pling apling learning Ligand X forms a complex with both cobalt and copper, each

Question

pling apling learning Ligand X forms a complex with both cobalt and copper, each of which has a maximum absorbance at 510 nm and 645 nm, respectively. A 0.226-g sample containing cobalt and copper was dissolved and diluted to a volume of 100.0 mL. A solution containing ligand X was added to a 50.0 mL aliquot of the sample solution and diluted to a final volume of 100.0 mL. The measured absorbance of the unknown solution was 0.502 at 510 nm and 0.367 at 645 nm, when measured with a 1.00-cm cell. The molar absorptivities of the cobalt and copper complexes at each wavelength are shown in the table below Wavelength A, nm 510 645 Molar Absorptivity (e,M-1cm-1) Co 37205 1267 5642 17880 What is the concentration of cobalt and copper in the final diluted solution? Number Number What is the weight percent of cobalt (FM- 58.933 g/mol) and copper (FM 63.546 g/mol) in the 0.226-g sample? Number Number

Explanation / Answer

Absorbance of a mixture is sum of all its components.

A = A1 + A2

A = ebC[Co] + ebC(Cu]

e = molar absorptivity

b = 1 cm

So, at 510 nm,

0.438 = 37728[Co] + 5721[Cu]

and, at 645 nm,

0.320 = 1285[Co] + 18140[Cu]

Solving the two equations we get,

Concentration in the final diluted solution would be,

[Co] = 9.03 x 10^-6 M

[Cu] = 1.70 x 10^-5 M

Now, weight percent of metals in original sample,

Molarity of metals in original solution would be,

[Co] = 2 x 9.03 x 10^-6 = 1.806 x 10^-5 M

[Cu] = 2 x 1.70 x 10^-5 = 3.40 x 10^-5 M

moles of metal in original solution becomes,

[Co] = 0.1 x 1.806 x 10^-5 = 1.806 x 10^-6 mols

[Cu] = 0.1 x 3.40 x 10^-5 = 3.40 x 10^-6 mols

Mass of metals in original solution would be,

[Co] = 1.806 x 10^-6 x 58.933 = 1.06 x 10^-4 g

[Cu] = 3.40 x 10^-6 x 63.546 = 2.16 x 10^-4 g

So the weight percent of metals in original sample is,

[Co] = 100 x 1.06 x 10^-4 g/0.218 g = 0.05%

[Cu] = 100 x 2.16 x 10^-4 g/0.218 = 0.1%

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