12 minutes reading time (2477 words)


    At the 1991 National Chemical Education Conference in Perth, participants were invited to carry out a demonstration. The following is a list of demonstrations carried out at the conference.


    Brief Description (what the audience sees)
    Using a red solution of oxyhaemoglobin, prepared from red blood cells, the reversible formation of violet deoxyhaemoglobin is demonstrated. Then cigarette smoke is bubbled through the solution forming red carboxyhaemoglobin; and nitrite (from bacon) and vitamin C is added to another solution to form red nitroxyhaemoglobin. We ask the audience to suggest an experiment to show that oxygen has been replaced in the latter two reactions.

    A specially modified overhead projector will be used to project the image of the solutions onto a screen.

    Chemicals Required

    • One sachet of reconstituted red blood cells (from the local Blood Bank)
    • 250mL 0.4M NH3
    • 5g FeSO4 (ground to a powder)
    • 5g tartaric acid
    • 1 packet of 20 cigarettes
    • 5g NaS2O4 (sodium dithionite)
    • 20g bacon and 20g ham
    • 5g NaNO2 (sodium nitrite)
    • 5g ascorbic acid

    Equipment Required
    • Special Overhead Projector adapted for "vertical demonstrations"
    • "Cigarette Smoking Simulator" - both items provided by us!
    • Normal overhead projector
    • Two projection screens
    • Approx. 2m by 2m table
    • Plastic gloves, paper towel

    1. A sample of red blood cells are lysed in aqueous ammonia to form oxyhaemoglobin.
    2. Deoxyhaemoglobin is produced reversibly, using freshly prepared Stokes Reagent (FeSO4 and tartaric acid) and shaking in air, and irreversibly, using sodium dithionite.
    3. The smoke of cigarettes is bubbled through oxyhaemoglobin solution.
    4. Sodium nitrite and ascorbic acid are added to another oxyhaemoglobin solution.
    5. Follow audience suggestion.

    Once the red blood cells (handled with gloves) are added to the ammonia the haemoglobin solution is safe. Apart from the cigarette smoke, which is contained, the demonstration is not hazardous.

    Chemical Principles

    • Le Chatelier's Principle
    • Importance of molecular size and shape in chemical recognition
    • Redox reactions with food additives

    Chem Matters Feb 84

    Submitted by
    Dr Roy Tasker, Faculty of Science and Technology, University of Western Sydney, Nepean, PO Box 10, Kingswood, NSW 2747, and
    Dr Harkirat S Dhindsa, Cumberland College of Health Science, The University of Sydney, PO Box 170, Lidcombe, NSW 2141

    A. Thermochromic Solids (1)
    A reversible colour change occurs on heating and cooling salts of the form M2HgI4 (M = Ag (I), Cu (I) )

    Preparation (M = Ag (I) )

    1. Dissolve 3.25g of Hg (NO3)2 in boiling water. Add 10% Kl solution until the initial precipitate of HgI2 dissolves to give a clear solution.
    2. Add 50 mL of a solution of 3.4g of AgNO3 and boil for a few minutes. Filter off the yellow Ag2HgI4 and dry in a desiccator.
    Preparation (M = Cu (I) )
    1. Prepare a solution of Cu (I) by adding 25 mL of a solution of 2.5g CuSO4 to a boiling solution of 6g Na2SO3.7H2O + 5g NaCl in 50mL of water. A greenish precipitate forms initially, but redissolves to give a yellow solution.
    2. Dissolve 1.6g of Hg (NO3)2 in 25 mL of boiling water. Form the HgI42- ion as in the Ag (I) prep, using 10% KI. To this solution, add the hot Cu (I) solution. A deep purple solid precipitates, which changes colour to red on filtering.

    The solids change colour (yellow → orange for Ag (I) and red → purple for Cu (I) ) when heated - smearing some of the solution on a filter paper and heating with a hot plate or hair dryer is a convenient way to demonstrate the colour change. The change is reversible, and can be repeated many times. Both solids are quite stable on storage.

    The colour change is due to a subtle change in structure. The β or low temperature form is tetragonal, with tetrahedral HgI42- units occupying the corners of the unit cell but th€e M+ ions occupy only 4 out of 6 face-centre positions (see diagram). In the (cubic) a and form -(high T) all 6 faces, as well as the corners are occupied at random by M or Hg in the ratio 2:1. In both α and β forms the positions of the iodides are fixed (cubic close packed). 

