1 SYNTHESIS OF AN ALUM M(I)T(III)(SO 4 ) 2. 12 H 2 O Potash (KOH) alum oil of vitriol (H 2 SO 4 ) 2.

SYNTHESIS OF AN ALUM M(I)T(III)(SO 4 ) 2. 12 H 2 O Potash (KOH) alum oil of vitriol (H 2 SO 4 ) 2

Examine how chemists plan and execute synthesis of desired substances OBJECTIVES Explore factors that contribute to quantity & purity of synthesized substances Learn how a desired substance is recovered from a mixture. Observe some chemical properties of aluminum and its compounds 3

Concepts: Actual YieldByproducts ConditionsLimiting Reagent Percent YieldProducts Reactants/Starting Materials ReactionsSeparation Process Synthetic PathwayTarget Theoretical YieldVerified Purity 4

Techniques: Gravity FiltrationVacuum Filtration Crystallization Quantitative Transfer Apparatus: Funnel Filter Paper Filter Flask Buchner Funnel Ice Bath Aspirator 5

cation ALUMS: a class of double sulfates in which one cation is monovalent and the second is trivalentBACKGROUND The most common alum is: Potassium Aluminum Sulfate, K Al (SO 4 ) 2.12 H 2 O. often called Alum M(I) M(I) represents monovalent cation e.g., K +, Na +, NH 4 +, etc. T(III) represents trivalent cation e.g., Al 3+, Fe 3+, Cr 3+, etc. 12 - number of molecules of water of crystallization - (dodeca hydrate) T(III)(SO 4 ) 2. 12 H 2 O cat cation: a positively charged ion; attracted to cathodes A monovalent cation is a positively charged ion with a charge of +1 A trivalent cation is a positively charged ion with a charge of +3 6

Natural Alum Deposit in Australia 7

Water purification Medicinal Food additive (pickling) Fabrics (mordant/marbling) Paper processing Filler Uses of Alum BUT 1500 B.C. Styptic pencil 8

Allom "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. An unusual use for Alum In addition to a cure for freckles, Wirzung discovered the pancreatic duct 9

ALUMINUM CHEMISTRY HHe LiBeBCNOFNe NaMgAlSiPSClAr KCaGaGeAsSeBrKr 10 Position of Al in periodic table suggests it may have interesting properties

ALUMINUM CHEMISTRY HHe LiBeBCNOFNe NaMgAlSiPSClAr KCaGaGeAsSeBrKr 11 Position of Al in periodic table suggests it may have interesting properties

ALUMINUM CHEMISTRY HHe LiBeBCNOFNe NaMgAlSiPSClAr KCaGaGeAsSeBrKr Position of Al in periodic table suggests interesting properties Aluminum metal reacts equally well with aqueous acids 2 Al (s) + 6 H + + 6 H 2 O 2 Al(H 2 O) 6 3+ + 3H 2 (g) and bases 2 Al (s) + 2 OH - + 6 H 2 O 2 Al(OH) 4 - + 3H 2 (g) Aluminum also reacts with atmospheric oxygen to form stable oxide binds to metal & protects it from further oxidation 12

In aqueous solution, Al 3+ is a moderately strong acid: In strongly basic solutions, Al 3+ 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, Al 2 O 3.nH 2 O. Al (H 2 O) 6 3+ Al (H 2 O) 5 (OH) 2+ + H + These compounds are very insoluble, gelatinous (colloidal) substances with large surface areas. 13

Potassium Aluminum Alum We have written the formula as K Al (SO 4 ) 2. 12 H 2 O, more accurate description of the substance is Aluminum is present in its acidic form Al(H 2 O) 6 3+ Not as Al 3+ or Al(OH) 4 - This affects our approach to synthesis! K(H 2 O) 6 + Al(H 2 O) 6 3+ (SO 4 = ) 2 3+3+ octahedron 14

All alums contain Aluminum A.True B.False 15

B = False CLASS tri ALUMS: a CLASS of double sulfates in which one cation is monovalent and the second is trivalent. M(I)T(III)(SO 4 ) 2. 12 H 2 O All alums contain Aluminum Aluminum is only one example of a trivalent cation. Others are Iron, Chromium, etc. 16

Our objective is to synthesize the alum, potassium aluminum alum, KAl(SO 4 ) 2 12H 2 O SYNTHESIS Product Starting materials reactions SYNTHESIS: The process by which a desired substance (Product) is produced from other substances (Starting materials) by one or more chemical reactions We use the observation that aluminum metal dissolves in strong bases. An obvious starting material for the synthesis of alum is a source of aluminum, Al. What might the others be? Starting Materials reactions Product 17 It would be helpful to have the Al in solution. How shall we do that?

