9 minutes reading time (1884 words)

    A COLOURFUL DEMONSTRATION OF THE HENDERSON-HASSELBALCH EQUATION

    M.A. HOOPER
    School of Chemistry,
    University of Sydney,
    NSW,2006

    The reaction of dry ice, CO2(s), with water or dilute sodium hydroxide solution has long been used to demonstrate the colours of universal indicator. The bubbling and clouds of condensing water vapour make it popular with students. By increasing the concentration of the sodium hydroxide solution stepwise by factors of ten, the ratio of weak acid to conjugate base in the equilibrium solution varies stepwise by factors of ten, and the pH of the equilibrium mixture varies stepwise by one unit. All the colours of the universal indicator are seen.

    THEORY
    The Henderson-Hasselbalch equation is merely the definition of Ka rewritten.

    Ka = [H+][A-]/ [HA]

    Taking logarithms of both sides and re-arranging gives:

    pH = pKa + log10 ([A-]/ [HA])

    The concentration of weak acid, HA= H2CO3, is kept constant by equilibrium with CO2(g) at one atmosphere pressure. Using the Henry's Law constant at 25°C, the equilibrium value of [H2CO3] is 0.0329 M. Dry ice is continuously subliming and bubbling through each solution. Any excess simply bubbles out of the solution. Temperature change is not large enough to affect the qualitative aspects of this demonstration. (See appendix.)

    The equilibrium concentration of the conjugate base, A- = HCO3-, is determined by the initial concentration of hydroxide ion. As CO2(g) is slowly added, acid- base equilibrium in the solution is rapid. The predominant base in solution changes from hydroxide ion to carbonate ion and finally to hydrogencarbonate ion.

    2OH-(aq) + CO2(g) → CO32-(aq) + H2O

    CO32-(aq)+ H2O + CO2(g) → 2HCO3-(aq)

    EQUIPMENT
    • 6 x 250 mL conical flasks (or tall thin beakers to improve the visual effects)
    • 500 mL beaker
    • tongs
    • hotplate
    • large universal indicator/pH chart

    REAGENTS
    • dry ice, solid CO2 (crushed, enough to fill a 500 mL beaker)
      CAUTION! CO2(s) sublimes at -78°C and can cause frostbite. Handle with tongs.
    • distilled water (100 mL)
    • sodium hydroxide solution, NaOH(100 mL, 0.0015M)
    • sodium hydroxide solution, NaOH(100 mL, 0.015M)
    • sodium hydroxide solution, NaOH(100 mL, 0.15M)
    • sodium hydroxide solution, NaOH(100 mL, 1.5M)

    • CAUTION! Concentrated sodium hydroxide solution reacts rapidly with skin and can cause serious burns. Wash the affected area immediately with running water.
    • sodium hydroxide solution, NaOH(100 mL, 5M)
      CAUTION! More concentrated solution would be a serious hazard without changing the final result, as explained below.
    • universal indicator

    PREPARATION
    • The solutions are prepared using a series of 10:1 dilutions. The exact volume of solution is not important since excess CO2(s) bubbles out of the reaction vessel. The relative concentrations of the solutions are important. Stopper the solutions immediately after preparing them. CO2 absorbed from air could change the pH of the dilute solutions.
    • Into each of the conical flasks place about 100mL of solution. Add universal indicator.
    • Just prior to the demonstration, heat the 0.15M, 1.5M, and 5M solutions until they are quite warm. (The heat absorbed by subliming CO2(s) is significant for these more concentrated solutions.)
    • Supply the dry ice in the 500 mL beaker.

    PROCEDURE
    • Excess dry ice is added to each solution. Solid carbon dioxide should remain when the colour change is complete.

    RESULTS
    ​Solution ​Final Colour PH
    ​distilled water ​red ​<4
    ​0.0015M NaOH
    ​orange ​5
    ​0.015M NaOH
    ​yellow ​6
    ​0.15M NaOH
    ​green ​7 ​initially warm
    ​1.5M NaOH
    ​blue ​8 ​initially very warm
    ​5M NaOH
    ​purple ​>9 ​​initially very warm

    This last result is what will probably be observed in practice. The true equilibrium value is calculated to be about 8, as described below. The final two results involve multiple competing equilibria. I would expect first year university students to be able to follow the calculations, but I would not expect most of them to be able to produce the calculations. I would not introduce the calculations except to a student who showed particular interest. Of course, the demonstrator should know in detail what is happening.

    pH of the water/ dry ice solution.

