Synthesis of Alum Prem Sattsangi Copyright 2009(Rev.)

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Presentation transcript:

Synthesis of Alum Prem Sattsangi Copyright 2009(Rev.)

#2 Synthesis of Alum Al    KAl(SO 4 ) 2.12H 2 O A Complex Salt FW = 474g/mol K 2 SO 4 + Al 2 (SO 4 ) 3  2KAl(SO 4 ) 2.12H 2 O Theoretical Yield: (BLB, p. 102) Limiting reagent (BLB, p. 100) Al (FW = 27.0) 1mol Al produces 1mol KAl(SO 4 ) 2.12H 2 O Calculate Theo. yield of Alum (g) from g Al. [Hint: Al(g)  Al (mol)  Alum (mol)  Alum(g)] 2 Simple salts

#3 Analytical Balance Tare, place foil, close door, record wt. Calculate Theoretical Yield Of Alum from g Al

#4 Cut Al foil in 16 squares Length wise Breadth wise

#5 Plastic Reaction vial with cut Aluminum foil

#6 Mark MP Bulb 1, 2, 3, cm b. Fill Micropipet with 1.5 M KOH to 2 cm mark c. Add to reaction vial a.

#7 Add more KOH (1cm) b. Add to reaction vial a. Fill KOH to 1 cm mark

#8 Start reaction, In fume hood, Label 250 mL beaker with Stn. # (a) Take hot tap water up to 2cm mark in the coffee cup. (c) FLOAT reaction vial in the coffee cup using forceps. (b) Support the coffee cup in 250 mL beaker (Stn.#__) STN# 4

#9 Oxidation Numbers HELP KEEP TRACK OF ELECTRONS Ox.No. Aluminum (foil) 0 Aluminum ion (Al 3+ ) +3 Loss of Electron = Oxidation Hydrogen (H 2 (g) 0 Hydrogen ion (H + ) +1 Hydride ion (H: - ) -1 Gain of electron = Reduction H in water [H  OH - ] +1 BLB p Ox.No. of elements in standard states = 0 Ox.No. of an ION = Charge on it LEO the lion says GER Al foil

#10 Allow REDOX reaction to finish ELECTRONS GAINED = ELECTRONS LOST Oxidation: (Loss of Electrons) Each electron lost = 1+ charge 2 Al  2 Al e - Aluminum [0 to +6 ] Reduction: (Gain of Electrons) 6 [ H  OH - ] + 6e -  3 H OH - Water [ +6 to 0 ] Hydrogen gas forming (FLAMMABLE) 2Al(s) + 2KOH(aq) + 6HOH(l)  2KAl(OH) 4 (aq) + 3H 2 (g) [Ox.Nos./atom or ion] 0 6 x +1= +6 2x+3 = +6 0

#11 KAl(OH) 4 (aq) Complex Finished reaction AMPHOTERISM: (BLB p ) KOH(aq) + Al(OH) 3 (s)  KAl(OH) 4 (aq) Two Simple hydroxides ADD, form a SOLUBLE complex. Black particles are impurities in “Al”

#12 Remove Black Particles Insert a wad of cotton in a glass pipet. clamp to a stand, wet, remove water, filter. May need 2 times filtration on the same pipet to get a clear filtrate. KAl(OH) 4 (aq)

#13 To KAl(OH) 4 (aq) colorless, Add Sulfuric acid  White ppt. Al(OH) 3 (s) Add 1/3 MP H 2 SO 4 Stir with a glass -rod REMEMBER: 2KOH (aq) + 2Al(OH) 3 (s)  2KAl(OH) 4 (aq) Colorless #10 Base + Acid  Salt + water [2KOH (aq) + H 2 SO 4 (aq)  K 2 SO 4 (aq) + 2HOH] A. 2KAl(OH) 4 (aq) + H 2 SO 4 (aq)  K 2 SO 4 (aq) + 2Al(OH) 3 (s) + 2HOH B. 2Al(OH) 3 (s) + 3H 2 SO 4 (aq)  Al 2 (SO 4 ) 3 (aq) + 6HOH Soluble, More reactive Colorless Solution  White ppt.

#14 White precipitate dissolves. Alum forms. (b)Heat on hot plate K 2 SO 4 (aq) + Al 2 (SO 4 ) 3 (aq)  2KAl(SO 4 ) 2.12H 2 O Simple salts ADD, form a COMPLEX salt (a)Stir with a glass rod.

#15 Crystallization of Alum Room temperature (15 minutes or more) SEEDING? Ice bath (10 minutes)

#16 Alum crystals Single Typical result

#17 Suction Filter Alum Water aspirator [Removes Mother liquor] Isolation of Alum: (a) Suction filter. (b) Rinse 2x with 15 drs 50% ethanol(0 o C) (c) Rinse 2x with 95% ethanol(0 o C) (d)Continue suction ~5 more minutes to dry the crystals.

