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Unformatted text preview: p the final desired volume, 100 ml of 0.2 M sucrose.
Applying this formula to the problem of diluting the K3Fe(CN)6 requires two minor
modifications since you do not have a numerical value for C1. This problem is solved by making C1 =
1 and C2 = to the desired degree of dilution that you wish to make (1/2, 1/5, 1/10, or whatever).
Determination of the concentration of K3Fe(CN)6 in your sample.
As stated earlier, the first step toward making this determination is to produce an abbreviated
absorption spectrum for K3Fe(CN)6 that shows its max. In order to do this, proceed as follows:
1) Obtain your solution of unknown concentration from your TA and record its code number in
your lab notebook.
2) Pipette 5.0 ml of your unknown into a test tube and measure absorbance at 10 nm intervals
from 390 nm to 450 nm, using the instructions described previously. Use water as your blank.
Record these readings in your lab notebook.
3) Should you find that your unknown is too concentrated (absorbance > 2), dilute your sample as
described above in a systematic manner until you are able to obtain a spectrum with a max
that has an absorbance of greater than 0.5 absorbance units. Once again, the solvent for the
K3Fe(CN)6 is water.
4) Once you have obtained readings over the specified range of wavelengths that meet the criteria
described above, graph the absorbances as a function of wavelength and identify the max.
5) Now, you are ready to calculate, using Beer's Law, the concentration of K3Fe(CN)6 in your
sample. The value of "A" for your calculation is obtained from the graph that you just made at
the max. The molar extinction coefficient for K3Fe(CN)6 at its max will be calculated
using Beer’s Law and the data you collected earlier for the 0.75 mM K3Fe(CN)6 solution.
6) Calculation of the concentration of K3Fe(CN)6 in the original sample (the one that was given
you by your TA) requires that you take into account the TOTAL dilution factor that was used.
(For example, if you diluted your original sample by a factor of 1/5th and you use the “A” value
obtained from this diluted sample for your Beer’s calculation, the resulting concentration will
be 1/5th that of the undiluted sample. Thus you must multiply the result of your Beer’s
calculation by 5 to get the concentration of the original sample.) Biology 05LA – Fall Quarter 2012 Lab 2 – page 6 In your lab notebook, you make sure that you include your graphs for this exercise, as well as the
the absorbance value for the max of the unknown K3Fe(CN)6 solution that you
the factor by which you diluted the unknown sample.
the value of E that you calculated.
the code number for your unknown sample and the concentration that you calculated for
Please show all calculations, including those for your dilutions. Learning Goals/Desired Outcomes.
1. Be able to define the following terms: absorbance, transmittance, absorption spectrum, path
length, molar extinction coefficient, and max.
2. Be able to describe the relationship between the color of a solution and the wavelength of light
absorbed by that solution.
3. Be able to list the 7 major components of a spectrophotometer and to provide a simple
description of the function of each.
4. Be able to demonstrate competence in the operation of the provided spectrophotometers.
5. Be able to explain why a blank is required for spectrophotometric studies and to describe how
the blank for a particular study is chosen.
6. Be able to describe how the value of E (the molar extinction coefficient) is determined for a
7. Be able to use Beer’s law to calculate the concentration of an unknown solution.
8. Be able to use C1V1 = C2V2 to make dilutions.
9. Be able to present spectrophotometric data in an appropriately labeled graph. Biology 05LA – Fall Quarter 2012 Lab 3 – page 1 LAB 3: DIFFUSION, OSMOSIS AND THE PERMEABILITY OF CELL MEMBRANES
The movement of materials between cells and their external environment as well as within the
cell is essential to life. For example, the raw materials for cellular metabolism must be imported and
the metabolic waste products resulting from this metabolism must be exported. Within the cell,
substrates for the cell’s enzymes must move into the active site of the enzyme and the soluble products
of these reactions then distributed throughout the cell. Given the fundamental importance of this
traffic, an understanding of the mechanisms whereby these materials move is essential.
Since both the internal and external cellular environments are aqueous, the materials to be
moved are dissolved in water and thus form a solution (remember that in an aqueous solution, water is
called the solvent and the materials dissolved in the water are called solutes). Solute movement may
be separated into two broad categories. One of these requires the direct output of energy by the cell
and is referred to as active transport. Here, a solute is moved against a concentration gradient of that
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- Fall '12