A 0.4668 g sample of pewter, containing tin, lead, copper, and zinc, was dissolv

Question

A 0.4668 g sample of pewter, containing tin, lead, copper, and zinc, was dissolved in acid. Tin was precipitated as SnO_2 4H_2 O and removed by filtration. The resulting filtrate and washings were diluted to a total volume of 300.0 mL. A 20.00 mL aliquot of this solution was buffered, and titration of the lead, cooper, and zinc in solution required 34.04 mL of 0.001519 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 34.00 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 25.77 mL of the EDTA solution. Determine the percent composition by mass of each metal in the pewter sample.

Explanation / Answer

Determination of concentration of Lead (Cu and Zn are masked with cyanide)

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 = 25.77 * 0.001519

M 1 = 1.30 *10-3 M

Total volume of the solution= 300mL

moles of lead present in 300mL = 1.30 *10-3 *300/1000 = 3.91 * 10-4 mole

Mass of lead =3.91 * 10-4 mole * 207.2 g/mol = 0.0811

% of Pb in the sample = (0.0811/0.4668)*100% = 17.4%

---------------------------------------------------------------------------------------

Determination of Zn

When Copper is masked with thiosukphate , titration of Lead + zinc requires 35 mL EDTA

apply the above formula

M1V1 = M2V2

M1 = 34 mL * 0.001519 M/ 25 mL = 2.07* 10-3 M

moles of lead + Zinc present in the 300 mL solution = 2.07 * 10-3 * 300/1000 = 6.197 *10-4

Moles of lead present in the sample = 3.91 * 10-4 mole

moles of zinc present in 300mL = 6.197 *10-4- 3.91 * 10-4 mole = 2.287 *10^-4 mole

weight of Zinc present = 2.287 *10-4 mol * 65.38 g/mol = 0.015 g

% of zinc = (0.015/0.4668) *100% = 3.21%

--------------------------------------------------------------------------------------------------------

Determination of Copper

Lead + zinc + copper requires 34.04 mL EDTA

apply the above formula

M1V1 = M2V2

M1 = 34.04 mL * 0.001519 M/ 20 mL = 2.59* 10-3 M

moles of lead + Zinc present in the 300 mL solution =7.76 *10-4

moles of lead present in 300mL = 3.91 * 10-4 mole

moles of Zinc present = 2.287 *10^-4 mole

moles of Copper present = [7.76 - (3.91 +2.287)] *10-4 = 1.563 *10-4 moles

weight of the copper = 1.563 *10-4 moles * 63.54 g/ mole = 0.0099 gm

% Cu = 0.0099/0.4668 * 100 = 21.27 %

----------------------------------------------------------------------------------------------------------------

Determination of the weight of Tin

weight of Tin = 0.4668g-weight of (lead + zinc + copper)

                   = 0.3608 gm

% Sn = 77.29 %

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.