Molar absorptivity data for the cobalt and nickel complexes with 2, 3-quinoxalin

Question

Molar absorptivity data for the cobalt and nickel complexes with 2, 3-quinoxalinedithiol are epsilon C_0 = 36400 and epsilon N_i = 5520 at 510 nm and epsilon_Co = 1240 and epsilon N_i = 17500 at 656 nm. a 0.503 -g sample was dissolved and diluted 50.0 mL. A 25.0-mL aliquot was treated to eliminate interferences; after addition of 2, 3-quinoxalinedithiol, the volume was adjusted to 50.0 mL. This solution had an absorbance of 0.530 at 510 nm and 0.341 at 656 nm in a 1.00-cm cell. Calculate the concentration in parts per million of cobalt and nickel in the sample. Concentration of cobalt = ppm Concentration of nickel = ppm

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 concentration of Cobalt in final aliquot (which OD is recorded) = C

      the concentration of Nickel in final aliquot (which OD is recorded) = N molar

#1. At 510 nm

Given, Total sample absorbance = 0.530

Absorbance due to Co alone = ECo x C x 1.0 cm = 36400 x C x 1.0 = 36400C

Absorbance due to Ni alone = ENi x N x 1.0 cm = 5520 x N x 1.0 = 5520N

Since the absorbance is additive, the sum of absorbance of species Co and Ni is equal to the total absorbance of the sample.

So,      

            36400C + 5520N = 0.530                          - equation 2

Note: The calculation is made assuming the stranded unit of variables as mentioned in equation 1. So, no unit is used in calculating absorbance of the two species.

#2. At 656 nm

Given, Total sample absorbance = 0.341

Absorbance due to Co alone = ECo x C x 1.0 cm = 1240 x C x 1.0 = 1240C

Absorbance due to Ni alone = ENi x N x 1.0 cm = 17500 x N x 1.0 = 17500N

Since the absorbance is additive, the sum of absorbance of species Co and Ni is equal to the total absorbance of the sample.

So,      

            1240C + 17500N = 0.341                          - equation 3

#3. Comparing (equation 2 x 1240) – (equation 3 x 36400)

            45136000 C + 6844800 N              = 657.2

     (-)[45136000 C + 637000000 N           = 12412.4]

                        - 630155200 N= - 11755.2

                        Or, N = 11755.2 / 630155200 = 1.865 x 10-5

Thus, concentration of Ni in final aliquot = N Molar = 1.865 x 10-5 M

Now, putting the value of N in equation 2-

            36400C + 5520 (1.865 x 10-5) = 0.530      

            Or, 36400C + 0.10297 = 0.530

            Or, C = (0.530 – 0.10297) / 36400 = 1.173 x 10-5

Thus, concentration of Co in final aliquot = C molar = 1.173 x 10-5 M

#4. Given, volume of final aliquot = 50.0 mL

Amount of Ni in final aliquot = [Ni] x volume of aliquot in liters

                        = 1.865 x 10-5 M x 0.050 L                           ; [1 L =1000 mL]

                        = 0.0000009325 mol                                   ; [1 M = 1 mol/ L]

                        = 0.0000009325 mol x (58.6934 g/ mol)   ; [Mass = mol x molar mass]

                        = 0.0000547315955 g

Amount of Co in final aliquot = [Co] x volume of aliquot in liters

                        = 1.175 x 10-5 M x 0.050 L                           ; [1 L =1000 mL]

                        = 0.0000005875 mol                                   ; [1 M = 1 mol/ L]

                        = 0.0000005875 mol x (58.9332 g/ mol)   ; [Mass = mol x molar mass]

                        = 0.000034623255 g

#5. Coming to aliquot preparation:

            Step 1: 0.503 g sample dissolved to a final volume of 50.0 mL . Say, it’s solution 1.

            Step 2: 25.0 mL of step 1 solution diluted to 50.0 mL. Say, it’s the aliquot. OD of this aliquot recorded.

Thus, amount of Co and Ni obtained in #4 is equal to the amount of these species in 25.0 mL of step 1 solution. Also, total amount of the species in 0.503 g sample is equal to their amount in 50.0 mL of step 1 solution.

#5A. Total amount of Co in sample =

(Vol. of soln.1 / Vol. of soln.1 used to make aliquot) x Mass of Co in aliquot

= (50.0 mL/ 25.0 mL) x 0.000034623255 g

            = 0.00006924651 g

            = 0.06924651 mg

ppm of Co in sample = mg of Co per kg sample

                                    = 0.06924651 mg / [(0.503 / 1000 kg)

                                    = 137.67 mg/ kg      

                                    = 137.67 ppm

#5B. Total amount of Ni in sample =

(Vol. of soln.1 / Vol. of soln.1 used to make aliquot) x Mass of Ni in aliquot

= (50.0 mL/ 25.0 mL) x 0.0000547315955 g

            = 0.000109463191 g

            = 0.109463191mg

ppm of Ni in sample = mg of Ni per kg sample

                                    = 0.109463191 mg / [(0.503 / 1000 kg)

                                    = 217.62 mg/ kg      

                                    = 217.62 ppm

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