In-Class Section        ______
Take-Home Section   ______
Combined Score        ______                  Name ________________________

Quantitative Analysis
Exam 3
In-Class Section
April 29, 2004

(3 pts.)  1.  What is one cause for deviation from Beer’s Law?

• at high concentrations, the solute molecules do not see the radiation
• chemical interactions
• non-monochromatic radiation
• stray light

(2 pts.)  2. The source in a atomic absorption spectrometer is abbreviated HCL. What do these three letters stand for?

Hollow Cathode Lamp

(8 pts.)  3.  State Beer’s Law (define each term)

A = abc

A => absorbance
a => absorptivity, a function of the substance being analyzed and the wavelength being used
b => path length of the light shinning through the sample
c => concentration of the solution being analyzed, units must agree with the units of absorptivity

(4 pts.)  4.  Distinguish the terms absorbance and molar absorptivity.

absorbance is a measure of the amount of light that is absorbed by the solution
molar absorptivity is a constant that is a function of what the analyte substance is and what wavelength is being used

(8 pts.)  5.  State the advantages and disadvantages of the inductively coupled plasma compared with conventional flames in atomic spectroscopy.

advantages
• higher temperature, thus greater sensitivity
• residence time of sample is longer
• many interferences are eliminated
• more than one element can be analyzed simultaneously
disadvantages
• more expensive instrument
• requires greater operator skill

(3 pts.)  6.  What is one of the causes of zone broadening in chromatography?

Zone broadening refers to the fact that the peaks become broader as their retention times increase. This is because:
• multiple flow paths
• longitudinal diffusion - flow away from high concentration ahead of and behind the zone center
• mass transfer to and from liquid stationary phase
• mass transfer in mobile phase

(4 pts.)  7. For what purpose is chromatography used?

Chromatography is used for the separation of mixtures.

(4 pts.)  8.  Define cathode and anode in terms of the type of electrode reaction is occurring at each.

cathode - the electrode where reduction takes place

anode - the electrode where oxidation takes place

(4 pts.)  9.  Describe one of the components of a spectrophotometer.

There are five components common to all spectrophotometers. They are:
• source of radiaiton - this is the part of the instrument that produces the radiation that is directed at the sample
• sample - the way that the sample is handled depends on the type of instrument
• monochromator - this takes the continuum of the radiation and selects a single wavelength of radiation or a narrow band of wavelengths to be analyzed by the detector
• detector - the device which monitors variations in the intensity of the radiation and converts the signal into an electrical signal
• signal processor and readout device - takes the electrical signal and processes it into a magnitude and format for the readout device which displays the information for the operator

Name ______________________

Quantitative Analysis
Exam 3
Take-Home Section
Due 4pm May 3, 2004

(12 pts.) 8.  Derive a potentiometric titration curve for 50.0 mL of 0.100 M Fe⁺² titrated with 25.0, 50.0, and 60.0 mL of 0.100 M Ce⁺⁴.

(a.) at 25.00 mL of 0.100 M Ce⁺⁴ added, V_FeM_Fe > V_CeM_Ce, pre-equivalence point

because [Fe⁺²] = [Ce⁺⁴] and V<_Ce = 1/2 V_Fe

then [ox] = [red]

thus E_cell = E_cell^o - (0.592/n)log([red]/[ox]) = E_cell^o - (0.592/n)log(1) =E_cell^o

E_Fe^o = 0.771 V

thus E_cell = E_Fe^o = 0.771 V

(b.) at 50.00 mL of 0.100 M Ce⁺⁴ added, V_FeM_Fe = V_CeM_Ce, equivalence point

E_cell = (n_FeE_Fe^o + n_CeE_Ce^o)/(n_Fe + n_Ce)

E_cell = (1E_Fe^o + 1E_Ce^o)/(1 + 1) = (0.771v + 1.72v)/2 = 1.25v

(c.) at 60.00 mL of 0.100 M Ce⁺⁴ added, V_CeM_Ce > V_FeM_Fe, post-equivalence point

[Ce⁺⁴] = ((V_CeM_Ce) - (V_FeM_Fe))/(V_Ce + V_Fe)

[Ce⁺⁴] = ((60.00mL*0.100M) - (50.00mL*0.100M))/((50.00 + 60.00)mL

[Ce⁺⁴] = 0.009091M

[Ce⁺³] = (V_FeM_Fe)/(V_Ce + V_Fe)

[Ce⁺³] = (50.00mL*0.100M)/((50.00 + 60.00)mL

[Ce⁺³] = 0.04545M

E_cell = E_Ce^o - (0.592/n)log([Ce⁺³]/[Ce⁺⁴]) = 1.72V - (0.592)log(0.04545M/0.009091M)V

E_cell = 1.72V - 0.04V = 1.68V

(8 pts.)  9. The molar absorptivity of an aqueous solution of the amino acid tryptophan (gfw = 204) is 5495 at 278 nm.  (a) Calculate the expected absorbance at 278 nm of a 4.3 x 10^-5M solution of tryptophan if a 1.50 cm cell was employed.  (b) What is the molar concentration of a tryptophan solution having an absorbance of 1.16 in a 0.96-cm cell?