    The acid-base equilibrium is:

    H2CO3(aq) + H2O ⇌ HCO3-(aq) + H3O+(aq)
    At 25°C, Ka1 of H2CO3 = 10-6.35M = Q = [H+].[HCO3-] / [H2CO3]
    Using [H2CO3(aq)] = 0.0329 M as described above, and [H+] = [HCO3-]
    [H+] = √(0.0329x10-6.35)M = 1.2x10-4M
    pH = 3.92. Using universal indicator, this solution appears red.

    pH of the 0.0015 M NaOH / dry ice solution.

    The Henderson-Hasselbalch equation applies if the concentrations are the equilibrium values, not the concentrations added

    pH = pKa + logl0([conjugate base]/[weak acid])
    = 6.35 + log10([HCO3-/[H2CO3])
    = 6.35 + log10(0.0015/0.0329)
    = 5.01 Using universal indicator, this solution appears orange.

    Calculating the carbonate ion concentration at equilibrium:

    [OH-]initial = [OH-]eqm + 2x[CO32-]eqm + 1x[HCO3-]eqm

    As a first approximation, the first two terms on the RHS are taken as negligible.

    [CO32-] = [HCO3-].pKa2/[H+] (Ka2=10-10.33M)
    = 0.0015M x 10-10.33M / 10-5.01M
    = 7x10-9M

    This concentration is indeed negligible. The conversion of hydroxide ion to hydrogencarbonate ion is effectively qualitative.

    pH of the 0.015 M NaOH / dry ice solution.

    [CO32-] at equilibrium is 7x10-7M, which is negligible.
    pH = 6.35 + log10(0.015/0.0929)
    = 6.01 Using universal indicator, this solution appears yellow.

    pH of the 0.15 M NaOH / dry ice solution.

    [CO32-] at equilibrium is 7x10-5M, which is negligible.
    pH = 6.35 = log10(0.15/0.0329)
    =7.01 Using universal indicator, this solution appears green.

    pH of the 1.5 M NaOH / dry ice solution.

    [CO32-] at equilibrium is 0.007 M, which is still an order of magnitude less than [H2CO3]. Precipitation of solid sodium hydrogencarbonate starts to affect the pH. If this reaction is ignored then

    pH = 6.35 = log10(1.499/0.0329) (underlined digits are not significant)
    = 8.01 Using universal indicator, this solution appears blue.

    The extra reaction to be considered is:

    Na+(aq) + HCO3-(aq) ⇌ NaHCO3(s) (K=KSO-1 = (1 / 1.5)M-2

    Qualitatively, the reaction will remove hydrogencarbonate ion, the conjugate base, from solution. As such it should lower the pH and will decrease formation of the carbonate ion. Let x be the amount of NaHCO3(s) precipitated per litre.

    x + [HCO3-]= 1.493 M
    [Na+] x [HCO3-] = 1.5M2 = (1.493M - x)2
    x = (1.493 - √(1.5)) mol L-1 = 0.27 mol L-1
    [HCO3-] = 1.493M - O.27 mol L-1 = 1.22 M
    PH = 6.35 + log10(1.22/0.0329)
    = 7.92 Using universal indicator, this solution appears blue.

    The phenolphthalein originally present in universal indicator is degraded radidly and irreversibly at high pH to a colourless prodict, leaving the blue colour of bromothumol blue. 

     Once equilibrium has been reached, a small amount of phenolphthalein (or more universal indicator) can be added without affecting the colour.

    pH of the 5 M NaOH / dry ice solution.

    In practice, the reaction mixture appears purple with universal indicator, indicating a pH above 9. So much solid sodium hydrogencarbonate has precipitated that the reaction mixture is a sludge. Universal indicator ( or phenolphthalein ) should again be added to replace that which reacted with concentrated sodium hydroxide. In theory, reaction should proceed until almost all the carbonate ion is consumed, giving a pH of about 8 and a blue colour, but only purple was observed.

    [CO32-] at equilibrium is calculated to be insignificant; [CO32-] = 0.005M. Precipitation of solid sodium hydrogencarbonate ( and sodium carbonate-10-water before equilibrium is reached ) must be considered. A short form of the calculation follows. Again let x be the amount of NaHCO3(s) precipitated per litre.

    x = [HCO3-] = 5M
    [Na+]x[HCO3-] = 1.5M2 = (5 M-x)2
    x = (5 - √(1.5))M = 3.78 mol L-1
    pH = 6.35 + log10(1.22/0.0329)
    = 7.92

    Using universal indicator, this solution appears blue if/when it eventually reaches equilibrium. Effectively, all the excess sodium hydroxide solution has been precipitated as sodium hydrogencarbonate, and the result is the same as for the 1.5 M solution. In practice, it is difficult to add enough dry ice to reach equilibrium without freezing the solution solid ( or taking so long that the class loses interest). The titration curve below shows why this is so.

