To analyze the chromium in an unknown sample 1.5623 f of the sample was pretreat

Question

To analyze the chromium in an unknown sample 1.5623 f of the sample was pretreated and diluted to the exact 100 mL mark in a volumetric flask. Two 20.0 mL samples of this solution were taken. In the first sample, a Cr-EDTA complex is developed and sufficient water is added to give a total volume of 100.0 mL. In the second sample, 10.0 mL of Cr(III) standard solution at to concentration of 15 mM was added before developing the Cr-EDTA A complex. Sufficient water is also added to give a total volume of 100.0 mL. The absorbance of each is measured in a 1.0 cuvette at 450 nm and is found to be 0.15 for the first solution and 0.30 for the second solution. Calculate the chromium (atomic wt 52.0 g/mol) in the original sample. Molar absorptivity of cobalt and copper complexes with a ligand L are epsilon Co = 605 and epsilon_Cu = 552 at 510 nm and epsilon_Cu = 750 at 625 nm sample containing Co and Cu was dissolved and dilute to 500.0 m A 25.0-mL aliquot was treated and completed with the ligand L. This solution had an of 0.646 at at 625 cell. Calculate the percentage of Co (Mw: 58.9 g/mol) and cu (Mw: 63.5 g/mol) in the original sample. (II)

Explanation / Answer

2) Analysis of chromium

20 ml aliquot diluted to 100 ml without standard gave aborbance = 0.15

IInd 20 ml aliquot with standard,

concentration of standard in diluted sample = 15 mM x 10 ml/100 ml = 1.5 mM

let x be the concentration of Cr in diluted sample

then,

x/(x + 1.5) = 0.15/0.30

0.30x = 0.15x + 0.225

x = 0.225/0.15 = 1.5 mM

concentration of Cr in original sample = 1.5 x 100/20 = 7.5 mM

grams of Cr = 7.5 x 52 x 0.1/1000 = 0.039 g

%Cr in original sample = 0.039 x 100/1.5623 = 2.50%

3) Absorbance of mixture is sum total of absorbances of all the species in solution

Absorbance = molar absorptivity x path length x concentration

let A be concentration of Co and B be concentration of Cu

then,

at 510 nm,

0.646 = 605[A] + 552[B]

0.462 = 121[A] + 750[B]

Solving,

[A] = 5.93 x 10^-4 M

[B] = 5.20 x 10^-4 M

%Co in sample = 5.93 x 10^-4 x 0.5 x 58.9 x 100/0.425 = 4.11%

%Cu in sample = 5.20 x 10^-4 x 0.5 x 63.5 x 100/0.425 = 3.88%

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