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  1. Potash (KOH) alum SYNTHESIS OF AN ALUM oil of vitriol (H2SO4) M(I) T(III) (SO4)2.12 H2O

  2. OBJECTIVES Examine how chemists plan and execute synthesis of desired substances Explore factors that contribute to quantity & purityof synthesized substances Learn how a desired substance is recoveredfrom a mixture. Observe some chemical properties of aluminum and its compounds

  3. Concepts: • Actual Yield Byproducts • Conditions Limiting Reagent • Percent Yield Products • Reactants/Starting Materials • Reactions Separation Process • Synthetic Pathway Target • Theoretical Yield Verified Purity

  4. Techniques: • Gravity Filtration Vacuum Filtration • Crystallization Quantitative Transfer • Apparatus: • Funnel Filter Paper • Filter Flask Buchner Funnel • Ice Bath Aspirator

  5. BACKGROUND ALUMS: a class of “double” sulfates in which one cationis monovalent and the second is trivalent  M(I) T(III) (SO4)2 . 12 H2O cation: a positively charged ion; attracted to cathodes M(I)represents monovalentcation e.g., K+, Na+, NH4+, etc. T(III)represents trivalentcation e.g., Al3+, Fe3+, Cr3+, etc. A monovalent cation is a positively charged ion with a charge of +1 12 - number of molecules of water of crystallization - (dodecahydrate) A trivalent cation is a positively charged ion with a charge of +3 The most common alum is: Potassium Aluminum Sulfate, KAl (SO4)2.12H2O. often called “Alum”

  6. Natural Alum Deposit in Australia

  7. Uses of Alum 1500 B.C. Water purification Medicinal Food additive (pickling) Fabrics (mordant/marbling) Paper processing – Filler Styptic pencil BUT

  8. An unusual use for Alum "For the Freckles which one getteth by the heat of the Sun: Take a little Allom beaten small, …… —Christopher Wirzung, General Practise of Physicke, 1654. In addition to a cure for freckles, Wirzung discovered the pancreatic duct

  9. ALUMINUM CHEMISTRY Position of Al in periodic table suggests it may have interesting properties

  10. ALUMINUM CHEMISTRY Position of Al in periodic table suggests it may have interesting properties

  11. ALUMINUM CHEMISTRY Position of Al in periodic table suggests interesting properties Aluminum metal reacts equally well with aqueous acids 2 Al (s) + 6 H+ + 6 H2O 2 Al(H2O)63+ + 3H2 (g) and bases 2 Al (s) + 2 OH- + 6 H2O 2 Al(OH)4- + 3H2 (g) Aluminum also reacts with atmospheric oxygen to form stable oxide – binds to metal & protects it from further oxidation

  12. In aqueous solution, Al3+ is a moderately strong acid: Al (H2O)63+ Al (H2O)5(OH)2+ + H+ In strongly basic solutions, Al3+exists as “aluminate” ion which we have written as Al(OH)4- In aqueous solutions of intermediate pH, aluminum forms insoluble compounds such as Al(OH)3, AlO(OH) & hydrated oxides, Al2O3.nH2O. • These compounds are very insoluble, gelatinous (colloidal) substances with large surface areas.

  13. Potassium Aluminum Alum We have written the formula as KAl(SO4)2.12 H2O, more accurate description of the substance is octahedron K(H2O)6+ Al(H2O)63+ (SO4=)2 3+ Aluminum is present in its acidic form Al(H2O)63+ Not as Al3+ or Al(OH)4- This affects our approach to synthesis!

  14. All alums contain Aluminum • True • False

  15. All alums contain Aluminum B= False ALUMS: a CLASSof “double” sulfates in which one cation is monovalent and the second is trivalent. M(I) T(III) (SO4)2.12 H2O Aluminum is only one example of a trivalent cation. Others are Iron, Chromium, etc.

  16. SYNTHESIS Our objective is to synthesize the alum, potassium aluminum alum, KAl(SO4)2 12H2O SYNTHESIS: The process by which a desired substance (Product) is produced from other substances (Starting materials) by one or more chemical reactions reactions Starting Materials  Product An obvious starting material for the synthesis of alum is a source of aluminum, Al. What might the others be? It would be helpful to have the Al in solution. How shall we do that? We use the observation that aluminum metal dissolves in strong bases.

