learning Map Ligand X forms a complex with both cobalt and copper, each of which

Question

learning Map 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 0.218-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 0.438 at 510 nm and 0.320 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, Mcm1) Co 37728 1285 Cu 5721 18140 What is the concentration of cobalt and copper in the final diluted solution? Number Number 0 What is the weight percent of cobalt (FM 58.933 g/mol) and copper (FM sample? 63.546 g/mol) in the 0.218-9 Number Number wt% CO2+- wt% Cu" = Previous Give Up & View Solution O Check Answer Next SExit Hint

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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