Ligand X forms a complex with both cobalt and copper, each of which has a maximu

Question

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 0.205-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 0.441 at 510 nm and 0.322 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. What is the concentration of cobalt and copper in the final diluted solution? [Co^2+] = [Cu^2+] = What is the weight percent of cobalt (FM = 58.933 g/mol) and copper (FM = 63.546 g/mol) in the 0.205-g sample?

Explanation / Answer

Ans. Beer-Lambert’s Law, A = e C L             - equation 1,              

where,

                       A = Absorbance

                       e = molar absorptivity at specified wavelength (M-1cm-1)

                        L = path length (in cm)

                        C = Molar concentration of the solute

Let the [Co2+] = C M , and [Cu2+] = U M    in the final aliquot.

#1. At 510 nm,

Total absorbance of the mixture = Abs of Co2+ + Abs of Cu2+

Or, 0.441 = (36614 M-1cm-1) x C M x 1.0 cm + (5552 M-1cm-1) x U M x 1.0 cm

Or, 0.441 = 36614 C + 5552 U

Hence, 36614 C + 5552 U = 0.441                       - equation 1

#2. At 645 nm,

Total absorbance of the mixture = Abs of Co2+ + Abs of Cu2+

Or, 0.322 = (1247 M-1cm-1) x C M x 1.0 cm + (17600 M-1cm-1) x U M x 1.0 cm

Or, 0.322= 1247 C + 17600 U

Hence, 1247 C + 17600 U = 0.322                       - equation 2

#3. Comparing (equation 1 x 1247) – (equation 2 x 36614)-

            45657658 C   + 6923344 U = 549.927

      (-) 45657658 C   + 644406400 = 11789.708

                                    - 637483056 U = - 11239.781

                                    Or, U = 11239.781 / 637483056 = 1.76315 x 10-5

Thus, [Cu2+ = U M = 1.76315 x 10-5 M

Now, Putting the values of U in equation 1-

            36614 C + 5552 x (1.76315 x 10-5) = 0.441

            36614 C = 0.441 – 0.09789 = 0.34311

            Or, C = 0.34311 / 36614 = 9.371 x 10-6

Thus, [Co2+] = C M = 9.371 x 10-6 M

#4. Mass of Co2+ in final aliquot = Vol. of aliquot in liters x Molarity x Molar mass

                                                = 0.100 L x (9.371 x 10-6 M) x (58.9332 g/ mol)

                                                = 0.0000552263 g = 5.52263 x 10-5 g

Mass of Cu2+ in final aliquot = Vol. of aliquot in liters x Molarity x Molar mass

                                                = 0.100 L x (1.76315 x 10-5 M) x (63.546 g/ mol)

                                                = 0.0001120411 g= 1.12041 x 10-4 g

The final aliquot is prepared from 50.0 mL out of total 100.0 mL original solution. So, the amount calculated is for 50.0 mL of original solution.

Now,

Total mass of Co2+ in 100.0 mL original sol. = (100.0 mL) x Mass of Co2+ in 50.0 mL

                                                            = 100 mol x (0.0000552263 g/ 50.0 mL)

                                                            = 0.0001104526 g

Total mass of Cu2+ in 100.0 mL original sol. = (100.0 mL) x Mass of Cu2+ in 50.0 mL

                                                            = 100 mol x (0.0001120411 g / 50.0 mL)

                                                            = 0.0002240822 g

Now,

Wt % Co2+ = (Mass of Co / Mass of sample) x 100

                        = (0.0001104526 g / 0.205 g) x 100

                        = 0.0539 %

Wt % Cu2+ = (Mass of Co / Mass of sample) x 100

                        = (0.0002240822 g / 0.205 g) x 100

                        = 0.1093 %

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