3. A student, following the procedure in this experiment, studied the response o
ID: 546075 • Letter: 3
Question
3. A student, following the procedure in this experiment, studied the response of red food dye to visible light and analyzed a stock solution of red food dye of unknown concentration. A series of dilutions of the stock solution of red food dye were made so that a Beer's law plot could be prepared. The solutions were prepared as follows: Solution 1: 1 drop of stock solution and 50.0 mL of water Solution 2: 5.0 mL of Solution 1 and 5.0 mL of water; Solution 3: 5.0 mL of Solution 2 and 5.0 mL of water; Solution 4: 5.0 mL of Solution 3 and 5.0 mL of water; Solution 5: 5.0 mL of Solution 4 and 5.0 mL of water. (1) Calculate the concentration in drops mL- of Solution 1, using Equation 6, and the other four solutions, using Equation 7. Record these concentrations in the table in (2)Explanation / Answer
(1) 1 drop of stock solution was added to 50.0 mL water. We will assume that the volume of the drop is insignificant as compared to 50.0 mL and hence, we will treat the final volume of solution 1 as 50.0 mL. The concentration of the food dye in solution 1 is (1 drop)/(50.0 mL) = 0.02 drop/mL (ans).
The final volume of solutions 2-5 is 10.0 mL. We will use the dilution equation
M1*V1 = M2*V2
where M1 = concentration of red dye in stock solution (treat solution 1 as the stock solution here); V1 = volume of stock solution taken; V2 = final volume of the dilute solution and M2 = concentration of red dye in the final solution.
Solution 2: (0.02 drops/mL)*(5.0 mL) = M2*(10.0 mL)
=======> M2 = 0.01 drops/mL (ans).
Solution 3: (0.01 drops/mL)*(5.0 mL) = M2*(10.0 mL) (treat solution 2 as the stock solution)
=======> M2 = 0.005 drops/mL (ans).
Solution 4: (0.005 drops/mL)*(5.0 mL) = M2*(10.0 mL) (treat solution 3 as the stock solution)
=======> M2 = 0.0025 drops/mL (ans).
Solution 5: (0.0025 drops/mL)*(5.0 mL) = M2*(10.0 mL) (treat solution 4 as the stock solution)
=======> M2 = 0.00125 drops/mL (ans).
(2) Determine the absorbance of each solution (correct to four decimal places) using the relation
A = 2 – log (%T)
where A = absorbance of the solution; %T = transmittance of the solution.
Solution
Concentration
drops/mL
%T
A = 2 – log (%T)
1
0.02
5.5
1.2596
2
0.01
22.6
0.6459
3
0.005
46.2
0.3353
4
0.0025
68.4
0.1649
5
0.00125
82.8
0.0820
Unknown
51.3
0.2899
(3) Plot absorbance vs concentration (drops/mL) as below.
The slope of the plot is 62.635 AU/(drops/mL) where AU = absorbance unit.
(4) Plug y = 0.2899 in the regression equation to obtain the concentration of the unknown sample.
0.2899 = 62.635*x + 0.0121
===> 62.635*x = 0.2778
===> x = 0.004435 0.00443
The concentration of the red dye in the unknown sample is 0.00443 drops/mL (ans).
Solution
Concentration
drops/mL
%T
A = 2 – log (%T)
1
0.02
5.5
1.2596
2
0.01
22.6
0.6459
3
0.005
46.2
0.3353
4
0.0025
68.4
0.1649
5
0.00125
82.8
0.0820
Unknown
51.3
0.2899
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