CHEM Lab 5 1. Experiment 5 Introduction to Separation Techniques II. Objectives

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Lab 5 1 CHEM 0011 Experiment 5 Introduction to Separation Techniques II Objectives 1. To learn about solutions, the concept of solubility and precipitation 2. To learn the centrifuge technique 3. To learn how to separate Pb 2+ and Ag + ions from solutions. Apparatus: 1. Centrifuge 2. Small test tubes 3. Test tube rack 4. Glass rod 5. Distilled water bottle 6. Pasteur pipets Solution: 1. 0.5 M lead (II) nitrate, Pb(NO 3 ) 2 solution 2. 0.1 M silver nitrate, AgNO 3, solution 3. 0.5 M potassium chromate, K 2 CrO 4, solution 4. Dilute hydrochloric acid, HCl Introduction There are many methods that one can use to separate one substance from another. The method that one would choose will be dependent on the physical properties of the substances involved. In this lab, we will focus on two separation techniques: 1. Precipitation 2. Centrifuge Precipitation If a solution contains ions, it is possible to select a reagent that will form an insoluble solid with the ions and separate them from the solution as a result of a chemical reaction. The insoluble solid is called a precipitate. During the process, the solution becomes murky and a suspension is formed. When the density of the precipitate is greater than the solution, over a period of time, the precipitate

Lab 5 2 settles to the bottom. A technique known as centrifuging accelerates the settling process. Separation of Pb (II) Ions The separation of Pb 2+ ions from an aqueous solution will be carried out using the following reaction. The precipitating reagent that provides the CrO 4 2- ions is potassium chromate, K 2 CrO 4. Separation of Ag + Ions The separation of Ag + ions from an aqueous solution will be carried out using the following reaction. The precipitating reagent that provides the Cl - ions is hydrochloric acid. CentrifugeTechnique Centrifugation uses centrifugal force to separate mixtures. It works on the principle that the more-dense components of the mixture will migrate away from the axis of the centrifuge, while less-dense components of the mixture will migrate towards the axis. When the mixture to be centrifuged is comprised of a solution and a precipitate, the end result of centrifugation will cause the precipitate to settle toward the bottom of the test tube.

Lab 5 3 The solution that is on top of the precipitate is known as the supernate. When the supernate is decanted, the precipitate, which contains the ions, is separated from the original solution. When using a centrifuge, always remember to balance the centrifuge by placing test tubes containing roughly the same volume of liquid into opposite compartments. An unbalanced centrifuge will cause the machine to operate poorly and vibrate violently. Decantation is a process used to separate the liquid from the mixture to be filtered. To decant a liquid from a solid, in one hand hold the test tube that has the mixture in it. Hold a glass-stirring rod in the other hand. Touch the lip of the test tube to the glass rod and pour the mixture to be filtered. Use the glass rod as a guide to pour slowly to ensure that the solid is not carried along. This also prevents the liquid from running back along the outside of the test tube. If small test tubes are used, then decantation is not as effective. In this case, a Pasteur pipet is used to remove the remaining solution. Practice similar concepts coming in lecture and lab (1) Phosphate ions can be removed from wastewater in sewage treatment by precipitation with aluminum (III) cations. Write the formula of the precipitate formed. (2) One cup of a particular clear chicken broth yields 4.308 g of silver chloride when excess silver nitrate is added to it. Assuming that all the chloride comes from sodium chloride, what is the mass of sodium chloride in the sample of chicken broth?

Lab 5 4 Part A Separation of Pb(II) Ions from Solution as Lead (II) Chromate Procedure: 1. Add about 12 drops of 0.5 M lead (II) nitrate, Pb(NO 3 ) 2 solution in a test tube. Record your observations on the data sheet. 2. Add a few drops of 0.5 M potassium chromate, K 2 CrO 4 solution to the test tube. Mix well. Record your observations on the data sheet. 3. Centrifuge and separate the precipitate from the solution. Remember to balance the centrifuge with a second test tube containing approximately the same volume of liquid. Place the test tube in opposite compartments. 4. Pipet out the supernate. Record your observations on the data sheet. 5. Repeat steps 1 to 4, but replace potassium chromate with dilute HCl solution. Part B Separation of Ag + Ions from Solution as Silver Chloride Procedure: 1. Add about 12 drops of 0.1 M silver nitrate, AgNO 3, solution in a test tube. Record your observations. 2. Add a few drops of dilute HCl solution to the test tube. Mix well. Record your observations on the data sheet. 3. Centrifuge and separate the precipitate from the solution. Remember to balance the centrifuge with a second test tube containing approximately the same volume of liquid. Place the test tube in opposite compartments. 4. Pipet out the supernate. Record your observations on the data sheet.

Lab 5 5 Observations Part A Separation of Pb(II) Ions from Solution as Lead (II) precipitates Step Record Your Observations 1 Add 12 drops of 0.5 M Pb(NO 3 ) 2 solution in a test tube. 2 - Add a few drops of 0.5 M K 2 CrO 4 solution to the test tube. 4 Centrifuge and remove the supernate. Step Record Your Observations 1 Add 12 drops of 0.5 M Pb(NO 3 ) 2 solution in a test tube.

Lab 5 6 2 - Add a few drops of dilute HCl solution to the test tube. 4 Centrifuge and remove the supernate. Part B Separation of Ag+ Ions from Solution as Silver Chloride Step Record Your Observations 1 Add 12 drops of 0.1 M AgNO 3 solution in a test tube. 2 - Add a few drops of dilute HCl solution to the test tube. 4 Centrifuge and remove the supernate.

Lab 5 7 Questions: 1. Write the name, formula, and colour for the Pb 2+ precipitate and Ag + precipitate obtained in the last step of Parts A and B. Precipitate with Name Formula Colour Pb 2+ Pb 2+ Ag + 2. Sodium chloride is reacted with excess silver nitrate yielding 0.9364 g of silver chloride. What are the moles of chloride?