  17. Since the desired product, alum, contains potassium, we could use the strong base, KOH to dissolve Al. This would formAl(OH)4- However, to convert Al to the form it has in alum, Al(H2O)63+,we will need to acidify the solution once the Alhas dissolved. Which acid shall we use? • H2SO4would provide both: • an acid environment, and • the sulfate anion required to make the alum! • So, reasonable starting materials could be: • Al, KOH and H2SO4

  18. The gas produced when Aluminum metal dissolves in strong bases is: • CO2 • H2 • O2 • N2

  19. The gas produced when Aluminum metal dissolves in strong bases is: Aluminum metal reacts equally well with aqueous acids 2 Al (s) + 6 H+ + 6 H2O 2 Al(H2O)63+ + 3H2 (g) and bases 2 Al (s) + 2 OH- + 6 H2O 2 Al(OH)4- + 3H2 (g) BH2

  20. Does mixing the three substances, in the stoichiometric ratio (and the correct order), guarantee forming the desired product? Could other products be formed? Stoichiometry: the accounting system used to keep track of quantitative relationships between basic chemical entities. If so, can we optimize the formation of alum? If so, how will we recover the alum? • 1KOH, 1Al and 2 H2SO4 Suppose we have an aqueous solution containing 1 M K+ and 1 M Al3+ and 2 M SO4= If we remove water by evaporation, what substance will precipitate from the solution first? Alum? Al2(SO4)3? K2SO4? K1Al1(SO4)2•12H2O. The answer lies in the solubilities of these salts which, in turn, depend on the temperature of the solution.

  21. Let’s look at the solubilities of these substances K2SO4 Temperature (oC)

  22. Al2(SO4)3 K2SO4 Temperature (oC)

  23. Alum Alum is most soluble Alum is least soluble Al2(SO4)3 K2SO4 Temperature (oC)

  24. So, if we lowerthe temperature • from: • a temperature at which all three components are soluble • to: • a low temperature, say, 0o C, • the primary material to come out of solution will be the least soluble substance at the low temperature, • namely, the desired product -- Alum.

  25. Note, however, that even at 0oC, the solubility of alum is not 0.0 g/mL It is about 5 g/100 mL Pre-lab problem 3 addresses the issue of what may be left in solution. The answer is posted on the course website. So, we will wish to limit the amount of water when we cool the solution. We are now ready to specify a full synthetic pathway. Including the isolation of the desired product

  26. Dissolve a sample of aluminum in excess aqueous KOH. This makes Al the limiting reagent (so far) • 2 Al + 2 K+ + 2 OH- + 6 H2O 2 K+ + 2 Al(OH)4-+ 3 H2(g) 2 KOH 2 KAl(OH)4 2. Add excess H2SO4until the solution is acidic. This will: • neutralize KOH in excess of the amount needed to • dissolve the aluminum, and • convert the aluminum into its acidic form Excess = more than the stoichiometric amount • Al (OH)4- + 4 H+ + 2 H2O Al(H2O)63+ 3. Cool the resulting solution to 0o C so that alum will precipitate. 4. Separate the solid by filtration.

  27. PROCEDURE • Weigh approximately 0.5 g(500/27.0 = 18.5 mmol) of aluminum on the top loading balance • 2. Dissolve aluminum in 25 mL of 1.5 M KOH. • (1.5 X 25 = 37.5 mmol) KOH solutions are CORROSIVE 3. Warm solution gently on a hot plate until aluminum begins to dissolve (Hydrogen is being evolved) in the hood 4. Cool & Filter BY GRAVITY Keep Solution! It contains Al(OH)4- and excess KOH

