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 middot 4H_2 O and removed by filtration. The resulting filtrate and washings were diluted total volume of 300.0 mL. A 20.00 mL aliquot of this solution was buffered, and titration of the lead, copper, 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. %o Cu % Zn % Pb % Sn

Explanation / Answer

Calculation for Pb

For the complexometric titration of Pb the 30 ml aliquot required 25.77 ml EDTA of strength 0.001519 M

So mmol of Pb in the main stock ie in 300 ml=300*25.77*0.001519/30=0.3914

So Pb=0.3914*207 mg=81 mg

Calculation for Pb+Zn

25 ml aliquot required EDTA=34 ml of strength 0.001519 M

So 300 ml will require 300*34*0.001519/25=0.6179

Thus mmol of Zn=0.6179-0.3914=0.2283

So Zn=0.2283*65 mg=14.83 mg

Calculation of Cu

20 ml aliquot containing mixture of Cu, Zn and Pb required EDTA=34.04 of strength 0.001519 M

So 300 ml will require 300*34.04*0.0001519/20=0.7756 ie mmol of Cu, Zn and Pb.

So mmol of Cu=0.7756-0.6197=0.1359=0.1359*63.5 mg=9.899 mg

So % Cu=(9.899/466.8)*100=2.12

% of Zn=(14.83/466.8)*100=3.17

% of Pb=(81/466.8)*100=17.35

% of Sn=(361.07/466.8)*100=77.35

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