Practical - solutions to UV-vis practice problems PDF

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Solutions to UV-vis practice problems...

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CHEM2001: UV/vis Practice Problems (with bonus content related to Crystal Field Theory) (As well as these questions, do the problems in the lecture notes.)

1.

A UV/vis spectrum of a 0.0001 M solution of a compound in methanol in a 0.5 cm pathlength quartz cuvette is recorded. The spectrum shows λmax = 235 nm with an absorbance of 1.05. 21,000 M-1 cm-1

(a)

Calculate εmax at 235 nm.

(b)

What absorbance would be measured if the solution was contained in a 0.1 cm quartz 0.21 cuvette?

(c)

What absorbance would be measured if the solution was diluted 1:1 with more 0.525 methanol.

2.

It is OK to use a glass cuvette or even a plastic cuvette to record a visible spectrum, but for a Short answers: No, because glass and UV spectrum the cuvette must be quartz. Why? plastic cuvettes absorb UV wavelengths. More detailed answers: No, glass or plastic cuvettes are not appropriate for UV spectra because they have significant absorbance in the UV region. Quartz cuvettes, although more expensive, are preferable because they provide a larger "spectral window". Quartz cuvettes can be used to record spectra over the range 200-2500 nm, which covers the entire visible spectrum and most of the UV spectrum. Quartz cuvettes absorb wavelengths below 190 nm. Plastic (polystyrene) and glass cuvettes typically absorb wavelengths below about 340 nm, so their usable range covers the visible region but very little of the UV region.

3.

A student records a UV/vis spectrum of p-nitroaniline from a benzene solution. The student is confused by the spectrum and takes it to a professor for advice. The professor is at first sad, then sighs, and sends the student away to record the spectrum from a solution in a different solvent.

4.

(a)

Give two likely reasons why the professor was sad that the student used benzene as a 1. Benzene is an aromatic compound and so absorbs UV solvent in this work. wavelengths, so if it is used a solvent in UV spectroscopy, its absorbance will mask absorbance by solute molecules. 2. Benzene is toxic (carcinogenic) and its use is tightly regulated for some purposes and banned for others. If someone asks you to do something with benzene, your responses should be: Why? Can I use toluene instead?

(b)

Suggest a solvent that the student could use to appease the professor's sadness. Any organic solvent that would dissolve p-nitroaniline but does not absorb widely in the UV wavelengths range would do. I would suggest ethanol, methanol, chloroform, ethyl acetate... but not benzene, toluene, chlorobenzene...

Calculate the molar extinction coefficient for 3-methylcyclohexanone at 280 nm and 320 nm

from the data in the figure below. Estimating A280nm as 0.43 gives ε280nm = 17 M-1 cm-1. Estimating A320nm as 0.05 gives ε320nm = 2.0 M-1 cm-1.

5.

You have several solutions of benzoic acid in ethanol, and you are required to use UV/vis spectroscopy to determine their concentrations. You have some 1 cm quartz cuvettes and a UV/vis spectrophotometer. The spectrophotometer works properly only for solutions that show an absorbance less than or equal to 1.35. The molar extinction coefficient for benzoic acid in ethanol at 273 nm is about 970 M-1 cm-1 . (a)

What is the maximum concentration of benzoic acid in ethanol that could be analyzed reliably in this experiment? Express your answer in: (i) mol/L; (ii) g/L; and (iii) % w/v. Note modifications to question to satisfy the pedants among us. The maximum concentration will correspond to the one that has absorbance = 1.35 at 237 nm. (i) 1.39 M (ii) 0.170 g L-1 (iii) 1.70 % w/v

(b)

6.

You need to make up three standard solutions to construct a standard curve so that you can estimate the concentrations of benzoic acid in the unknown solutions. Suggest amounts of benzoic acid (expressed in mg) that you could dissolve in ethanol and make up to a final volumes of 100 mL so that you would have three standard solutions to cover the range of concentrations that can be analyzed. To answer this question, I would aim to have three concentrations that evenly distributed between somewhere near 0 M (0 mg / 100 mL) and somewhere a little bit less than 1.39 M (i.e., 0.170 g L-1 = 170 mg/L = 17 mg/100 mL) so a reasonable answer would be 1.5 mg, 7 mg, and 13 mg.

The following absorbances were measured for three solutions containing A and B separately and for a mixture of A and B, all using the same cuvette. Calculate the concentrations of A and B in the mixture.

