A 1.728 g sample of a Pb/Cd alloy was dissolved in acid and diluted to exactly 5

Question

A 1.728 g sample of a Pb/Cd alloy was dissolved in acid and diluted to exactly 500 mL. Titration of a 25.00 mL aliquot at a pH = l0 in an NH4 + /NH3 buffer required 43.60 mL of standard 0.01080 M EDTA and involved both cations (Pb2+ and Cd2+). The cadmium ion was then masked by using an HCN/NaCN buffer to bring the pH of a second 25.00 mL aliquot to 10; 32.70 mL of the 0.01080 M EDTA standard was used in this titration and involved only the Pb2+. Calculate the respective percentages of lead and cadmium in the alloy.

Explanation / Answer

Given,

For titration (I)

molality of EDTA = 0.01080 M

Volume of EDTA = 43.60 mL = 0.04360 L

Thus, moles of EDTA = M*V = 0.01080 * 0.04360

                                     = 4.71*10-4 mol (involved both cations, hence Pb moles + Cd moles)
For titration (II)

molality of EDTA = 0.01080 M

Volume of EDTA = 32.70 mL = 0.03270 L

Thus, moles of EDTA = M*V = 0.01080 * 0.03270

                                     = 3.53*10-4 mol (involved only Pb+2, hence Pb moles )
Cd moles = (4.71*10-4 mol) - (3.53*10-4 mol) = 1.18*10-4 mol
grams of Pb in sample = 3.53*10-4 *207.2 * 500 / 25 = 14628.3*10-4 gm = 1.46283 gm = 1.463 gm
grams of Cd in sample = 1.18*10-4 * 112.4 * 500 / 25 = 2652.6*10-4gm = 0.26526 gm = 0.2653 gm
Pb % = (1.463 gm Pb / 1.728 gm sample) * 100 = 84.7 % Pb+2
Cd % = 100 - 84.7 = 15.3 % Cd+2

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