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Chem.125-126: Feb. 5 - 10Experiment 2 Session 2

Preparation• Pre-lab prep and reading for E2, Parts 3-5

Experiment 2 Session 2Electrons and Solution Color

Three hour lab• Complete E2 (Parts 1 - 5)• Prepare discussion presentation• Prepare team report.• Give team report to GSI at the end of labor turn in by grace deadline to GSI’s atriumlevel mailbox.

Electrons will move from a lower to an availablehigher energy level if the provided energy = exactlythat needed for a possible energy level transition

DEMO

Background Information: Energy and electrons

If electrons move from a higher to a lower energylevel, the difference in energy will be released.

λ 400 Violet - Blue - Green - Yellow - Orange - Red λ 800

Visible Light and Energy

•The shorter the wavelength (λ), the greater is its energy•The shorter the wavelength (λ), the higher is its frequency

Light is a form of energy

Light source →

Wavelength and Energy

Balloon containing H2 and Cl2

DEMO1. Expose a balloon containing H2 and Cl2 to red light (≅ λ 650).2. Expose a balloon containing H2 and Cl2 to blue light (≅ λ 450).

Upon exposure to blue light, the bonds between H atoms and Clatoms are broken. When the atoms of H and Cl recombine to formmore stable HCl there is a release of energy and the balloon bursts.

• The shorter the wavelength, the greater its energy

Part 3. Solution Color and Light Interaction

• Determine if salt solution color is predictable based onthe cations interaction with visible light versus:

• Plot an absorption spectrum for each teamassigned salt solution

- placement of the cation’s element in the periodic table?- the cation’s electron configuration?- the cation’s ionic radius?

2

1A VIIIA

1H

1s1 IIA IIIA IVA VA VIA VIIA

2He1s2

3Li

2s1

4Be

2s2

5B

2s22p1

6C

2s22p2

7N

2s22p3

8O

2s22p4

9F

2s22p5

1 0Ne

2s22p6

1 1Na

3s1

1 2Mg

3s2 IIIB IVB VB VIB VIIB VIIIB ! VIIIB IB IIB

1 3Al

3s23p1

1 4Si

3s23p2

1 5P

3s23p3

1 6S

3s23p4

1 7Cl

3s23p5

1 8Ar

3s23p6

1 9K

4s1

2 0Ca

4s2

2 1Sc

3d14s2

2 2Ti

3d24s2

2 3V

3d34s2

2 4Cr

3d54s1

2 5Mn

3d54s2

2 6Fe

3d64s2

2 7Co

3d74s2

2 8Ni

3d84s2

2 9Cu

3d1 04s1

3 0Zn

3d1 04s2

3 1Ga

4s24p1

3 2Ge

4s24p2

3 3As

4s24p3

3 4Se

4s24p4

3 5Br

4s24p5

3 6Kr

4s24p6

3 7Rb

5s1

3 8Sr

5s2

3 9Y

4d15s2

4 0Zr

4d25s2

4 1Nb

4d35s2

4 2Mo

4d55s1

4 3Tc

4d55s2

4 4Ru

4d75s1

4 5Rh

4d85s1

4 6Pd

4d10

4 7Ag

4d1 05s1

4 8Cd

4d1 05s2

4 9In

5s25p1

5 0Sn

5s25p2

5 1Sb

5s25p3

5 2Te

5s25p4

5 3I

5s25p5

5 4Xe

5s25p6

5 5Cs

6s1

5 6Ba

6s2

5 7

La*

5d16s2

7 2Hf

5d26s2

7 3Ta

5d36s2

7 4W

5d46s2

7 5Re

5d56s2

7 6Os

5d66s2

7 7Ir

5d76s2

7 8Pt

5d96s1

7 9Au

5d1 06s1

8 0Hg

5d1 06s2

8 1Tl

6s26p1

8 2Pb

6s26p2

8 3Bi

6s26p3

8 4Po

6s26p4

8 5At

6s26p5

8 6Rn

6s26p6

8 7Fr

7s1

8 8Ra

7s2

8 9Ac#

6d17s2

1 0 4 +

6d27s2

1 0 5 +

6d37s2

1 0 6 +

6d47s2

1 0 7 +

6d57s2

1 0 8 +

6d67s2

1 0 9 +

6d77s2

+ Element synthesized,

but no official name assigned

Pre-transition.