However, to convert Al to the form it has in alum, Al(H 2 O) 6 3+, we will need to acidify the solution once the Al has dissolved. H 2 SO 4 would provide both: an acid environment, and the sulfate anion required to make the alum! So, reasonable starting materials could be: Al, KOH and H 2 SO 4 Since the desired product, alum, contains potassium, we could use the strong base, KOH to dissolve Al. This would form Al(OH) 4 - 18 Which acid shall we use?

The gas produced when Aluminum metal dissolves in strong bases is: A. CO 2 B. H 2 C. O 2 D. N 2 19

The gas produced when Aluminum metal dissolves in strong bases is: B H 2 Aluminum metal reacts equally well with aqueous acids 2 Al (s) + 6 H + + 6 H 2 O 2 Al(H 2 O) 6 3+ + 3H 2 (g) and bases 2 Al (s) + 2 OH - + 6 H 2 O 2 Al(OH) 4 - + 3H 2 (g) 20

Suppose we have an aqueous solution containing If we remove water by evaporation, what substance will precipitate from the solution first? The answer lies in the solubilities of these salts which, in turn, depend on the temperature of the solution. Does mixing the three substances, K 2 SO 4 ?Alum?Al 2 (SO 4 ) 3 ? 1 M K + and 1 M Al 3+ and 2 M SO 4 = 1 KOH, 1 Al and 2 H 2 SO 4 K 1 Al 1 (SO 4 ) 2 12 H 2 O. Stoichiometry: the accounting system used to keep track of quantitative relationships between basic chemical entities. guarantee forming the desired product? in the stoichiometric ratio (and the correct order), Could other products be formed? If so, how will we recover the alum? If so, can we optimize the formation of alum? 21

Temperature ( o C) K 2 SO 4 Lets look at the solubilities of these substances 22

Temperature ( o C) K 2 SO 4 Al 2 (SO 4 ) 3 23

Temperature ( o C) K 2 SO 4 Al 2 (SO 4 ) 3 Alum Alum is most soluble Alum is least soluble 24

So, if we lower the temperature from: a temperature at which all three components are soluble to: a low temperature, say, 0 o 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 0 o C, the solubility of alum is not 0.0 g/mL We are now ready to specify a full synthetic pathway. isolation Including the isolation of the desired product It is about 5 g/100 mL So, we will wish to limit the amount of water when we cool the solution. 26 Pre-lab problem 3 addresses the issue of what may be left in solution. The answer is posted on the course website.

1.Dissolve a sample of aluminum in excess aqueous KOH. 2. Add excess H 2 SO 4 until the solution is acidic. This will: This makes Al the limiting reagent (so far) neutralize KOH in excess of the amount needed to dissolve the aluminum, and convert the aluminum into its acidic form Al (OH) 4 - + 4 H + + 2 H 2 O Al(H 2 O) 6 3+ 3. Cool the resulting solution to 0 o C so that alum will precipitate. Excess = more than the stoichiometric amount 4. Separate the solid by filtration. 2 Al + 2 K + + 2 OH - + 6 H 2 O 2 K + + 2 Al(OH) 4 - + 3 H 2 (g)2 KAl(OH) 4 2 KOH 27

1.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 PROCEDURE 3. Warm solution gently on a hot plate until aluminum begins to dissolve (Hydrogen is being evolved) 4. Cool & Filter BY GRAVITY Keep Solution! It contains Al(OH) 4 - and excess KOH in the hood 28

