3. A student prepares 5 calibration standards of a red dye by mixing a stock dye
ID: 228870 • Letter: 3
Question
3. A student prepares 5 calibration standards of a red dye by mixing a stock dye solution (M = 3.8 x 10-2 M) with water in the volumes presented in the table below. Order the solutions from lowest (1) to highest (5) absorbance. This image is a table with the following header columns- Position Number, Tube Label, Volume of Distilled Water (mL), Voluume of Standard Red Dye (mL), and Total Volume (mL). Data is presented as follows: Position Number- 1, Tube Label- A, Volume of Distilled Water (mL)- 3.0, Volume of Standard Red Dye (mL)- 7.0, Total Volume (mL)- 10.0; Position Number- 2, Tube Label- B, Volume of Distilled Water (mL)- 0.0, Volume of Standard Red Dye (mL)- 10.0, Total Volume (mL)- 10.0; Position Number- 3, Tube Label- C, Volume of Distilled Water (mL)- 10.0, Volume of Standard Red Dye (mL)- 0.0, Total Volume (mL)- 10.0; Position Number- 4, Tube Label- D, Volume of Distilled Water (mL)- 8.0, Volume of Standard Red Dye (mL)- 2.0, Total Volume (mL)- 10.0; Position Number- 5, Tube Label- E, Volume of Distilled Water (mL)- 5.0, Volume of Standard Red Dye (mL)- 5.0, Total Volume (mL)- 10.0.
Explanation / Answer
Present the table as below.
Position Number
Tube Label
Volume of distilled water (mL)
Volume of standard Red dye (mL)
Total volume (mL)
Concentration of Red dye in the dilute solution (M) **
1
A
3.0
7.0
10.0
(3.8*10-2)*(7.0)/(10.0) = 2.66*10-2
2
B
0.0
10.0
10.0
(3.8*10-2)*(10.0)/(10.0) = 3.8*10-2
3
C
10.0
0.0
10.0
(3.8*10-2)*(0.0)/(10.0) = 0.0
4
D
8.0
2.0
10.0
(3.8*10-2)*(2.0)/(10.0) = 7.6*10-3
5
E
5.0
5.0
10.0
(3.8*10-2)*(5.0)/(10.0) = 1.9*10-2
** Use the dilution equation
M1*V1 = M2*V2
where M1 = 3.8*10-2 M is the concentration of the Red dye in the stock solution; V1 = volume of the Red dye solution taken in mL; M2 = concentration of the Red dye in the dilute solution and V2 = final volume of the dilute solution = 10.0 mL.
Therefore,
M2 = M1*V1/V2 = (3.8*10-2 M)*V1/(10.0 mL)
The Beer’s law is given as
Abs = *(concentration of light absorbing substance)*l
where Abs = absorbance of the solution due to the light absorbing substance; = molar absorptivity of the substance and l = path length of the solution.
We see that and l remaining constant, Abs is directly proportional to the concentration of the light absorbing substance. Therefore, the higher the concentration of the light absorbing substance, the higher is the absorbance. Since the concentrations of the Red dye in the 5 samples listed above varies as C < D < E < A < E, hence, the order of the absorbances (lowest-highest, 1 = lowest, 5 = highest) are
C – 1
D – 2
E – 3
A – 4
B – 5
Position Number
Tube Label
Volume of distilled water (mL)
Volume of standard Red dye (mL)
Total volume (mL)
Concentration of Red dye in the dilute solution (M) **
1
A
3.0
7.0
10.0
(3.8*10-2)*(7.0)/(10.0) = 2.66*10-2
2
B
0.0
10.0
10.0
(3.8*10-2)*(10.0)/(10.0) = 3.8*10-2
3
C
10.0
0.0
10.0
(3.8*10-2)*(0.0)/(10.0) = 0.0
4
D
8.0
2.0
10.0
(3.8*10-2)*(2.0)/(10.0) = 7.6*10-3
5
E
5.0
5.0
10.0
(3.8*10-2)*(5.0)/(10.0) = 1.9*10-2
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