The following information is to be used for the next two questions. Molar absorp
ID: 521101 • 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 0.262-g sample was dissolved and diluted to 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 0.595 at 510 nm and 0.318 at 656 nm in a 1.00-cm cell.
Calculate the concentration of cobalt (in ppm) in the initial solution prepared by dissolving the sample.
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, 0.595 = (5520 M-1cm-1) x N M x 1.0 cm + (36400 M-1cm-1) x C M x 1.0 cm
Or, 0.595 = 5520N + 36400 C
Hence, 5520 N + 36400 C = 0.595 - equation 1
#2. At 656 nm,
Total absorbance of the mixture = Abs of Ni + Abs of Co
Or, 0.318= (17500 M-1cm-1) x N M x 1.0 cm + (1240 M-1cm-1) x C M x 1.0 cm
Or, 0.318 = 17500 N + 1240 C
Hence, 17500 N + 1240 C = 0.318 - equation 2
#3. Comparing (equation 1 x 17500) – (equation 2 x 5520)-
96600000 N + 637000000 C = 10412.5
(-) 96600000 N + 6844800 C = 1755.36
---------------------------------------------------------------------------------------------------------------------------------
630155200 C = 8657.14
Or, C = 8657.14 / 630155200 = 1.374 x 10-5
Therefore, [Co] in the final aliquot = C M = 1.374 x 10-5 M
Putting the values of C in equation 1-
5520 N + [36400 x (1.374 x 10-5)] = 0.595
Or, 5520 N = 0.595 - 0.500136 = 0.094864
Or, N = 0.094864 / 5520 = 1.719 x 10-5
Therefore, [Ni] = N M = 1.719 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
= 1.719 x 10-5 M x 0.050 L
= (1.719 x 10-5 mol/L) x 0.050 L
= 8.595 x 10-7 mol
= 8.595 x 10-7 mol x (58.6934 g/ mol)
= 5.045 x 10-5 g
Final aliquot is prepared from 25.0 mL of original 50.0 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 = 5.045 x 10-5 g
Now,
Amount of Ni in 50.0 mL original sample = 2 x Amount of Ni in 25.0 mL original sample
= 2 x 5.045 x 10-5 g
= 1.009 x 10-4 g
= 0.1009 mg
Therefore, amount Ni in total original 50.0 mL solution = 0.1009 mg
Also, Amount of Ni in 0.262 g sample = amount Ni in total original 50.0 mL solution = 0.1009 mg
Now,
[Ni] = mg of Ni / Mass of sample in kg
= 0.1009 mg / 0.000262 kg
= 385.11 mg/ kg
= 385.11 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
= 1.374 x 10-5 M x 0.050 L
= 6.87x 10-7 mol
= 6.87x 10-7 mol x (58.9332 g/ mol)
= 4.049 x 10-5 g
Final aliquot is prepared from 25.0 mL of original 50.0 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 = 4.049 x 10-5 g
Now,
Amount of Co in 50.0 mL original sample = 2 x Amount of Co in 25.0 mL original sample
= 2 x 4.049 x 10-5 g
= 8.098 x 10-5 g
= 0.0809 mg
Therefore, amount Co in total original 50.0 mL solution = 0.0809 mg
Also, Amount of Co in 0.262 g sample = amount Co in total original 50.0 mL solution = 0.0809 mg
Now,
[Co] = mg of Co / Mass of sample in kg
= 0.0809 mg / 0.000262 kg
= 308.78 mg/ kg
= 308.78 ppm
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