  28. How to conduct a gravity filtration Ring Iron Ring Stand Triangle

  29. 5. When cool, add 10 mLof 9 M H2SO4 (10 X 9 = 90 mmol) S L O W L Y Sulfuric acid is VERY CORROSIVE H2SO4does several things Neutralizes excess KOH 2 K++2 OH-+SO4= + 2 H+2H2O+SO4=+2 K+ Converts Al(OH)4- to Al(OH)3 Al(OH)4- +SO4= + H+H2O+SO4= + Al(OH)3(s) Converts Al(OH)3to Al(H2O)63+ 2 Al(OH)3(s) + 6 H++3SO4= +6 H2O  2 Al(H2O)63 + +3 SO4= white precipitate forms white precipitate dissolves

  30. 6. Cool in ice bath. Be sure bath level is as high as solution. Let temperature become as low as possible 7. If crystals don’t form, reduce volume of solution by 10%by heating on hot plate and repeat 6. 8. Vacuum filter Keep Precipitate!

  31. How to conduct a Vacuum filtration Iron Ring Stand Clamp Holder Buchner Funnel Aspirator Extension Clamp Thick walled rubber tubing Filter Flask

  32. What is the minimum volume of 1.5 M KOH required to consume 0.40 g of aluminum (~15 mmol)? • 10 mL • 20 mL • 30 mL • Need to know the molar mass of KOH

  33. 2Al + 2KOH + 6 H2O  2 KAl(OH)4 + 3 H2(g) • 2Al + 2KOH-+ 6 H2O  2 KAl(OH)4- + 3 H2(g) 2 mmol of Al require 2mmol KOH 15 mmol of Al require 15 mmol KOH KOH is 1.5 M Y mL X 1.5 mmol / mL = 15mmol A 10 mL We use 25 mL Y =15 / 1.5 = 10 mL

  34. The Filter Flask The Buchner Funnel

  35. The Aspirator VACUUM WATER

  36. Filter Paper

  37. The temperature at which alum is crystallized should be ______to maximize the yield of product. • As low as possible • approximately equal to 70 oC • Above 70 oC

  38. Alum Alum is most soluble Alum is least soluble Al2(SO4)3 K2SO4 Temperature (oC)

  39. 9. Wash product carefully 10. Dry product in oven (Pay attention to schedule) 11. Weigh product and measure volume of filtrate

  40. WHY DO WE MEASURE VOLUME OF FILTRATE? Alum is soluble in water even at 0oC. Some alum will remain in solution after we filter the precipitate. Assuming solubility is 5.5 g/100 mL & measured volume allows calculation of maximum amount of alum that we cannot recover due to its remaining in solution. E.g., if we used a total volume of water (including the wash volume) of 35 mL, we can expect to have 35 mL X 5.5 g/100 mL = 1.9 g of alum still in solution

  41. CALCULATIONS - YIELD Stoichiometery of KAl(SO4)2.12 H2O K : Al : SO4 = 1 : 1 : 2 If we begin with 0.55g of Al 0.55g ------------------ 27 g / mol Al If allAlwere converted to alum, we would get 20 mmol alum Theoretical yield = 20 mmol X 474.3 mg/mmol = 9.5 X 103 mg = 9.5 g Molar mass of alum which you have calculated as part of the pre-lab = 20 mmolAl

  42. Suppose we produce only 7.6g of Alum. What is our percent yield? 100 X Actual yield Pct yield = --------------------- Theoretical yield 100 X 7.6 = ------------- = 80% 9.5 If you calculate a pct yield > 100%, your sample is probably not dry. Return it to the oven until the yield is under 100% YOUR PRODUCT, DATA SHEET & POST-LAB QUESTIONS are due at the end of the laboratory

  43. WHY WE COLLECT YOUR PRODUCT “Mercury, cinnabar, pyrites, alum of excellent quality, borax, black pepper - each one part,…… are to be kept in a stone bowl which is to be deposited in a heap of cow-dung. After one year, a liquid emerges out of it. This (liquid) is divine as well as flawless, and is to be compounded with mercury admixed with pure gold as ‘seed’.” Alum possesses the capability of transmuting a thousand times its weight of all metals intoVI, 241-245) gold (Rasopanisat, XVI, 241-245) A Medieval Indian alchemical text .

  44. Quiz 1 will be given during the first 15 min in Lab. It will cover the first three exercises: SUSB-003, SUSB-037, and SUSB-009 No Electronic Devices except Calculators are permitted.