The following information is to be used for the next two questions. Molar absorp
ID: 520700 • Letter: T
Question
The following information is to be used for the next two questions. Molar absorptivity data for the cobalt and nickel complexes with 2,3-quinoxalinedithiol are Co = 36400 and Ni = 5520 at 510 nm and Co = 1240 and Ni = 17500 at 656 nm. A 0.343-g sample was dissolved and diluted to 150.0 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 0.786 at 510 nm and 0.468 at 656 nm in a 1.00-cm cell.
1. Calculate the concentration of cobalt (in ppm) in the initial solution prepared by dissolving the sample.
2.Calculate the concentration of nickel (in ppm) in the initial solution prepared by dissolving the 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 [Ni] = N molar , and [Co] = C molar - in the finally diluted aliquot whose OD is taken.
It’s assumed that the unit of molar absorptivity is in terms of M-1cm-1 and path length is 1.0 com in all cases.
#1. At 510 nm,
Total absorbance of the mixture = Abs of Ni + Abs of Co
Or, 0.786 = (5520 M-1cm-1) x N M x 1.0 cm + (36400 M-1cm-1) x C M x 1.0 cm
Or, 0.786 = 5520N + 36400 C
Hence, 5520 N + 36400 C = 0.786 - equation 1
#2. At 656 nm,
Total absorbance of the mixture = Abs of Ni + Abs of Co
Or, 0.468 = (17500 M-1cm-1) x N M x 1.0 cm + (1240 M-1cm-1) x C M x 1.0 cm
Or, 0.468 = 17500 N + 1240 C
Hence, 17500 N + 1240 C = 0.468 - equation 2
#3. Comparing (equation 1 x 17500) – (equation 2 x 5520)-
96600000 N + 637000000 C = 13755.0
(-) 96600000 N + 6844800 C = 2583.36
630155200 C = 11171.64
Or, C = 11171.64 / 630155200 = 1.773 x 10-5
Therefore, [Co] in the final aliquot = C M = 1.773 x 10-5 M
Putting the values of C in equation 1-
5520 N + [36400 x (1.733 x 10-5)] = 0.786
Or, 5520 N = 0.786 - 0.630812 = 0.155188
Or, N = 0.155188/ 5520 = 2.811 x 10-5
Therefore, [Ni] = N M = 2.811 x 10-5 M
#4. Final volume of aliquot = 50 mL = 0.050 L
Amount of Ni in final aliquot = [Ni] in final aliquot x Vol. of final aliquot
= 2.811 x 10-5 M x 0.050 L
= (2.811 x 10-5 mol/L) x 0.050 L
= 1.4055 x 10-6 mol
= 1.4055 x 10-7 mol x (58.6934 g/ mol)
= 8.2494 x 10-5 g
Final aliquot is prepared from 25.0 mL of original 150.0 mL in which the solid sample is dissolved.
Therefore, amount of Ni in 25.0 mL original solution = Amount of Ni in 50.0 mL final aliquot = 8.2494 x 10-5 g
Now,
Amount of Ni in 150.0 mL original sample =
(150 mL/ 25 mL) x Amount of Ni in 25.0 mL original sample
= 0.0004949614422 g
= 1.009 x 10-4 g
= 0.49496 mg
Therefore, amount Ni in total original 150.0 mL solution = 0.49496 mg
Also, Amount of Ni in 0.343 g sample = amount Ni in total original 150.0 mL solution = 0.49496 mg.
Mass of sample = 0.343 g = 0.000343 kg
Now,
[Ni] = mg of Ni / Mass of sample in kg
= 0.49496 mg / 0.000343 kg
= 385.11 mg/ kg
= 1443 ppm
#5. Final volume of aliquot = 50 mL = 0.050 L
Amount of Co in final aliquot = [Co] in final aliquot x Vol. of final aliquot
= 1.773 x 10-5 M x 0.050 L
= 8.865 x 10-7 mol
= 8.865 x 10-7 mol x (58.9332 g/ mol)
= 5.2244 x 10-5 g
Final aliquot is prepared from 25.0 mL of original 150.0 mL in which the solid sample is dissolved.
Therefore, amount of Co in 25.0 mL original solution = Amount of Co in 50.0 mL final aliquot = 5.2244 x 10-5 g
Now,
Amount of Co in 150.0 mL original sample =
(150 mL/ 25 mL) x Amount of Co in 25.0 mL original sample
= 0.0003134656908 g
= 0.313466 mg
Therefore, amount Co in total original 150.0 mL solution = 0.313466 mg
Also, Amount of Co in 0.343 g sample = amount Co in total original 150.0 mL solution = 0.313466 mg
Now,
[Co] = mg of Co / Mass of sample in kg
= 0.313466 mg / 0.000343 kg
= 913.89 mg/ kg
= 914 ppm
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.