# 2.4: UV-Visible Spectroscopy- Solutions.

- Page ID
- 189757

__Exercise 2.1.1__

You observe a colored object. Estimate the wavelength of light that was absorbed by the object.

a) 700 nm (just an estimate, but maybe somewhere between 620 and 750 nm)

b) 425 nm c) 540 nm d) 580 nm

__Exercise 2.1.2:__

a) 590 nm b) 400 nm c) 500 nm d) 300 nm

__Exercise 2.1.3__

a) 590 nm (strong); that's all b) 400 nm (strong); 600 nm (weak)

c) 500 nm (strong); 350 nm (weak) d) 300 nm (strong); 450 nm (weak)

__Exercise 2.1.4__

Remember, here we are observing the photons directly, rather than the onew complementary to the absorbed photons.

copper > barium > sodium > calcium > lithium

__Exercise 2.2.1:__

- 270 nm (strong, MLCT); 500 nm (weak, d-d)
- < 200 nm* (very strong, MLCT); 275 nm (strong, MLCT); 400 nm (weak, d-d) *can't see exactly where because the peak is so tall that we can't see the top
- < 200 nm* (very strong, MLCT); 275 nm (strong, MLCT); 660 nm (weak, d-d)
- < 250 nm (strong, MLCT); 300 nm (strong, MLCT); 620 nm (weak, d-d)

__Exercise 2.2.2:__

- absorbs blue-green; we see orange-red
- absorbs violet; we see yellow
- absorbs red; we see green
- absorbs orange; we see blue

__Exercise 2.2.3:__

a) \(A= \varepsilon c l = 14 L g^{-1} cm^{-1} \times 0.025 g L^{-1} \times 1 cm = 0.35\)

a) \(A= \varepsilon c l = 14 L g^{-1} cm^{-1} \times 0.011 g L^{-1} \times 1 cm = 0.15\)

__Exercise 2.2.4:__

- \(A = \varepsilon c l \: or \: c = \frac{A}{\varepsilon l} = \frac{0.75}{14 L g^{-1} cm^{-1} \times 1cm} = 0.054 g L^{-1}\)
- \(A = \varepsilon c l \: or \: c = \frac{A}{\varepsilon l} = \frac{0.21}{14 L g^{-1} cm^{-1} \times 1cm} = 0.015 g L^{-1}\)

__Exercise 2.2.5:__

Based on our color wheel it absorbs both yellow and green; we see a mix of violet and red. (KMnO_{4} is really a deep purple.)

__Exercise 2.2.6:__

KMnO_{4} has MW = 158 g mol^{-1}

\(14 \frac{L}{g \: cm} \times 158 \frac{g}{mol} = 2212 \frac{L}{mol \: cm}\)

__Exercise 2.2.7:__

a) \(A = \varepsilon c l = 2200 L mol^{-1} cm^{-1} \times 0.00015 mol L^{-1} \times 1cm = 0.33\)

b) \(A = \varepsilon c l = 2200 L mol^{-1} cm^{-1} \times 0.00009 mol L^{-1} \times 1cm = 0.20\)

__Exercise 2.2.8:__

- \(A = \varepsilon c l \: or \: c = \frac{A}{\varepsilon l} = \frac {0.47}{2200 L mol^{-1} cm^{-1} \times 1 cm} = 2.1 \times 10^{-4} mol L^{-1}\)
- \(A = \varepsilon c l \: or \: c = \frac{A}{\varepsilon l} = \frac {0.89}{2200 L mol^{-1} cm^{-1} \times 1 cm} = 4.1 \times 10^{-4} mol L^{-1}\)

__Exercise 2.3.1:__

a) blue b) orange

__Exercise 2.3.2:__

Add another 30 or 40 nm and you get to 270 or 280 nm.

__Exercise 2.3.3:__

a) about 250 nm b) about 280 nm

__Exercise 2.3.4:__

a)

b) yellow

__Exercise 2.3.5:__