An experiment that has always fascinated me is observing what happens when CO₂ is bubbled into limewater (which is a saturated solution of calcium hydroxide).^1-2 This experiment can be carried out by blowing bubbles of exhaled breath (which contains roughly 4% CO₂)³ through a straw into limewater. A series of reactions convert the calcium ions in the limewater into calcium carbonate:^{1-2, 4-8}

CO₂(g) ⬌ CO₂(aq)             Eq. 1

CO₂(aq) + H₂O(l)  ⬌ H⁺(aq) + HCO₃^-(aq)           Eq. 2

Ca²⁺(aq) + 2 HCO₃^-(aq) ⬌ CaCO₃(s) + H₂O(l) + CO₂(aq)    Eq. 3

These reactions are instrumental in the formation of CaCO₃ in many marine organisms.^4-8 In the reaction between CO₂ and limewater, the formation of solid CaCO₃ causes the mixture to gain a milky white appearance as the reaction proceeds. A curious effect happens if one uses dry ice (solid CO₂) as the source of CO₂ instead of exhaled breath. In this case, the CaCO₃(s) that forms at first dissolves as the CO₂ from the dry ice continues to bubble through the mixture (Video 1).

Video 1: Seashell chemistry, Tommy Technetium YouTube Channel (accessed 5/30/2023)

 

Notice that the system of Equations 1-3 can be simplified by adding them together (Eq. 1 + twice Eq. 2 + Eq. 3). Doing so yields the following:

Ca²⁺(aq) + CO₂(g) + H₂O(l) ⬌ CaCO₃(s) + 2 H⁺(aq)        Eq. 4

While different than other presentations of this system,^4-8 I have found Equation 4 to be quite helpful in explaining to students how the limewater alternately forms or dissolves CaCO₃, depending upon the conditions of the experiment. Inspection of Equation 4 indicates that both the presence of CO₂ and H⁺ will impact which way the reaction will proceed. That is, the principle of Le Chatelier assures us that increased CO₂ pressure should favor the formation of CaCO₃, while increased H⁺ should favor its dissolution. But here’s the rub: dissolved CO₂ concentrations increase with CO₂ pressure (Equation 1), and increased CO₂(aq) increases H⁺ concentration (Equation 2). Therefore, the presence of CO₂(g) introduces competing effects into the system: CO₂(g) is required to form CaCO₃, but too much CO₂(g) can cause the acidity to get too high, dissolving the CaCO₃.  

 

Going further

I have quantitatively analyzed Equation 4 with my students to provide further understanding of how the reaction behaves upon addition of dry ice to limewater. Specifically, we look at how the reaction first forms CaCO₃, which then later dissolves. To start off, using Gibbs energies of formation (Table 1), we calculate the Gibbs energy of the reaction outlined in Equation 4 under standard conditions (

) to be +36 kJ mol^-1. The positive value of

 indicates that CaCO₃ will not form under standard conditions.

 

Table 1: Standard Gibbs Energies of formation of substances in Equation 4.⁹

Substance	DGfo / kJ mol-1
Ca2+(aq)	-554
CO2(g)	-395
H2O(l)	-237
CaCO3(s)	-1129
H+(aq)	0

But the reaction is not carried out under standard conditions. To find the Gibbs energy of the reaction under the nonstandard conditions (

) of the experiment, we use the following:

         Eq. 5

where R is the gas constant (8.314 J mol^-1 K^-1), T is temperature, and Q for Equation 5 is:

        Eq. 6

Substitution of Equation 6 into Equation 5 yields the following, which we use to calculate the Gibbs energy of the reaction:

               Eq. 7

When the dry ice is first placed into the limewater, a good estimate for the concentrations of all species is [H⁺] = 5 x 10^-13 M, [Ca²⁺] = 0.0108 M, and P_CO2 = 1 bar (see Appendix). Plugging these values, T = 298 K, and

= + 57 kJ mol^-1 into Equation 7 yields

 = -72 kJ mol^-1. The negative sign for  

indicates the reaction outlined in Equation 4 is expected to be spontaneous when the dry ice is first added to the limewater, forming CaCO₃(s) – consistent with observations. However, after the dry ice has been allowed to bubble through the limewater for quite some time, the pH drops. A good estimate for the [H⁺] at this point is 1.2 x 10^-4 M (see Appendix). Inserting this value, [Ca²⁺] = 0.0108 M, P_CO2 = 1 bar, T = 298 K, and = + 57 kJ mol^-1 into Equation 7 gives

 = +23 kJ mol^-1. In this case the Gibbs energy for the reaction is positive, and the reaction is spontaneous in the reverse direction. The CaCO3 dissolves!