     Another interesting thermochromic solid is (C2H5NH2)2CuCl4 (2). This is a salt of CuCl42-, which is green at room temperature but becomes orange when heated. This change is also reversible. The green form is square-planar, while in the orange form the CuCl42- ion forms a distorted tetrahedron.

    B. Mercury Tornado (3).

    A colourless solution is rapidly stirred on a mechanical stirrer. Addition of a drop of colourless liquid causes an orange 'tornado' to appear and gradually dissipate.

    Solution A: 2g Hg (NO3)2/100mL of 0.1M HNO3

    Solution B: 16g KI/100mLof water.

    Place a large beaker with about 700mL of water on a mechanical stirrer. Add 7mL of A and mix. Then add B dropwise. An orange precipitate forms, and after a few mLs the precipitate persists. Continue adding until the solution becomes clear again, then add an extra mL of B. If the solution is stirred rapidly, to create a vortex, adding a drop or two of A the vortex will cause an orange tornado, to appear and gradually dissipate. If, after several additions of A, the solution turns orange, adding a mL of B will restore the original clear solution.

    The initial orange precipitate is the kinetically favoured product, HgI2 (whiqh is orange and insoluble) giving way to the thermodynamically favoured HgI42- ion (colourless and soluble).

    C. "Old Nassau" Reaction (4,5,6).
    When three colourless solutions are mixed, the mixture suddenly turns orange after a few seconds. After a few more seconds the solution suddenly turns black. By varying proportions of the solutions other sequences can be produced.

    Solution A: 15g KIO3/L

    Solution B: 15g NaHSO3/L + few mLs starch solution

    Solution C: 3g HgCl2/L

    Mix equal volumes of A and B. Solution turns black. Dilution delays colour development. Mix equal volumes of C + B + A (in that order). Solution turns orange, then black. Mix C + A + 2B. Colour sequence is clear, orange, clear. The first reaction is just the iodine clock (Landolt) reaction (steps 1 to 3 in the scheme below). In the second reaction the iodine being slowly formed in step 1 reacts quickly to form the insoluble orange HgI2, (step 4). Only when all the Hg2+ is consumed does I2 form via step 2, and hence the iodine-starch colour forms and masks the orange colour. In the third reaction, enough HSO32- is added to reduce all the I2 formed (via step 3).


    1. IO3- + 3HSO3- → I- + 3SO42- + 3H+
    2. 5I- + IO3- + 6H+ → 3I2 + 3H2O
    3. I2 + HSO3- → 2I- + SO42- + 3H+
    4. Hg2+ + 2I- → HgI2 (orange)
    5. I2 + starch → blue-black complex

    1. Palmer, W.G,, "Experimental Inorganic Chemistry", p. 192
    2. Van Oort, M.J.M., J. Chem. Ed.,65,84 (1988)
    3. Lippincott, W.T., "Source Book for Chemistry Teachers", p.7
    4. Alyea, H. N., J. Chem. Ed., 54, 167 (1977)
    5. Moss, A., J. Chem. Ed., 55, 244 (1979)
    6. Lambert, J.L., Fine, G.T., J. Chem. Ed.,61, 1037 (1984)

    Submitted by
    J. Hughes
    Department of Applied Chemistry
    R.M.I.T., Melbourne.

    Brief Description (what the audience sees)
    Very spectacular and quick. In a large cylinder, on addition of a small amount of potassium dichromate a purple foam grows and pours out of the cylinder and then changes back to orange.

    Chemicals required (specify formulas, concentrations, quantities)

    • 50mLs of 100 vol H2O2 Hydrogen peroxide
    • Squirt of detergent
    • Spatula of potassium dichromate

    Equipment required
    • 250mL measuring cylinder
    • Spatula
    • Large tray
    • 50mL 100 vol H2O2
    • Dishwashing detergent
    • Potassium dichromate

    Place a generous squirt of detergent in the bottom of a cylinder. Add 50mL of H2O2 and mix well. Add a spatula load of potassium dichromate and leave in a tray to catch overflow. Hot foam develops quickly.