#18 Suction Filter Alum (a)Water aspirator [Removes Mother liquor] (b) 50% ethanol [Removes Al 3+ (aq), K + (aq), SO 4 2- (aq)] (c) 95% ethanol. [Removes water] (d) Continue suction ~5 minutes. [To air dry the crystals]

#19 Dry on a pre-weighed paper Transfer Alum on PAPER. Cover with a Funnel. Allow to dry. Weight of paper_____ g

#20 Weigh crystals, Calculate %Yield Wt of crystals + paper = g Wt of paper = – g Wt of crystals = g Exp. Yield = g %Yield = Exp. Yield x 100 Theo. Yield (Slide #2) Please calculate %Yield now: =

#21 Summary of Equations (i) REDOX (Change in oxidation no.) 2Al(s) + 2KOH(aq) + 6HOH(l)  2KAl(OH) 4 (aq) + 3H 2 (g) (ii) Complex (Add the reagents) Al(OH) 3 (s) + KOH(aq)  KAl(OH) 4 (aq) K 2 SO 4 (aq) + Al 2 (SO 4 ) 3 (aq)  2 KAl (SO 4 ) 2.12H 2 O (iii) Acid-Base (Water is formed) 2KAl(OH) 4 (aq) + H 2 SO 4 (aq)  K 2 SO 4 (aq) + Al(OH) 3 (s) + 2HOH 2Al(OH) 3 (s) + 3H 2 SO 4 (aq)  Al 2 (SO 4 ) 3 (aq) + 3H 2 O

#22 Qualitative Analysis of Alum KAl(SO 4 ) 2.12H 2 O 1. Test for pH, pink color (Al 3+ ion) 2. Test for SO 4 2- ion, [with BaCl 2 (aq)  white ppt) 3. Flame test for K + ion (LILAC COLOR) and evolution of H 2 O(g) and SO 3 (g) [PUFFS UP]. 4. Test for absence of sulfate, [with BaCl 2 (aq)] (after THOROUGHLY heating the alum). 5. Test for presence of Al 3+, 0.15 M KOH.  gelatinous ppt Al(OH) 3 (s). Add 1.5 M KOH to ppt  dissolves, Al(OH) 4 (aq), Forms SOLUBLE Complex

#23 Prepare saturated Alum solution. Take 2 mm mound of Alum, KAl(SO 4 ) 2.12H 2 O, with 3 drops of water on the plastic sheet. Crush alum with a glass rod.

#24 pH Test for Al 3+ Alum, KAl(SO 4 ) 2.12H 2 O, contains Al 3+ ion. Al 3+ ion has: a. High charge. b. Small radius. Such cations, may rip apart the water molecule to produce H +. Al 3+ + HOH  AlOH 2+ + H +

#25 Chemical Test for Al M KOH  Gel (soluble in 1.5 M KOH) In the 24 well tray, mix 1 dr. 1.5 M KOH + 9 drs. Distilled water to make 0.15M KOH Alum contains Al drop Alum(aq) + 1 drop 0.15 M KOH  Gel Al KOH(0.15M)  Al(OH) 3 (gel) + 3K + Add 1 drop 1.5 M KOH to gel  Dissolves. Al(OH) 3 (gel) +KOH(1.5 M)  KAl(OH) 4 (aq) Simple hydroxides ADD  COMPLEX Insol(s)

#26 Test for SO 4 2- ion in Alum KAl(SO 4 ) 2.12H 2 O 1 drop Alum(aq) + 1 drop BaCl 2 (aq)  white ppt, BaSO 4 (s) Sulfate present. SO 4 2- (aq) + BaCl 2 (aq)  BaSO 4 (s) + 2 Cl - (aq) Alum(aq)

#27 Flame test sample prep. Place some Alum on a watch glass. Heat nail. Touch Alum with Red Hot Nail

#28 Evolution of gases and Lilac flame, K + ion. Slowly bring close to the flame. Watch it swell as gases H 2 O(g) and SO 3 (g) evolve. Bring it into the flame. Lilac color, “K + ” ion.

Alum 2KAl(SO 4 ) 2 12H 2 O 24H 2 O (g) 2SO 3 (g) Al 2 O 3 (s) insoluble K 2 O (s) soluble + H 2 O 2KOH #29 Heating Alum (Chemistry) Base HEAT

#30 Tests in heated Alum sample 2KAl(SO 4 ) 2.12H 2 O  24H 2 O(g) + 4SO 3 (g) + Al 2 O 3 (s) + K 2 O(s) Crush it in a ~2 cm pool of distilled water. Remove Alum RESIDUE

#31 Tests with HEATED Alum sample Check pH, Basic. WHY? K 2 O(s) + H 2 O(g)  2KOH(aq) Filter over cotton. Test with BaCl 2 (aq) No SO 4 2-, No ppt. CAUTION: Any Un- decomposed alum may give positive test with BaCl 2.

Acknowledgements Experiments: Ms. Virginia Roll Power Point Assistance: Mr. Bill Hager Christopher Byers