My apartment complex pool was closed today due to a resident dumping a large quantity of dry ice into the pool. In the notification from my landlord, they stated:

Dry ice is frozen carbon dioxide and not only imbalances the pool, but can cause burns due to the below freezing temperatures.

My understanding of dry ice being added to water is that it would sublimate and carbonate the water, causing a minor temperature decrease, and creating a low-hanging cloud of carbon dioxide gas on top of the surface (since it's heavier than air).

However, it would seem that if the dry ice indeed does cause "below freezing temperatures", that would just freeze the water; so, unless you already in the water, you shouldn't get a freeze-burn.

I've done a bit of searching, but most material talks about burns from solid dry ice, not from adding it to water.

  • One thread on Reddit suggests that freeze-burns are more possible from dry ice gas if the skin is wet, but there are no citations.
  • CCOHS states,

    Direct contact with the liquefied [carbon dioxide] gas can chill or freeze the skin (frostbite).

    ... but refers to liquefied gas, not gas in the gaseous state.

  • Wikipedia doesn't list any particular safety concerns aside from handling of liquid or solid carbon dioxide.

So, here's what I'm wondering: How does dry ice affect the pool's pH, and to what extent? How does it affect the temperature, and to what extent? And what other side effects (mostly harmful ones) would arise from the addition of dry ice to a swimming pool?

  • 1
    $\begingroup$ (1) Negligibly (if at all) given the small amount of dry ice relative to that of a buffered system in equilibrium with the atmosphere (small bump in the positive direction until equilibrium is reestablished); (2) Negligibly (if at all) given the small amount of dry ice relative to that of a buffered system in thermal equilibrium with the surroundings (small bump in the negative direction until thermal equilibrium is reestablished); and (3) None unless you happen to contact it with your bare skin. $\endgroup$ Aug 5, 2017 at 0:17
  • 1
    $\begingroup$ @ToddMinehardt For (3), you're referring to gas-to-bare-skin contact, or solid-to-bare-skin contact? $\endgroup$
    – Cat
    Aug 5, 2017 at 0:18
  • $\begingroup$ Bare skin to solid contact. Sorry for not being clearer. That said, you wouldn't want to be exposed (bare or not) to something like a jet of carbon dioxide gas (really, any gas) blasting at you, but that's far from you scenario. What I'm seeing here is a larger version of a standard intro chemistry demonstration of dry ice in water that "boils" the water at room temperature, and is frequently seen in Halloween and similar mock-ups of "boiling" liquids in laboratory apparatus. $\endgroup$ Aug 5, 2017 at 0:23
  • $\begingroup$ There are only two problems. First if someone comes into contact with the solid dry ice then they can suffer frostbite burns. There would be no "liquid" dry ice (carbon dioxide) at the pressure of the atmosphere. The second problem is that carbon dioxide is heavier than air. so if the pool were in a recessed area the carbon dioxide could "pool" displacing the air (and hence the oxygen in it). Thus swimmers could suffocate. $\endgroup$
    – MaxW
    Aug 5, 2017 at 4:20
  • $\begingroup$ I would ask the land lord to walk on the ice formed on the below freezing surface of the pool. Sounds like someone who wants to inconvenience people to try and force social pressure onto some pranksters because some other mother Grundies take it on themselves to complain about pranksters having fun. Ask him to calculate how much pocket money it would cost to purchase sufficient dry ice to freeze the pool, then ask if that is likely to occur. $\endgroup$
    – KalleMP
    Feb 16, 2019 at 23:46

1 Answer 1


Solubility of $\ce{CO2}$ in water is $\pu{1.45 g/L}$ at SATP. Apparent dissociation constant for carbonic acid at the same conditions is $K_\text{a}(\ce{H2CO3}) = \pu{4.47e−7 mol L^{-1}}$.

Now lets speculate how much dry ice could be disposed and what the pool capacity might be, guessing that the water warmed back up. It's hard to tell which pool size your apartment complex (hopefully not this one) might have, so I'm going to propose some values (marked in $\mathrm{\color{red}{red}}$) which look plausible to me.

Assuming $m(\ce{CO2}) = \color{red}{\pu{10 kg}}$, and the pool volume of approx. $V = \color{red}{\pu{5e5 L}}$ (olympic pool is $\pu{2.5e6 L}$, I took $\frac{1}{50}$ of that), one notices that in order to completely dissolve $\pu{10 kg}$ of $\ce{CO2}$, $\pu{6.9e3 L}$ of water is required (based on solubility data), which means that we have more than enough water in the pool to completely dissolve all $\ce{CO2}$.

For the weak acid

$$[\ce{H+}] \approx \sqrt{K_\text{a} C} = \sqrt{\frac{K_\text{a} m}{M V}} = \sqrt{\frac{\pu{4.47e−7 mol L^{-1}} \cdot \pu{e4 g}}{\pu{44 g mol ^{-1} \cdot \pu{5e5 L} }}} = \pu{1.43e-5 mol L^{-1}}$$

so the pool water will indeed be acidified to the level of

$$\mathrm{pH} = -\log[\ce{H+}] = 4.85,$$

assuming the values I've willfully established, and that all carbon dioxide completely dissolved in water. You can put your values into the equation and argue with your landlord further.

  • 1
    $\begingroup$ There's no way you're hitting pH of 4.85 in system buffered by chemicals in the pool (basic salts) as well as atmospheric carbon dioxide levels. At worst, pH is 5.65. The scenario you're describing is on that does not include time: equilibrium will very quickly be reestablished. $\endgroup$ Aug 5, 2017 at 1:44
  • 4
    $\begingroup$ @ToddMinehardt You are absolutely right, I took the idealized boundary simplest case because there is too many variables to take into account; it's more like a model for OP to get initial idea of what's causing the pH shift. $\endgroup$
    – andselisk
    Aug 5, 2017 at 5:48
  • $\begingroup$ Another factor that will limit the amount of CO<sub>2</sub> that is dissolved in the water is the loss of gas as it boils off and bubbles escape the water's surface. The constraint of complete dissolution was named in the calculation to establish it as a theoretical limit. The system is not closed and the large surface area will help to reestablish equilibrium. $\endgroup$
    – BalooRM
    May 16, 2020 at 17:46

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