    Protect benches with a tray to catch foam. Perform in a well ventilated area as the hot detergent small needs to be removed.

    Chemical Principles
    Catalysts and Exothermic Reactions.
    H2O2(aq) + H2O2(aq) → 2H2O(l) + O2(g)

    The Australian Science Teachers Journal; May 1990 Vol 36 No 2.
    Marlene Bevan and Chris Commons
    Scotch College

    Submitted by:
    Rosemary Phillips
    BARDON QLD 4065

    Brief Description
    In a large cylinder a solution initially colourless changes through yellow to dark blue then back to colourless. This will continue for 15-20 minutes.

    Chemicals required

    • 13g Potassium iodate (KIO3)
    • 1g Manganese sulphate-1-hydrate MnSO4.1H2O
    • 5g Malonic acid CH2(COOH)2
    • 2.5g vitex
    • 3mL conc sulphuric acid H2SO4
    • 70mL of 100 vol H2O2

    Equipment required
    • 1 x 500mL Measuring Cylinder
    • 1 x 1L beaker
    • 1 x 500mL beaker
    • 1 x 100mL Measuring Cylinder
    • 1 x 10mL Measuring-Cylinder

    To 850mL of water add 13g KIO3, 1g MnSO4.1H2O, 5g malonic acid, 2.5g vitex and 3mL of conc H2SO4. Quickly transfer mixture to a measuring cylinder or tube.

    Handle all chemicals with care, particularly the conc H2SO4.

    Chemical principles
    This is a particularly entertaining demonstration and is enhanced by the use of a long tube. The mechanism of the reaction is quite complex and not fully understood.

    1. H+ + KIO3 → HIO3 + K+
    2. 5H2O2 + 2HIO3 → I2 + 6H2O + 5O2(g)
    3. I2 + CH2(COOH)2 → ICH(COOH)2 + I- + H+
    4. I2 + ICH(COOH)2 → I2C(COOH)2 + I- + H+
    5. I2 + H2O2 → 2HIO
    6. HIO + H2O2 → I- + O2 + H+ + H2O
    As I2 is generated (2) the blue colour is formed with vitex and as it is reduced to I- the blue colour disappears.

    I have used this demonstration for years and have lost the original reference but it was photocopied from another teacher originally and has always proved fascinating for students and parents at demonstrations.

    Submitted by:
    Rosemary Phillips
    BARDON QLD 4065

    Brief description (what the audience sees)
    A bottle of colourless liquid which when shaken turns deep blue. On standing the solution fades and becomes colourless again. The procedure can be repeated many times. Students become involved in reminding the demonstrator that another 'shake' is needed.

    Chemicals required (specify formulas, concentrations, quantities)

    • Glucose - 12 g
    • Sodium hydroxide - 12g
    • Methylene blue - 1 mL of 1% solution

    Equipment required
    • One litre conical or volumetric flask.

    • Dissolve glucose in 900 mL water. 1 hour before demonstration, add the sodium hydroxide.
    • Then add 1mL of methylene blue solution or enough to give a deep blue colour.
    • Stopper the bottle and leave to stand. The blue colour will fade.
    • Commence demonstration by shaking the flask or by pouring the solution from the flask into a beaker - from a height which will enable the mixture to agitate and dissolve air.
    For comparison the Colourful Cycle can also be presented. The chemicals required are identical but Indigo Carmine - 50 mL of 1% alcoholic solution, is used as the indicator.

    The colour change involved here is from -

    Lime Green → Red-Orange → Yellow

    The solution on standing becomes yellow. On shaking it becomes red-orange.

    By doing both demonstrations in the same presentation you can also discuss the use of indicators and their suitability for an experiment.

    No special precautions.

    Chemical Principles

    • Redox reaction
    • Colour and oxidation state
    • Reducing action of glucose
    • Oxygen solubility in water - leads onto the discussion of why fish can live in water, biological oxygen demand etc.
    Chemlcal Equations
    Me = Colourless Methylene Blue
    MeBl = Blue Methylene Blue
    1. O2(g) → O2(aq)
    2. O2(aq) + Me(aq) → MeBl(aq)
    3. C6H12O6(aq) + OH-(aq) ↔ C6H11O6-(aq) + H2O
    4. C6H11O6(aq) + MeBl(aq) → Me(aq) + OH-(aq) + C6H12O6(aq)

    Chemistry Demonstrations, Tony Sperring. Published by the Science Teachers' Association of New South Wales, 1990.

