[MathisFun]

Here is the background information:

Equipment,
Spectrophotometer (colorimeter)
20 mL vials x 6
10.00 mL graduated cylinder
25.00 mL volumetric flask and stopper x 1
5.00 mL volumetric pipet and pipet bulb

Reagents,
Food coloring
Distilled water
Unknown solution

A spectrophotometer is used to read a set of 6 vials:
1, distilled water
2-5, a set of diluted solutions from the same stock solution of decreasing concentration
6, an unknown solution

The appropriate wavelength used to measure the absorbance is given at 430 nm.

12 mL of distilled water is placed into one of the glass vials, about 2/3 full. It is placed into the colorimeter, and its absorbance is measured. Its concentration and absorbance is listed as 0.000.

It is removed from the colorimeter. 12 mL of solution #1 is transferred into a glass vial. Its absorbance is read. This procedure is repeated for solutions #2 through #4.

Here is the read absorbances:
Sample – Concentration – Absorbance – Trans - Scatter
Distilled water – 0.000 – 0.000 – 100.000 – 0.000
Solution 1 – 0.200 – 0.091 – 81.132 – 0.000
Solution 2 – 0.400 – 0.173 – 67.170 – 0.000
Solution 3 – 0.600 – 0.262 – 54.742 – 0.000
Solution 4 – 0.800 – 0.340 – 45.736 – 0.000

A linear regression line and equation are generated as a result of the measured absorbances of the solutions. Here is a link to the graph:
http://i48.tinypic.com/2csffdg.jpg

A 10.00 mL graduated cylinder is used to measure out 10.00 mL of an unknown colored solution. A 5.00 mL pipet is used to transfer the unknown solution into a 25.00 mL volumetric flask and diluted to the mark. Since this is an unknown solution, it will not be a part of the calibration curve in the linked graph above (this will be addressed later). Some of this unknown solution is transferred to a numbered glass vial and placed in the colorimeter.

Quick question before I continue, this phrase “is transferred to a numbered glass vial” does that mean one of the vials being solution #1-4, and not a new vial?

The “X” on the graph represents the unknown:
http://i48.tinypic.com/2csffdg.jpg
The Absorbance of the unknown solution is listed as: 0.179

Another quick question, based on the compiled data of the absorbances of distilled water and solutions #1-4 above, is there a formula to determine the absorbance of the unknown solution without it being given?

I’m aware of this:
A = log (100/%T), where %T is the percent transmittance. But given the above data, I don’t see how I can compute %T.

There is another formula: A = Ebc
A = absorbance || E = epsilon, molar absorptivity constant, units M-1 cm-1
B = path length, units in cm || c = concentration, units of molarity, M (mol x L-1)
But like the previous formula, I don’t see the connection with the available data.

The equation in the graph, representing the best-fit line, is used to calculate the concentration of the unknown: Y = 0.4252x + 0.0029.

--Use the appropriate info from the graph and your unknown absorbance reading to calculate the following concentration, diluted unknown concentration.
(note: to determine the correct number of significant figures in the concentration of your diluted unknown, you will first need to determine the ‘number of significant figures’ to keep for the slope, and the ‘digits of precision’ to keep for the y-intercept of your graph’s equation.)
http://i48.tinypic.com/2csffdg.jpg

y = 0.4252x + 0.0029

The slope is the average of the x and y values. Since x = M (concentration), which is 3 significant figures, and y = Abs (absorbance), with the lowest output listed as 0.091, for solution #1, x (3 sig fig) divided by y (2 sig fig), would yield a value with 2 significant figures for its answer, therefore 0.4254x becomes 0.43x.

The y-intercept is the value, when x = 0. It is the values for y, which is Absorbance, with each one listed as 3 digits of precision. Therefore, 0.0029 becomes 0.003.

The Absorbance of the unknown solution is listed as 0.179, which is represented as ‘y’ in the equation.

The equation changes to: 0.179 = 0.43x + 0.003
X = (0.179 – 0.003)/0.43 = 0.4093, which becomes 0.41 M diluted concentration of unknown solution (2 significant figures)

--Use the dilution factor with which your unknown solution was prepared to calculate the concentration of your original stock unknown.

I will use the following formula: M₁V₁ = M₂V₂, converted to M₁ = ((M₂V₂)/V₁), and mL converted to L. The values for V₁ and V₂ were taken from the earlier paragraph, in which, a 5.00 mL pipet was used to transfer the unknown solution to a 25.00 mL volumetric flask.

((0.41 M)(0.02500 L))/0.00500 L = 2.05, becomes 2.1 M stock concentration of unknown solution.

--Why is the data of your unknown data point (unknown solution) not included in the creation of the calibration curve in the graph?

Because the data used in the graph was used to generate an equation representing the best-fit line to calculate the concentration of the unknown solution. If it was used, it would nullify itself, and another solution of unknown concentration would have to be used in its place.

[Hunter2]

Quick question before I continue, this phrase “is transferred to a numbered glass vial” does that mean one of the vials being solution #1-4, and not a new vial?

Its a new one.

Another quick question, based on the compiled data of the absorbances of distilled water and solutions #1-4 above, is there a formula to determine the absorbance of the unknown solution without it being given?

No. The absorbance is measured. You want to know the concentration.

More check this:

http://en.wikipedia.org/wiki/Beer%E2%80%93Lambert_law

[MathisFun]

Thanks for setting me straight with the vials and the unknown's absorbance. Since the absorbance was provided, and I plugged it into the best-fit equation, it was determined that the dilution comes out to, 0.41 M dilution concentration, as illustrated:

0.179 = 0.43x + 0.003
X = (0.179 – 0.003)/0.43 = 0.4093, which becomes 0.41 M diluted concentration of unknown solution (2 significant figures)

My concern is if I was able to use this dilution value, and properly calculate the concentration of the original stock unknown, since in another thread, I goofed with computing the proper concentration levels. Here is my assessment:

The values for V₁ and V₂ were taken from the earlier paragraph, in which, a 5.00 mL pipet was used to transfer the unknown solution to a 25.00 mL volumetric flask.

Using (M₁)(V₁) = (M₂)(V₂) becomes (M₁) = ((M₂)(V₂))/(V₁)
((0.41 M)(0.02500 L))/0.00500 L = 2.05, becomes 2.1 M stock concentration of unknown solution.

Also, was I correct in answering this question:
Why is the data of your unknown data point (unknown solution) not included in the creation of the calibration curve in the graph?

Graph's link:
http://i48.tinypic.com/2csffdg.jpg

The unknown's absorbance is represented as an 'X' on the graph, at 0.179. 
Because the data used in the graph was used to generate an equation representing the best-fit line to calculate the concentration of the unknown solution. If it was used, it would nullify itself, and another solution of unknown concentration would have to be used in its place.

Not too confident about my response.