Spectrophotometry

Metal-Ligand Complexes: Ni+2/en

section V-A

last edited: 9/3/01 (Web edition)

Spectrophotometry / Background

The Lambert-Beer Law states that the absorption of light in a thin layer of sample is proportional to the intensity of the light (I), the concentration of the species (c), and the path length (dx).

• dI = - k I c dx
Integrating yields the common form of Beer's law
• -log10[ I/I° ] = 2.303 kcl
The term -log10[ I/I° ] is called A, the Absorbance and the constant 2.303k is renamed e, the molar extinction coefficient (Greek epsilon.) By convention the units of concentration are moles/liter and the path length, l, is in cm. The Absorbance has no units but is often written as a number followed by the letter A or the symbol AU (absorbance units.)
• A = e c l
The most common application is the determination of solution concentration by measuring the Absorbance. The molar extinction coefficient is either found in the literature or it is determined from measurements of Absorbance on solutions of known concentration. The path length, l, is generally the width of a cuvette and most cuvettes are a standard 1.00 cm.

Most solutions satisfy Beer's law with a high degree of accuracy. The most common failure occurs when the species being observed participates in an equilibrium or is unstable. The failure is not really in Beer's law, but in the fact that the "c" we use is not the actual concentration in solution. If Beer's law isn't satisfied exactly, a calibration curve (Absorbance vs. concentration) still permits determining concentration.

Absorbance is a log unit. Absorbances of 0.1, 1.0, 2.0, and 3.0 mean that the sample absorbs 10%, 90%, 99% and 99.9% of the incident light, respectively. At high absorbances (above 1.5) the detector must accurately measure very low levels of light, and detector noise and stray light become important errors. At low absorbances (below 0.1) the detector must accurately report very minute changes in the light intensity.

As a practical point, absorbance readings between 0.1 < Abs < 1.5 tend to be more accurate and one should try to use solutions within this range. Readings below 0.01 or above 2.5 generally have very little accuracy or validity.

Reports:

In this case it makes more sense to submit separate reports for Part I and Part II. They both refer to the spectrum of inorganic complexes, but are otherwise unrelated. The report for part I is typically 1-2 pages in length. Part II is more involved and will require mastery of spreadsheets and finding a way to present spreadsheet results.

Experimental: Part I

SPECTROMETRY OF THE Fe+3 + SCN- COMPLEX

Most inorganic species do not absorb strongly and direct spectrophotometric methods are not very useful. Even worse, other colored species present in a sample can interfere. A common technique is to force the species of interest to react with a selected reagent to form a strongly colored species which can be measured. Ideally, the reagent is selective and only produces one colored complex. It is also important that the reaction go to completion so that all of the species of interest contributes to the final measurement.

We will examine the use of SCN- ion as a reagent for the measurement of iron. As we will see this is not an example of a good spectrophotometric reagent. The complex has a tendency to fade (an undesirable characteristic.) The equilibrium constant for the formation is relatively small, and under some conditions a modest fraction of the Fe+3 remains uncomplexed. In practice we can use the deviations from Beer's law to estimate the equilibrium constant for this complex. (The reagent 1,10 phenanthroline is a much better choice for an iron determination.)

• Prepare solutions using the following Stock Solutions (already prepared)
• Fe+3 0.0010M (in 0.5 M HCl)
• SCN- , 0.50 M (either KSCN or NH4SCN)
• KNO3 0.5 M
• The KNO3 is an inert species. It serve to keep the total ionic strength of the solutions the same. This is desirable because the apparent equilibrium constant is slightly dependent on the ionic strength.
• Mixtures-- place the following in a 100 ml volumetric flask and dilute each to volume with distilled water. Do not add the Fe3+ and dilute to volume until you are ready to record the spectrum.
• a. 5.00 ml Fe+3, 25.0 ml SCN-
• b. 5.00 ml Fe+3, 10.0 ml SCN-, 10 ml KNO3
• c. 5.00 ml Fe+3, 5.0 ml SCN-, 15 ml KNO3
• d. 5.00 ml Fe+3, 3.0 ml SCN-, 17 ml KNO3
• e. 5.00 ml Fe+3, 1.0 ml SCN-, 19 ml KNO3
• f. 5.00 ml Fe+3, 0.5 ml SCN-, 10 ml KNO3