    Submitted by:
    Mrs Nola Shoring
    COOK ACT 2614

    Brief description (what the audience sees)
    Turn copper into a silver metal
    Turn copper into a gold metal.

    Chemicals required (specify formulas, concentrations, quantities)

    1. Mercury (drop), conc HNO3 + water
    2. Zn dust (5g) 50mL 6M NaOH

    Equipment required
    • Beaker (100mL)
    • Evaporating Dish
    • Long Tweezers
    • Bunsen/Mat
    • Brillo Pad
    • Paper Towels

    1. Dissolve drop of Hg in 25mL conc HNO3, dilute a little. Place copper foil in mixture for 5-10 minutes
    2. Dissolve Zn in hot NaoH. Place copper metal in mixture for 3-4 minutes. Remove, dry and heat in Bunsen for a few seconds.

    This experiment is quite safe to perform in an open lab. (Some NO2 fumes on dissolving Hg)

    Chemical principles

    1. Activity series (redox) Cu(s) + Hg2+(aq) → Hg(l) + Cu2+
    2. Amphoteric nature of Zinc → Na2ZnO2 solution.

    Tested Demonstrations in Chemistry (Algea & Dutton)

    Submitted by
    Arthur J Davies
    Salisbury Campus, University of SA
    Smith Road

    Brief Description (what the audience sees)
    The demonstration illustrates the different diffusion rates of hydrogen, air and carbon dioxide. A porous pot attached to a U tube manometer is first immersed in an atmosphere of hydrogen. The audience sees that the pressure inside the pot increases. The atmosphere of hydrogen is removed and the pressure inside the pot then falls below atmospheric.

    This demonstrations may be repeated with carbon dioxide which produces an opposite, but less marked effect.

    Chemicals required (specify formulas, concentrations, quantities)

    • Cylinder of hydrogen with 2m of delivery tube attached
    • Cylinder of carbon dioxide with 2m of delivery tube attached.
    (Gases may be industrial quality. Purity is immaterial.)

    Equipment required

    • One 2L beaker
    • One 1L beaker
    • One large U tube manometer with coloured water (the one that I use has arms that are 12mm diameter and about 900 mm long.
    • One porous pot with one hole stopper fitted with 10cm glass tube.
    • One metre rubber delivery tube to connect manometer to pot.

    Submitted by
    David Clift
    Victoria College
    8. WHY ARE SOLUTIONS OF IRON (III) SALTS YELLOW?Why it is that hydrated iron (III) nitrate is a pale grey-mauve colour and yet if we dissolve iron (III) nitrate in water, the solution is yellow? The simple answer is that the Fe3+ ion is yellow. But what does this mean? We can simply demonstrate (but perhaps not prove) that the yellow colour is not due to the hexaqua ion Fe(H2O)63+, but to hydroxy species present as a result of reactions that can be represented by:
    1. [Fe(H2O)6]3+ ⇌ [Fe(H2O)5(OH)]2+ + H+(aq)
    2. [Fe(H2O)5(OH)]2+ ⇌ [Fe(H2O)4(OH)2]+ + H+(aq)

    The Demonstration
    1. Dissolve some iron(III) nitrate in water. The resulting solution is yellow.
    2. Add a little nitric acid solution and the solution becomes colourless.

    The Chemistry
    Yellow hydroxy species are neutralised by the addition of acid - i.e. the reverse of reactions (1) and (2) to form the colourless hexaqua ion. Solid Fe(NO3)3.9H2O contains the hexaqua ion.

    Don't use iron(III) chloride to make the solution and don't acidify with hydrochloric acid, because the chloro-substituted species are bright yellow and the more hydrochloric acid that you add, the more yellow the solution will become.

    Additional Note
    The solution made by dissolving iron(III) salts in water can be shown (pH meter) to be quite acidic (pH 1.5-2). Reaction (1) makes Fe(H2O)63+ a stronger weak acid than acetic acid.

    Submitted by
    Dr R B Bucat
    School of Chemistry
    University of Western Australia 

    The Chemistry of Acid Mine Drainage
    Radioactivity and the Environment


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