0.001 M 0.01 M

A B Mixture

Absorbance 475 nm 670 nm 0.90 0.20 0.15 0.65 1.65 1.65

This question is an exercise in algebra and solving simultaneous equations. To answer it, first work out ε for A and B at both wavelengths using the data for the individual compounds. Next, set up an equation to express the absorbance at 475 nm for the mixture in terms of [A] and [B], and set up a similar equation for absorbance at 670 nm, so that you have two equations and two unknowns, which can be solved. We don't know the path length, but if we express ε in terms of path length, then the path length cancels itself out in the two equations that we set up. Answers: [A] = 1.49 x 10-3 M, [B] = 0.0208 M 7.

The structures of phthalocyanine and a copper phthalocyanine complex, and the electronic absorption spectrum of the copper complex, are shown below and on the next page. Phthalocyanine is a planar organic compound. Phthalocyanines and their metal complexes are intensely coloured, and about 25% of artificial organic pigments are phthalocyanine derivatives. The copper complex is known commercially as CI Pigment Blue 15 (where CI stands for Colour Index).

N NH

N N

N N

N HN

N N

N Cu

N

N N

N N

(a)

What is the coordination geometry of the copper centre in the copper phthalocyanine complex? Square planar

(b)

What is the oxidation state of the copper centre in the complex?

2+

(c)

What is the d-electron count for the copper centre in the complex?

d9

(d)

Is the complex diamagnetic or paramagnetic? Give a reason for your answer. Must be paramagnetic since there is an odd number of d electrons.

(e)

What colour is the copper complex? Explain your reasoning. An answer such as the complex is blue because it is named "CI Pigment Blue 15" is not acceptable. Strong absorbance in of orange/red wavelengths means that the colour reflected (not absorbed)

will be blue/green (mostly blue). The colour wheel from the lecture notes suggests blue/green.

(8)

(f)

If the hydrogen atoms of the benzenoid rings of phthalocyanine are replaced by chlorine atoms, the resulting compound (hexadecachlorophthalocyanine, also known as CID5463091) is a green dye. Does this compound absorb light at higher or lower energy than the copper complex? Looking at the colour wheel from lectures, we see if the compound appears green the complementary colour (ie the colour absorbed) is red, ie longer wavelengths than red/orange, therefore lower energy absorbance.

(g)

The molar extinction coefficient for the copper phthalocyanine complex is about 2.15 x 106 M-1 cm-1. If the electronic absorption spectrum was recorded using a 1 cm pathlength cell, and the absorbance at 630 nm is 0.75, what is the concentration of the complex in the solution? Express your answer in (i) mol/L 3.5 x 10-7 mol/L and (ii) mg/mL MW = 576.07, so 3.5 x 10-7 mol/L = 2 x 10-4 g/L = 0.0002 mg/mL. apologies for the typo in the path length. If path length was 1 m, (i) 3.5 x 10-5 mol/L, (ii) 0.02 mg/mL

(h)

The compound [Cu(en)2(H2O)2]I2 (en = ethylene diamine, H2NCH2CH2NH2) dissolves in water to give solutions that contain the complex [Cu(en)2(H2O)2]2+, which has a molar extinction coefficient ε = 64 M-1 cm-1 at λ = 545 nm. What mass of [Cu(en)2(H2O)2]I2 would you need to dissolve in 1 mL to have a solution that had an absorbance of 0.75 at 545 nm for a path length of 1 cm? Apologies of path length omitted. MW = 474. Mass = 5.5 mg/m

Which of the following compounds has absorbance at the longest wavelength in the UV/vis spectrum. Give reasons for your answer. (a) ethylene vs 1,3-butadiene vs 1,3,5-hexatriene (1,3,5-hexatriene highest number of conjugated double bonds means smallest energy gap between π and π* orbitals) (b) glucose vs β-carotene (β-carotene has some conjugated double bonds, so has π ---> π* transition, whereas glucose has no conjugated double bonds, therefore only has higher energy transitions possible) (c) [Fe(CN)6]3- and [Fe(CN)6]4- (the Fe2+ complex will have smaller crystal field splitting of energies of t2g and eg orbitals (Δo) due to weaker attraction of ligands compared to Fe3+) (d) [Co(NH3)6]3+ and [Rh(NH3)6]3+ (the Co3+ complex will have smaller crystal field splitting of energies of t2g and eg orbitals (Δo) due to the expanded d orbitals of Rh3+ having better overlap with ligand orbitals than the less expanded orbitals of Co3+) (Apologies for typo in the question...not easy to answer if comparing Ru to Co, straightforward to answer for Rh and Co in same group of periodic table)

(e) [Co(CN)6]4- and [Co(H2O)6]2+ (the [Co(H2O)6]2+ complex will have smaller crystal field splitting of energies of t2g and eg orbitals (Δo) due H2O being a weaker field ligand than CN-)...