What are the side effects of adding dry ice to a swimming pool? 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?
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.