Prepare the solutions listed and measure the absorbance at 480 nm. Measure quickly and at a consistent time (say 1 or 2 minutes after mixing.) Caution-- make one solution and measure it before working on the next mixture. For the first solution, measure the absorbance several times over an hour or two to see if the reading is constant.

Analysis

Plot the Absorbance vs. [SCN-]. A simple minded application of Beer's Law would call for all readings to be the same since there is the same amount of iron and an excess of SCN-. In practice this reaction requires a large excess of SCN- before you can assume that all the iron has been complexed.

Extrapolate your Absorbance vs. concentration plot to estimate the absorbance in the presence of an excess of SCN-; from that value, estimate the extinction coefficient of the complex. For your solutions, determine the degree to which the iron is complexed and estimate the equilibrium constant. (These will not be very precise, but you should be able to get the correct magnitude for Keq. More precise calculations would require activity coefficients and other corrections.)

(One source, Sime, Physical Chemistry, gives a value of Keq=139.) II. The

Stoichiometry and Spectra of Complex Ions

You will determine the absorption spectra of the three complexes formed when Ni+2 reacts with ethylenediamine. (We will abbreviate ethylenediamine as en in the rest of these notes.) This will be complicated since most of the solutions will contain a mixture of two species.

You will also use the absorption data to make a Job's Law plot and demonstrate the formation of all three Ni2+/en complexes. Job's method consists of seeing how specific properties (like the spectra) change as we vary the ratio of the two reactants, holding the sum of the two species constant.

#### Solutions

The starting solutions are:
• stock Ni: 0.25M Nickel Nitrate (in water)
• stock en: 0.25M ethylenediamine (in water)
The mixtures can be made by putting Ni2+ solution into a buret. For example, delivering 3.00 ml of Ni3+to a 10.00 m; volumetric flask and diluting to volume with en solution will produce a solution with XNi=0.30. Ten ml. is an adequate sample size for these determinations. The solutions will keep. They can be prepared in one period and measured in the next period.

the mixtures should have

• X(Ni)=0.1,
• X(Ni)= 0.2
• X(Ni)= 0.225
• X(Ni)= 0.25
• X(Ni)=0.275
• X(Ni)= 0.3
• X(Ni)= 0.325
• X(Ni)= 0.35
• X(Ni)=0.375
• X(Ni)= 0.4
• X(Ni)= 0.45
• X(Ni)= 0.5
• X(Ni)=0.6
• X(Ni)= 0.7
• X(Ni)= 0.8
• X(Ni)=0.9

Use the HP8452a Diode Array spectrophotometer to record the spectrum of each solution (the mixtures, the Ni2+ stock, and the en stock.)

• Use water and the blank.
• It is desirable to change the integration time to 10 seconds (this reduces the noise level.)
• Store each spectrum as a data file. You may have too many spectra to plot on a single graph, so use some judgment when you use the hardcopy option.
• Adopt a naming system for the files.
• The goal is to have a file name that others are not likely to use and that you are likely to be able to remember. It is also helpful to the instructor since there is a point at which we must clean up the old files on the hard drive.
• A suggestion is to call them NImddcc where m=month (1-12), dd=day (01-31), and cc=conc*100 (45 would be XNi=0.45; use 99 for pure Ni and 00 for pure en.) This guarantees a recognizable and unique name.
• The data system automatically assigns a .WAV to the file name. Before you quit for the day, copy the files to a 3.5" floppy.
Data Analysis will be performed by reading the spectral data into a spreadsheet. This will eventually be a massive set of data: each spectrum has over 300 data points and each of the 10 solutions will eventually generate 2-3 columns of numbers.

