % = 0.010gm/0.5412 *100 =1.85 %

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mass % of Sn = 100- (12.34 +2.31 + 1.85) =83.5 %

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