A 0.4557 g sample of p containing tin, lead, copper, and zinc, was dissolved in

Question

A 0.4557 g sample of p 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 200.0 mL. A 20.00 mL aliquot of this solution was buffered, and titration of the lead, copper, and zinc in solution required 35.10 m of 0.001531 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 34.62 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 25.39 mL ofthe EDTA solution. Determine the percent composition by mass of each metal in the pewter sample. Cu Zn Number "2% Pb Sn.

Explanation / Answer

Determination of concentration of Lead

M1V1 = M2V2 where M1 and V1 is the molarity and volume of the pewter sample solution. M2 and V2 is the molarity and volume required for EDTA respectively

30 * M1 = 25.39 * 0.001531

M 1 = 1.29*10^-3 M

Total volume of the solution= 200mL

moles of lead present in 200mL = 1.29*10^-3 M *200/1000 = 2.59 * 10-4 mole = 0.054 gm

amount of lead in the sample = 0.054 gm

% Pb = (0.054 gm/0.4557 ) *100 % = 11.8%

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Determination of Zn

Lead + zinc requires 34.62 mL EDTA

apply the above formula

M1V1 = M2V2

M1 = 34.62 mL * 0.001531 M/ 25 mL = 2.12 * 10-3 M

moles of lead + Zinc present in the 200 mL solution = 2.12 * 10-3 * 200/1000 = 4.24 *10-4

moles of lead + zinc present in 300mL = 4.95 * 10-4

moles of Zinc present = (4.95-2.59) * 10-4 = 2.36 *10-4

weight of Zinc present = 2.36 *10-4 mol * 65.38 g/mol = 0.0154 g

% Zn = (0.0154 g/0.4557) *100% =3.38%

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Determination of Copper

Lead + zinc + copper requires 35.10 mL EDTA

apply the above formula

M1V1 = M2V2

M1 = 35.10 mL * 0.001531 M/ 20 mL = 2.69 * 10-3 M

moles of lead + Zinc present in the 200 mL solution = 2.69 * 10-3 * 200/1000 = 5.37 *10-4

moles of lead present in 200mL = 2.59 * 10-4 mole

moles of Zinc present = 2.36 *10-4

moles of Copper present = [5.37 - (2.36 +2.59)] *10-4 = 0.42 *10-4 moles

weight of the copper = 0.42 *10-4 moles * 63.54 g/ mole = 0.0027 gm

% Cu = 0.0027/0.4557) *100% = 0.59 %

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Determination of the weight of Tin

weight of Tin = 0.4557g-weight of (lead + zinc + Tin)

                   = 0.3836 gm

% Tin = 0.3836/0.4557 * 100 = 84.12 %

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