Analysis of the Data -- Job's Method
• (530, 546, 578, 622, and 640 nm. )
This involves a lot of arithmetic. Start by copying the data at these five wavelengths into a smaller block of data.

For each absorbance reading (each mixture at each wavelength) compute

• Y = Amix -(1-XEN)* ANi
[where ANi is measured at the same wavelength as the solution.]

The quantity Y would be zero if no reaction occurred (we assume the absorbance of EN is negligible.) Thus Y is a measure of the deviation from Beer's law at a given wavelength due to reactions and equilibrium. For each wavelength, prepare a graph of Y vs Xen. If a complex with the formula Ni(en)n2+ is formed, the Y plot (at a suitable wavelength) will show a maximum (or minimum.)

• Ymax occurs when n= Xen/(1-Xen)
Which complexes of Ni/en can you show exist?

(initial version: 9/8/91, revised 08/2/2001)

(The lab Web site or CD-ROM will contain additional information and examples on the Job's Method.)

Analysis of the Data-- Spectra of Ni(en)n

Again, this will involve a lot of calculations. However, you must begin by recognizing the underlying problem. Few, if any, of the solutions you analyzed will contain a single chemical species. The spectra are, in general, the spectra of mixtures. Your goal is to extract the spectra of the three Ni(en)n complexes.

There are two extreme cases where we can extract the spectrum of a single species. The obvious case is pure Ni2+ ion. Less obvious is the Ni(en)32+ species. This is the only Ni species present when Xen >0.8 ; the equilibrium constants are large enough that we can assume complete reaction with an excess of reagent.

You need to convert the Absorbance data for these solutions into the molar extinction coefficient at each wavelength.

The solutions in the 0.1<Xen<0.45 region are dominated by two species: Ni2+ and Ni(en)2+ . It is a good approximation to assume that the concentration of each is fixed by stoichiometry:

• [Ni(en)2+ ] = 0.25*Xen (all en is complexed)
• [Ni2+ ] = 0.25 * (1- Xen) (what's left unreacted)
• You can now subtract the known Ni2+ contribution to each spectrum and evaluate the molar extinction coefficient (spectrum) of Ni(en)2+ .
For solutions in the 0.67 < Xen <0.75 region we have two species present: Ni(en)22+ and Ni(en)32+. Again, you can assume that stoichiometry will fix the amounts of each, with little free en or Ni(en)2+ present. A little algebra gives the concentrations:
• a= [Ni(en)22+] b=[Ni(en)32+]
• 2a + 3b = 0.25 * Xen conserving en
• a + b = 0.25 *( 1-Xen) conserving Ni
• b = 0.75*Xen - 0.50
• a = 0.75 - Xen
• By subtracting the contribution of Ni(en)32+ , you can calculate the spectrum of Ni(en)22+
For solutions in the 0.5 < Xen < 0.67 region we expect to have two species present: Ni(en)2+ and Ni(en)22+ . You can subtract the contribution of Ni(en)2+ and confirm your earlier spectrum for Ni(en)22+ .

Compare your spectra with those in Gmelin, Handbuch der Anorganische Chemie. A copy of the relevant sections can be found in the lab (or on Web site and Class CD-ROM.)

Validity of the Assumptions:

The stability constants (data from Gmelin) for the three Ni2+ / en species are:

• K1 = [Ni(en)2+] / [Ni2+] x [en] = 3.3x107
• K2 = [Ni(en)22+] / [Ni2+] x [en]2 = 1.9x106,
• K3 = [Ni(en)32+] / [Ni2+] x [en]3 = 1.8x104
Verify that the assumptions are valid. Equilibrium calculations will be complex (nonlinear) but both Excel and MathCAD can handle the problem.

Spectrum below from Gmelin, Handbuch der Anorganische Chemie