Color versus Periodic Table Position

Transition Post-transition

• Compare the color data of salt solutions containing cationsfrom transition and pre and post transition families

See p. 54, lab manual for metalion outer electron configurations

Beer-Lambert Law A λ = ε c l

absorptivity factor • concentration • path length

Absorbance at λ =

• Concentration and path length are held constantwhile taking the sample’s absorbance readings forthe absorption spectrum (Part 3)

Absorbance differences across wavelengths are due to?1. Differences in the absorptivity coefficient ( ε )2. Differences in the concentration of the sample.3. Differences in the path length of the sample holder.4. All the above.

Spectrum of 0.10 M _________

Recording a spectrum• Calibrate (0 absorbance and 100% transmission) the

spectrophotomer with the blank every time you changethe λ before taking the absorbance of your sample

• Use the same sample and holder (cuvette) so thatsample concentration and path length are constant

Light Absorbance vs. Transmission

Abs = 0 100% light transmitted

Abs = 1 10% light transmitted

ABSORBANCE = -LOG TRANSMITTANCE

• Absorbance reading values = 0 to 1.0 for minimal error

Absorbance vs. Transmission Spectrum

The identity of a solution can be determined from itsabsorption (or transmission) spectrum.

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• Record the solution color and the wavelength andcolor of the wavelengths of absorption andtransmission max

Absorption Spectrum (Part 3)

Transmission λmax

Absorptionλmax

Solution Color and Light Interaction

DEMO

Q. What wavelength colors does a solution ofNiSO4 transmit?

NiSO4 transmits primarily blue, yellow, and green wavelengths

Which statement below is correct?1. Color of Abs λ max = blue-purple.2. The sample is green.3. ε is greater at λ 500 than at λ 400.

00.10.20.30.40.50.60.7

400 450 500 550 600 650 700

Abs

orba

nce

Wavelength ! (nm)

Vio

let

Blue

Yel

low

Gre

en

Ora

nge

Red

Beer-Lambert Law and Path Length

Beer-Lambert Law and Path Length

Path length and light absorbance are directlyproportional at a fixed wavelength and concentration.

DEMO

Path length changes will result in a proportional change inabsorbance values if sample concentration is constant. Absorbance readings will alter proportionately across allwavelengths; the spectrum pattern will not alter.

Path Length and Absorption Spectra

Fig. Absorption spectra at different sample path lengths

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Beer-Lambert Law and Path Length

• Different spectrophotometers have different path lengths!• Don’t change spectrophotometers in the middle of ananalysis regarding concentration (Parts 4 and 5)!

Part 4. Concentration and Light Absorbance

- Write a mathematical expression to express thepattern between the concentrations and theabsorbance values of your team assigned sample

• Plot a calibration curve (plot of absorbance versusconcentration at a specific wavelength) for your teamassigned sample

• Successfully use the calibration curve to determinethe unknown concentration of your team assignedsample (Part 5)

Beer-Lambert Law A λ = ε c l

absorptivity factor • concentration • path length

Absorbance at λ =

• Wavelength and path length are held constantwhen producing a calibration curve (Part 4)

Beer-Lambert Law and Concentration

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5 6 7 8

Abso

rban

ce

Concentration (mM)

Absorbance is proportional to concentration at aconstant wavelength (λ )and constant path length

DEMO

Beer-Lambert Law and Concentration Calibration Curve and Path Length

= 1/2pathlength

Calibration curve absorbance readings will alter if thepath length is altered

Concentration of unknown samples (Part 5) will be inerror if the path length is altered

0.50.40.30.20.10.00.0

0.5

1.0

1.5

2.0

2.5

[Plastocyanin], mM

Abso

rbance

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Absorbance and Path Length

Answer:• Use a constant sample holder (curvette) while

taking absorbance readings• Don’t switch spectrophotometers!

Path length must be fixed for a calibration curve orabsorbance readings will be in error.

