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This is my current understanding: Ocean acidification is the result of atmospheric $\ce{CO2}$ dissolving in the ocean's water. When this occurs the $\mathrm{pH}$ of the ocean decreases (from the increase in $\ce{H3O+}$ ions produced from the dissociation of $\ce{H2CO3}$). When there is an increase in temperature, the solubility of carbon dioxide decreases which results in a lower concentration of carbonic acid (the formation of carbonic acid is from the reaction of $\ce{CO2}$ with $\ce{H2O}$ which is exothermic, so the reverse reaction is favoured). So since the concentration of carbonic acid is lower, the concentration of $\ce{H3O+}$ ions decreases (lower concentration of acid dissociates in the water). Therefore, the $\mathrm{pH}$ of the ocean increases as there are fewer $\ce{H3O+}$ ions present in the water.

Is my understanding correct, or have I made a mistake in my explanation? If you can help me out I would be very grateful :)

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You have to compare the size of these counteracting effects.

The temperature increase slightly decreases the acidification caused by increased carbon dioxide partial pressure.

The shift of reaction constant is minimal.

From the Wikipedia carbon dioxide data page, Water solubility section, we can see for temperature 15 resp 17 $^\circ$C V/V solubility 1.019 resp. 0.956. That is 6.2% lower solubility.

Increased partial $\ce{CO2}$ pressure from 0.03 to 0.04% gives 33%, what is significantly more than by 6.2% decreased solubility.

Aside of that, higher $\ce{CO2}$ concentration and lower sea water $\mathrm{pH}$ are already observed.

This is directly affecting carbonate related equilibriums, particularly in context of forming of mollusk shells.

Some mollusks are already observed having troubles in forming aragonite - the more soluble form of calcium carbonate.

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    $\begingroup$ @Night Writer Well, in my understanding, in given context, the raise of temperature, increase of CO2 And sea acidification cannot be separated. $\endgroup$ – Poutnik May 4 at 14:27
  • $\begingroup$ Ah, yes, I see my mistake! $\endgroup$ – Buck Thorn May 4 at 16:13

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