Manganese and chromium in steel can be determined simultaneously by absorption spectral method. Dich — Analytical Chemistry Chemistry Question
Spectrophotometry
Manganese and chromium in steel can be determined simultaneously by absorption spectral method. Dichromate and permanganate ions (Cr2O7 2– and MnO4 – ) in the solution of H2SO4 (c = 1 mol dm-3) absorb light at 440 nm and 545 nm. At these wavelengths, molar absorptivity of MnO4 – solution is 95 dm3 mol-1 cm-1 and 2350 dm3 mol-1 cm-1, respectively and that of Cr2O7 2- is 370 dm3 mol-1 cm-1 and 11 dm3 mol-1 cm-1, respectively.
A steel sample, weighing 1.374 g, was dissolved and Mn and Cr in the resulting solution oxidised to MnO4 – and Cr2O7 2–. The solution was diluted with a H2SO4 solution (c = 1 mol dm-3) to 100.0 cm3 in a volumetric flask. The transmittances of this solution were measured with a cell of 1.0 cm path length and with the solution of H2SO4 as blank. The observed transmittances at 440 nm and 545 nm were 35.5 % and 16.6 %, respectively.
Calculate from these data the percentage of Mn and Cr in the steel sample. Assume that Beer’s law is valid for each ion and that the absorption due to one ion is unaffected by the presence of the other ion.
Model Answer
Denote the molar absorptivity of MnO4– at 440 nm and 545 nm by ε1 and ε2 and that of Cr2O7 2– by ε3 and ε4:
ε1 = 95 dm3 mol-1 cm-1, ε2 = 2350 dm3 mol-1 cm-1
ε3 = 370 dm3 mol-1 cm-1, ε4 = 11 dm3 mol-1 cm-1
The absorbance A is related to % transmittance T by
A = 2 − log T
From the values given for the sample solution
A440 = 2 – log 35.5 = 0.45
A545 = 2 – log 16.6 = 0.78
Now if one denotes the molar concentrations of MnO4− and Cr2O7 2– in the steel sample solution by c1 and c2 respectively, we have
A440 = ε1 c1 + ε3 c2
A545 = ε2 c1 + ε4 c2
Using the given data, we get
c1 = 0.0003266 mol dm–3
c2 = 0.001132 mol dm–3
The amount of Mn in 100 cm3 of the solution:
0.0003266 mol dm–3 × 54.94 g mol–1 × 0.1 dm3 = 0.001794 g
(0.001794 / 1.374) × 100 = 0.13 % Mn in steel sample
The amount of Cr present in 100 cm3 of the solution:
0.001132 mol dm–3 × 2 × 52.00 g mol–1 × 0.1 dm3 = 0.0118 g
(0.0118 / 1.374) × 100 = 0.86 % Cr in steel sample
Cobalt (II) forms a single complex CoL3 2+ with an organic ligand L and the complex absorbs strongly at 560 nm. Neither Co(II) nor ligand L absorbs at this wavelength. Two solutions with the following compositions were prepared:
Solution 1 [Co(II)] = 8×10–5 and [L] = 2×10-5
Solution 2 [Co(II)] = 3×10–5 and [L] = 7×10-5
The absorbances of solution 1 and solution 2 at 560 nm, measured with a cell of 1.0 cm path length, were 0.203 and 0.680, respectively. It may be assumed that all the ligand in solution 1 is consumed in the formation of the complex.
From these data calculate:
i. molar absorptivity of the complex CoL3 2+
ii. stability constant for the formation of the complex CoL3 2+.
Model Answer
In solution 1 all the ligand is consumed in the formation of the complex, thus:
[CoL3 2+] = (2 × 10–5) / 3 = 0.667 × 10–5
Absorptivity of the complex CoL3 2+ is
ε = 0.203 / (0.667 × 10–5 mol dm-3 × 1.0 cm) = 3.045 × 104 dm3 mol-1 cm-1
If the concentration of the complex CoL3 2+ in solution 2 is c
c = 0.68 / (3.045 × 104 dm3 mol-1 cm-1 × 1cm) = 2.233 × 10–5 mol dm-3
[Co2+] = [Co2+]total – [CoL3 2+] = 3×10–5 – 2.233×10–5 = 0.767×10–5
Similarly,
[L] = [L]total – 3 [CoL3 2+] = 7×10–5 – 3 × 2.233×10–5 = 0.300×10–5
The reaction of complex formation is
Co2+ + 3 L = [CoL3 2+]
The stability constant K is given by
K = [CoL3 2+] / ([Co2+][L]3) = 1.08 × 1017