Quantitative Analysis of Lead in Hair Color

The lead (II) acetate is used to darken hair by reacting with the sulfur present in the hair color product and in the amino acids cysteine and methionine. These amino acids are incorporated into the protein structure of hair. The product of this reaction is black lead (II) sulfide. You will calculate the amount of lead (II) acetate present by measuring the amount of an insoluble lead (II) compound formed when a sample of the hair color product is reacted with potassium chromate.

Two procedures are available for you to use, you may use whichever one you like. Both procedures are gravimetric procedures where an insoluble compound of lead will be formed and collected on filter paper. From the mass of the lead compound formed and the balanced formula equation the amount of lead (II) acetate present in the hair color product can be calculated. In procedure I the lead (II) acetate will be reacted with potassium chromate to form insoluble lead (II) chromate. In procedure II the lead (II) acetate will be reacted with potassium iodide to form insoluble lead (II) iodide. All measurements and calculations made should be carried out to the proper number of significant digits.


Procedure I
1.
Measure out about 10 cc of hair color product. Be sure to record the exact volume used. The exact mass of the solution will also be needed. This will require the massing of the graduated cylinder both before and after the hair color product is placed into it.

2. Place the hair color product into a 150 mL beaker and add about 15 cc of 0.01 M potassium chromate solution. Warm the solution gently over a very low flame for about 2 minutes. Do not let the solution get above 50 degrees Celsius.

3. Mass out a piece of filter paper and filter the solution. Make sure that no yellow particles pass through the filter paper. If yellow particles pass through the filter paper re-filter the solution.

4. If the filtrate is not a clear yellow solution (this is different from a cloudy yellow which indicates that particles of lead (II) chromate are in the filtrate ) add 10 mL more of 0.01 M potassium chromate solution to the filtrate. Heat the solution as you did in step 2 and filter using the same filter paper as in step 3. This is necessary because a clear colorless filtrate at this point indicates that all of the lead (II) acetate might not have reacted. A clear yellow filtrate at this point indicates that there is an excess of potassium chromate, therefore, all of the lead (II) acetate must have reacted.

5. Rinse the precipitate on the filter paper with 10 mL portions of distilled water until the filtrate is clear and colorless. A clear colorless filtrate is desired at this point to insure that all excess potassium chromate has been rinsed off the filter paper. Carefully open the filter paper and place it on a paper towel to dry overnight. Determine the mass of the filter paper when it is thoroughly dry.


Procedure II
1.
Measure out about 10 cc of hair color product. Be sure to record the exact volume used. The exact mass of the solution will also be needed. This will require the massing of the graduated cylinder both before and after the hair color product is placed into it.

2. Place the hair color product into a 150 mL beaker and add about 15 cc of 0.01 M potassium iodide solution.

3. Mass out a piece of filter paper and filter the solution. Make sure that no yellow particles pass through the filter paper. If yellow particles pass through the filter paper re-filter the solution.

4. After the solution has filtered add a few drops of 0.01M potassium iodide to the filtrate. If a yellow precipitate forms add an additional 5 cc of 0.01M potassium iodide to the filtrate, warm gently over a very low flame for about 2 minutes as you did in step 2. Cool and re-filter using the same filter paper. This is necessary to insure that all of the lead (II) acetate reacted. Repeat this step as often as necessary to insure that all of the lead (II) acetate reacted.

5. Rinse the precipitate on the filter paper with 10 mL portions of distilled water. Test the filtrate after the addition of each 10 mL portion of distilled water with a few drops of 0.05 M silver nitrate until no precipitate of silver iodide forms. This is necessary to insure that all excess potassium iodide has been rinsed off the precipitate of lead (II) iodide on the filter paper. Carefully open the filter paper and place it on a paper towel to dry overnight. Determine the mass of the filter paper when it is thoroughly dry.

PROCEDURE III:

REAGENTS

METHOD

To save time, the atomic absorption spectrophotometer will be warmed up and optimized by your laboratory instructor. If you are interested, you may return at a later time to learn how this is done. Your laboratory instructor will explain the proper method of operation of the instrument in performing this analysis, as it varies to some extent from one instrument to another. After this has been done, proceed as follows:

  1. Beginning with the standard of lowest lead concentration, aspirate each standard in turn. Record the absorbance value for each standard, or the meter reading, depending on the instrument you are using. Allow sufficient aspiration time for the pen to reach maximum deflection and trace a flat-topped peak. Following aspiration time of each standard, place the aspirator tube in a beaker of distilled water and wait for the recorder pen to return to the baseline and the absorbance digital readout to come to 0.000.
  2. When you have completed the aspiration of all of the standard solutions, aspire your lead unknowns in the same manner, recording the absorbance values as before. Aspirate the distilled water until the readout again reads 0.000.
  3. Shut down the instrument according to the instructions of your laboratory instructor.

TREATMENT OF THE DATA:

  1. Make a plot of the absorbance values for each standard solution versus the concentration of lead in that standard on a linear graph paper. Draw a line of best fit. Determine the lead concentration in the unknown by drawing a line from the absorbance value of your unknown to the calibration curve on your graph. Draw a second line from the intersection of the absorbance line with the curve to the concentration axis of your graph. The calibration curve must pass through the intersection of the two axes of the plot, as zero concentration would have zero absorbance. For instruments that do not have an absorbance mode of readout, plot the meter needle deflections versus concentration to produce the same result.
  2. Make a second plot of the peak height for each standard lead solution versus concentration, and find the concentration of your unknown solution from its peak height.
  3. Report both concentration values for your unknowns on the Data Sheet. Include your graphs with the Data Sheet and answered questions when you hand in your experiment.

 

Questions
1.
Determine the density of the hair color product.
2. Determine the mass of precipitate produced and collected on the filter paper. The precipitate will be either lead (II) chromate if procedure I was used, or lead (II) iodide if procedure II was used.
3. Write the balanced equation for the reaction that took place based on the procedure you followed.
4. From the mass of the precipitate produced and the balanced equation for the reaction, determine the amount (in grams) of lead (II) acetate that must have been present in the sample of hair color product used.
5. Determine the percentage of lead acetate in the hair color product.
6. Determine the number of mols of lead acetate present in 1000 mL (1 liter) of the hair color solution.

 

Teacher Comments:
The hair color solution I have the students analyze is Grecian Formula 16. The solution contains elemental sulfur which needs to be filtered off prior to the analysis. I usually do this before the laboratory procedure and save the filter paper to show the students and explain why this filtration is necessary.

After the first year I did this lab, I wrote the company and asked if they would share with me the actual percentage of lead (II) acetate. They indicated in their response that the product contained between 0.29 and 0.34% lead (II) acetate by weight. Because of the small amount of lead (II) acetate present, a balance capable of massing to 0.001 grams is necessary.

The profit margin calculation is very crude because it assumes that the only cost of production is the cost of the chemicals. However, it does serve to give students some insight into what profit margins are and how they are determined.

Final Note: Because of the expense of Grecian Formula 16, I usually buy one bottle, filter off the sulfur, and combine it with a 0.32% lead acetate solution that I mix up in the lab. One bottle is purchased so that the solution to be analyzed will have the correct properties, namely odor.

Both procedures have their advantages and disadvantages. I have found that Procedure I works the best.

REFERENCES:

http://chem.lapeer.org/Chem1Docs/LeadAnal.html