A 1.879-g sample of a Pb Cd alloy was dissolved in acid and diluted to exactly 2
ID: 992336 • Letter: A
Question
A 1.879-g sample of a Pb Cd alloy was dissolved in acid and diluted to exactly 250.0 mL in a volumetric flask. A 50.00-mL aliquot of the diluted solution was brought to a pH of 10.0 with an NH_4^+ NH_3 buffer: the subsequent titration involved both cations and required 28.89 mL of 0.06950 M EDIA. A second 50.00-mL aliquot was brought to a pH of 10.0 with an HCN NaCN buffer, which also served to mask the Cd^2+. 22.79 mL of the EDTA solution were needed to titrate the Pb^2+. Calculate the percent Pb and Cd in the sample. Percentage of Pb =_________% Percentage of Cd =________%Explanation / Answer
22.79 ml (v2) edta of concentration .06950 M (s2) is requir to tritrate Pb2+ 50 ml (v1 ) of strenght s1.
v1 s1 = v2 s2
s1 = .0316 M
1000 ml of solution contain .0316 mole
250 solution contain 7.9 *10-3 mole
molecular weight of Pb = 208 gm
250 solution contain 7.9 *10-3 mole * 208 = 1.6432 gm = 87.45 %
22.79 ml (v2) edta of concentration .06950 M (s2) is requir to tritrate Pb2+ 50 ml (v1 ) of strenght s1.
v1 s1 = v2 s2
s1 = .0316 M
1000 ml of solution contain .0316 mole
250 solution contain 7.9 *10-3 mole
molecular weight of Pb = 208 gm
250 solution contain 7.9 *10-3 mole * 208 = 1.6432 gm = 87.45 %
(28.89 - 22.79) = 6.1 ml (v2) edta of concentration .06950 M (s2) is requir to tritrate Cd2+ 50 ml (v1 ) of strenght s1.
v1 s1 = v2 s2
s1 = 8.4 *10-3 M
1000 ml of solution contain 8.4 *10-3 mole
250 solution contain 2.1 *10-3 mole
molecular weight of Pb = 112 gm
250 solution contain 2.1 *10-3 mole * 112 = .2374 gm = 12.69 %
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.