    In the following diagram, the vertical axis shows the concentrations of various species or the amount of precipitate per litre of solution. Several simplifying assumptions were made. Solutions were assumed to be ideal. The mole fraction of water was taken as 1. Water loss due to CO2bubbling through the solution was ignored. Coating of one solid by another (hindering reaction] was ignored. No first year university student shroud attempt this calculation. Reaction occurs in about six well defined stages.

    1. Formation of carbonate ion .

    2OH-(aq) + CO2(g) → H2O + CO32-(aq)

    2. Formation of solid sodium carbonate-10-water, which is the stable form below 32°C. Since this removes sodium ions from solution, the carbonate ion concentration increases.

    2OH-(aq) + CO2(g) + 2Na+(aq) + 9H2O → Na2CO3.10H2O(s)
    2OH-(aq) + CO2(g) → H2O + CO32-(aq)

    3. Formation of the hydrogencarbonate ion. After the hydroxide ion supply is exhausted, carbonate ion reacts as the next strongest base. Some solid sodium carbonate dissolves to restore that equilibrium.

    CO32-(aq) + CO2(g) + H2O → 2HCO3-(aq)
    Na2CO3.10H2O(s) → 2Na+(aq) + CO32-(aq) + 10H2O

    4. Formation of solid sodium hydrogencarbonate from solid sodium carbonate-10-water. This is as far as I usually take the reaction. The solution appears purple.

    NaCO3.10H2O(s) + CO2(g) → 2NaHCO3(s) + 9H2O

    5. Formation of solid sodium hydrogencarbonate from carbonate ion in solution. This increase in hydrogencarbonate ion concentration precipitates more sodium hydrogencarbonate.

    2Na+(aq) + CO32-(aq) + CO2(g) + H2O → 2NaHCO3(s)
    Na+(aq) + HCO3-(aq) → NaHCO3(s)

    6. Finally, the concentration of H2CO3 in solution builds up when it has nothing left to react with, until it reaches equilibrium with the gas at one atmosphere pressure. I have not observed the reaction reaching this stage when it should appear blue. If universal indicator is added to the solid adhering to the walls of the conical flask, it appears blue-green. 

    APPENDIX Temperature changes and equilibrium constants

    Henry's Law constants are given for CO2 dissolving in water at various temperatures, at 1 atm (total pressure). This will affect the final concentration of H2CO3(aq). The pKa values should increase at lower temperature. These trends will affect the pH in opposite directions. They are not large enough to alter the main point of this demonstration. 

    ​Temperature ​Henry's Law constants
    ​Vapour pressure
    ​Autoionisation of water
    ​0°C
    ​3.35 g kg-1 = 0.0761 molal
    ​p(H2O) = 4.58 torr
    ​pKW = 14.94
    ​20°C
    ​1.69 g kg-1 = 0.0384 molal
    ​p(H2O) = 17.53 torr
    ​pKW = 14.17
    ​25°C
    ​1.45 g kg-1 = 0.0329 molal
    ​p(H2O) = 23.76 torr
    ​pKW = 14.00

    A FINAL WRINKLE

    I often comment to my students that the Henderson-Hasselbalch equation suggests that the colour change of an indicator should occur over about 2 pH units, from pKa-1 where the acid species predominates to pKa+1 where the base species predominates. With universal indicator the series of colour changes is over about six units of pH. I then ask why universal indicator produces the colours of the rainbow in order. The responses my students produce to this question make fascinating reading.

    REFERENCES

    1. Nicholson, L., J. Chem. Educ., 1989, 66, 725-726.
    2. Vogel, A.I., Practical Organic Chemistry, 3rd ed, Longman, London, 1956, p984.

    ABOUT THE AUTHOR
    Malcolm Hooper studied at the University of Sydney, where he completed his Ph.D. in 1982. He worked for three years at Macquarie University on a project for the Gun Propulsion Research Group at Salisbury, South Australia, and since then at the University of Sydney where he is an Associate Lecturer.

    Research interests: chemical education, computer modelling, thermodynamics and equilibrium.

    Other interests: Bushwalking, canyoning, RACI Chemical Education group - NSW Secretary from 1990 to 1992 and actively involved in the High SChools Titration Competition. 

    IMMUNOSUPPRESSION IN TRANSPLANTATION
    Chaos in Chemistry
     

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