 

Conclusion:

This demonstration provides a rich assortment of chemical topics to discuss, including chemical thermodynamics, chemical equilibria, solubility product constants, and acid-base chemistry. I generally use this demonstration to illustrate to students how to calculate the Gibbs energy of a reaction under nonstandard conditions (Equation 5). When doing so I give students estimates for [H⁺], [Ca²⁺], and P_CO2, but don’t go through the calculations that justify these estimated values. This demonstration also provides a fantastic demonstration for the potential deleterious effects of increased CO₂ concentrations in Earth’s atmosphere due to the burning of fossil fuels. As the atmospheric CO₂ has risen, so also has the acidity of Earth’s oceans. This increase in ocean acidity has the potential to cause great stress to aquatic organisms that depend upon CaCO₃.^4-8     

 

Appendix:

These calculations assume that the dry ice supplies a constant P_CO2 of 1 bar pressure, because this is the vapor pressure of CO₂(s) at the temperature of dry ice (-78.5 °C).

Limewater is a saturated solution of Ca(OH)₂. Using K_sp = 5.0 x 10^-6 for Ca(OH)₂,¹⁰ it can be shown that [Ca²⁺] = 0.0108 M and [OH^-] = 0.215 M in limewater:

Let x = 

. Then:

x = 

 = 0.0108 M

2x = 

M

When dry ice is bubbled into water for a long period of time, the acidity increases due to increased concentration of dissolved CO₂ (Equations 1 and 2). The equilibrium constants for these reactions are:^11-12

CO₂(g) → CO₂(aq)                                                                 K_H = 0.034                              Eq. 1   

CO₂(aq) + H₂O(l) → H⁺(aq) + HCO₃^-(aq)                             K_a = 4.25 x 10^-7                      Eq. 2

From Equation 1:

From Equation 2:

Let x = the amount of CO2 that reacts to form H⁺:

If we assume that x is very small relative to 0.034:

Solving for x, we have:

x = [H⁺] = 1.2 x 10^-4 M

Notice that x is 0.4% of 0.034, so our assumption was justified.

 

References:

1. Shakhashiri, B.Z. Chemical Demonstrations: A Handbook for Teachers of Chemistry; vol. 1, University of Wisconsin Press, Madison, WI, 1983; pp. 327-377.
2. Bell, J. A. Every Year Begins a Millennium. J. Chem. Educ. 2000, 77, 1098-1102.
3. Tsoukias, N. M.; Tannous, Z.; Wilson, A. F.; George, S. C. Single-exhalation profiles of NO and CO2 in humans: effect of dynamically changing flow rate. Appl. Physiol. 1998, 85, 642-652.
4. Feely, R. A. et. al. Impact of Anthropogenic CO2 on the CaCO3 System in the Oceans. Science 2004, 305, 362-366.
5. Weston, R. E. Jr.; Climate Change and its Effects on Coral Reefs. J. Chem. Educ. 2000, 77, 1574-1577.
6. Buth, J. M. Ocean Acidification: Investigation and Presentation of the Effects of Elevated Carbon Dioxide Levels on Seawater Chemistry and Calcareous Organisms. J. Chem. Educ. 2016, 93 , 718-721.
7. Silverstein, T. P. Rising Atmospheric Carbon Dioxide Could Doom Ocean Corals and Shellfish: Simple Thermodynamic Calculations Show Why. J. Chem. Educ. 2022, 99 , 2020-2025.
8. Bozlee, B. J.; Janebo, M. A Simplified Model to Predict the Effect of Increasing Atmospheric CO₂ on Carbonate Chemistry in the Ocean. J. Chem. Educ. 2008, 85, 213-217.
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10. CRC Handbook of Chemistry and Physics, 99^th ed.; CRC Press.
11. NIST Chemistry WebBook, SRD 69, Carbon Dioxide. https://webbook.nist.gov/cgi/cbook.cgi?ID=C124389 (accessed 5/30/2023)
12. Chemistry 102, Prof. Shapley. http://butane.chem.uiuc.edu/pshapley/GenChem1/L25/web-L25.pdf (accessed 5/30/2023)