Iron Ring Stand Ring Triangle 29 How to conduct a gravity filtration

H 2 SO 4 does several things Neutralizes excess KOH 2 K + + 2 OH - + SO 4 = + 2 H + 2H 2 O + SO 4 = + 2 K + Converts Al(OH) 4 - to Al(OH) 3 Al(OH) 4 - +SO 4 = + H + H 2 O + SO 4 = + Al(OH) 3 (s) Converts Al(OH) 3 to Al(H 2 O) 6 3+ 2 Al(OH) 3 (s) + 6 H + + 3SO 4 = + 6 H 2 O 2 Al(H 2 O) 6 3 + + 3 SO 4 = 5. When cool, add 10 mL of 9 M H 2 SO 4 (10 X 9 = 90 mmol) white precipitate forms white precipitate dissolves Sulfuric acid is VERY CORROSIVE S L O W L Y 31

8. Vacuum filter Keep Precipitate! 6. Cool in ice bath. Be sure bath level is as high as solution. Let temperature become as low as possible 7. If crystals dont form, reduce volume of solution by 10% by heating on hot plate and repeat 6. 32

Iron Ring Stand Thick walled Thick walled rubber tubing Aspirator Extension Clamp Clamp Holder Filter Flask Buchner Funnel 33 How to conduct a Vacuum filtration

A.10 mL B.20 mL C.30 mL D.Need to know the molar mass of KOH What is the minimum volume of 1.5 M KOH required to consume 0.40 g of aluminum (~15 mmol)? 34

2 Al + 2 KOH + 6 H 2 O 2 KAl(OH) 4 + 3 H 2 (g) 2 mmol of Al require 2 mmol KOH 15 mmol of Al require 15 mmol KOH Y mL X 1.5 mmol / mL = 15 mmol Y = 15 / 1.5 = 10 mL We use 25 mL KOH is 1.5 M A A 10 mL 2 Al + 2 KOH - + 6 H 2 O 2 KAl(OH) 4 - + 3 H 2 (g) 35

The Buchner Funnel The Filter Flask 36

WATER VACUUM The Aspirator 37

Filter Paper 39

The temperature at which alum is crystallized should be ______ to maximize the yield of product. A.As low as possible B.approximately equal to 70 o C C.Above 70 o C 40

Temperature ( o C) K 2 SO 4 Al 2 (SO 4 ) 3 Alum Alum is most soluble Alum is least soluble 41

9. Wash product carefully filtrate 11. Weigh product and measure volume of filtrate 10. Dry product in oven (Pay attention to schedule) 42

WHY DO WE MEASURE VOLUME OF FILTRATE? Alum is soluble in water even at 0 o C. Some alum will remain in solution after we filter the precipitate. 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 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. 43

Stoichiometery of KAl(SO 4 ) 2.12 H 2 O K : Al : SO 4 = 1 : 1 : 2 If we begin with 0.55 g of Al 0.55 g ------------------ 27 g / mol Al If all Al were converted to alum, we would get 20 mmol alum Theoretical yield = 20 mmol X 474.3 mg/mmol = 9.5 X 10 3 mg = 9.5 g CALCULATIONS - YIELD Molar mass of alum which you have calculated as part of the pre-lab = 20 mmol Al 44

YOUR PRODUCT, DATA SHEET & POST-LAB QUESTIONS are due at the end of the laboratory 100 X Actual yield Pct yield = --------------------- Theoretical yield Suppose we produce only 7.6 g of Alum. What is our percent 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% 45

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. WHY WE COLLECT YOUR PRODUCT Alum possesses the capability of transmuting a thousand times its weight of all metals intoVI, 241-245) gold (Rasopanisat, XVI, 241-245) http://ignca.nic.in/ps_04014.htm http://ignca.nic.in/ps_04014.htm A Medieval Indian alchemical text. 46

in Lab. 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. 47

NEXT LECTURE IDENTIFICATION OF HOUSEHOLD CHEMICALS READ SUSB-004 & DO PRE-LAB A Test Exercise 48

1 SYNTHESIS OF AN ALUM M(I)T(III)(SO 4 ) 2. 12 H 2 O Potash (KOH) alum oil of vitriol (H 2 SO 4 ) 2.

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