Q. How do you hold the path length constant?

Preparation of Calibration Curve (Part 4)

1) Prepare a set of solutions of known and accurateconcentration by diluting the team prepared andassigned 0.1 M solution

M1V1 = M2V2

The wavelength of max absorbance is typically chosen. Why?

Preparation of Calibration Curve2) Refer to the samples absorption spectrum to choose awavelength for the calibration curve

Changes in absorbance with changes in concentration aremaximum and the calibration curve line has a maximum slope.

1.20

0.80

0.40

0.00

Abso

rban

ce

Wavelength (nm)

250 350 450 550 650 750 0.50.40.30.20.10.00.0

0.5

1.0

1.5

2.0

2.5

[Plastocyanin], mM

Absorbance

Spectrum of 0.16 mM Plastocyanin Calibration curve at 600nm

Calibration Curve Wavelength

Q. Will the slope of the linear line of a calibration curve produced at550 nm be greater or less than the slope of the line of the abovecalibration curve produced at 600 nm? _______________Less

Wavelength of Calibration Graph?

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Abso

rpti

on

0.8

0 0.1 0.2 0.3 0.4 0.5

[M+] (Molar)

0

0.2

0.4

0.6

0.8

1

1.2

Abso

rpti

on

400 450 500 550 600 650 700

Purp

le

Blu

e

! (nm)

Gre

en

Yel

low

Ora

nge

Red

Q. A 0.4 M solution of M+ has the absorption spectrum onthe left. Circle the wavelength of its calibration graph: 425 500 550 600 650

Preparation of Calibration Curve

3) Calculate the slope of the linear line of yourcalibration curve. You are done!

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5 6 7 8

Abs

orba

nce

Concentration (mM)

∆y = 1.15 – 0.0 = .15∆x 7.5 – 0.0

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Calibration Curve Slope

Q. What does the slope of the calibration curverepresent in the Beer-Lambert expression Aλ = εlc?

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5 6 7 8

Abso

rban

ce

Concentration (mM)

Answer:• Aλ = εlc is the same as y = mx + b• The slope of the line = εl

Concentration of Unknown (Part 5)

• Read the absorbance of your solution of unknownconcentration at the wavelength of your calibration graph

• Calculate the concentration of your unknown.

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5 6 7 8

Abso

rban

ce

Concentration (mM)

Fig. Calibration Curve of XY(aq) at 600 nm

Slope = 0.15Abs/mM

Concentration of Unknown (Part 5)

Q. A sample of unknown concentration of XY(aq) has Abs >1.2 at a λ of 600nm, What is its concentration (mM)?

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5 6 7 8

Abso

rban

ce

Concentration (mM)

Fig. Calibration Curve of XY(aq) at 600 nm

Slope = 0.15Abs/mM

Unknown Concentration Determination

Fig. Calibration Curve Deviation

• The Beer-Lambert law only applies at low concentrations!• Do NOT extrapolate the linear line of a calibration curve!

• Determine the concentration of a diluted sample of theunknown if the unknown’s absorbance reads above anAbs = 1 or outside the linear line of the calibration curve

“Eyeball” the graph ONLY for an approximateconcentration.

Unknown concentration determination

• Use the slope and Beer-Lambert law to determinean exact concentration

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5 6 7 8

Abso

rban

ce

Concentration (mM)

What is the unknown concentration?

The diluted sample has an absorbance at λ of 600nm = 0.57.

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5 6 7 8

Abso

rban

ce

Concentration (mM)

Fig. Calibration Curve of XY(aq) at 600 nm

Slope = 0.15Abs/mM

Abs 0.57 = 0.15Abs/mM • c 3.8 mM = c

Abs λ600 = elc

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What is the unknown concentration?

Q. A diluted sample of unknown concentration = 3.8 mM. Youprepared the diluted sample by adding 6.0 mL of water to 2.0mL of the sample of unknown concentration. What is theunknown concentration (mM)?

M1V1 = M2V2 M1•2.0 mL = 3.8 mM • 8.0 mL

M1 = 15.2 mM

a) 3.8 mMb) 7.2 mMc) 11.4 mMd) 15.2 mM

Questions?Contact nkerner@umich.edu