A = a b c

A = (5495cm^-1M^-1)(1.5cm)(4.30x10^-5M)

A = 3.5x10^-1 = 0.35

c = A/(b c)

c = (1.16)/(5495cm^-1M^-1)(0.96cm)

c = 0.00022 M

(8 pts.)  10. An unknown sample of Cu⁺² gave an absorbance of 0.567 in an atomic absorption analysis.  Then 2.00 mL of solution containing 100.0 ppm (= mg/L) Cu⁺²  was mixed with 95.0 mL of unknown, and the mixture was diluted to 100.0 mL in a volumetric flask.  The absorbance of the new solution was 0.710.  What is the [Cu⁺² ] in the unknown?

C_x = (A_x/(A_t - A_x))((C_s V_s)/( V_x)

C_x = (( 0.567/(0.71 - 0.567))( 100.0 ppm*2.00mL/100.0mL)

C_x = (0.567/0.143) (2.00 ppm)

C_x = 3.97*2.00 ppm = 7.93 ppm

C_x = (7.93 mg/L)(1 g/1000 mg)(1 mol/63.54 g) = 0.000125 M

(5 pts.)  11.  Describe how the pH electrode operates.

A typical pH combination electrode like was used in the laboratory is composed of a glass electrode and a reference electrode. The following is the cell diagram for a combination pH electrode.

Ag(s)/AgCl(s)/Cl-1(aq)//H+(aq, outside):glass membrane:H+(aq, inside), Cl-1(aq)/AgCl(s)/Ag(s)

The pH-sensitive part of the electrode is the thin glass membrane in the spherical tip. The inner reference electrode is in contact with the hydrogen ion concentration on the inside of the electrode. The outer reference electrode is in contact with the hydrogen ion concentration on the outside of the electrode. The two reference electrodes measure the difference in potential across the glass membrane resulting from different concentrations of hydrogen ions on either side of the membrane.

(5 pts.)  12. Describe how the standard hydrogen electrode operates.

A standard hydrogen electrode is the standard against which all other electrode systems are compared. It is assigned the potential of 0.00 volts. Because of its importance, you should know how it works.

It is composed of a tube through which hydrogen gas at 1 atm of pressure can pass over the piece of platinum which acts as the catalytic surface. To improve its operation, finely divided particles of platinum are plated onto the surface to increase active surface area.

As the gas bubbles over the platinum metal the 1 M solution of hydrogen ions sweep over the surface of the platinum. This makes this electrode system reversible because hydrogen ions can be converted into hydrogen gas molecules absorbed on the surface of the platinum by accepting electrons from the metal surface.

Or hydrogen gas molecules can give up electrons to the platinum metal surface and move away as hydrogen ions.

(3 pts.)  13. Describe the purpose of the mobile phase in chromatography.

The mobile phase causes the sample mixture to move through the instrument. The mobile phase is a non-reactive gas in glc (gas-liquid chromatography, gc) and a non-reactive liquid in llc (liquid-liquid chromatography, lc).

(3 pts.)  14.  Describe the purpose of the stationary phase in chromatography.

The statinary phase interacts with the components in the mixture. These interactions are not the same with each component, thus the components move through the instrument at different rates. This produces the separation of the components.

(4 pts.)  15.  For an inexpensive photometer which displays both absorbance and %T scales, why is it advisable to take the readings in %T and convert to absorbance?

The %T scale is linear while the absorbance scale is exponential. Therefore, it is very hard to interpolate values on the absorbance scale compared to the %T scale.

(6 pts.)  16.  Give sources appropriate for each of the following regions of electromagnetic radiation:

(a.) ultraviolet region

deuterium lamp

(b.) visible region

tungsten lamp

(c.) infrared region

Nernst glower or glowbar or incandescent wire

(3 pts.) 17.  Convert the percent transmittance data to an absorbance:  12.3

A = -log(%T/100)

A = -log( 12.3/100)

A = 0.910

(3 pts.) 18.  Convert the following absorbance in terms of percent tranmittance: 0.113

%T = (10^-A)*100 = (10^-0.113 )*100

%T = 77.1

I neither received not gave any assistance on this examination nor did I see anyone else give or receive assistance on this examination.

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