Quantitative Analysis of Lead

Discussion
In this experiment you will analyze a commercial hair color product for lead (II) acetate.

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. 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.

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.
7. Determine the profit margin on one bottle of the hair color solution (Grecian formula 16). Use the following data:




      Bottle size ---------------------------- 4 fl.oz. (118 mL)
      Price per bottle ---------------------- $6.25.
      Cost of reagent grade lead acetate ---- ???? look it up.


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.


 
Questions? Comments??
Revised on: 06/07/2007 at 05:26:23