The ocean is becoming acidic. This is not a natural evolutionary change but happens through human negative effect on the environment. What element or compound could be used to create a more base ocean? How much of it would be needed? How could it be distributed?

  • $\begingroup$ Calcium carbonate has been used to remediate acid rain damage in lakes and ponds. The scale needed in the ocean might make this unviable. $\endgroup$ Commented Aug 24, 2014 at 14:10
  • $\begingroup$ It is possible to get a coarse estimate for the amount of material needed to directly neutralize all the acid coming from anthropogenic $\ce{CO_2}$ absorbed by the ocean. It turns out that using the material with the highest ability to neutralize acid per mass (lithium metal), we'd need to throw ~100 billion metric tons of the pure substance into the ocean. Compare this with the entirety of the iron mining industry, which currently produces about 4 billion metric tons of iron/steel per year. There is no way we can access the amount of matter required for a simple neutralization reaction. $\endgroup$ Commented Aug 24, 2014 at 19:30

1 Answer 1


The ocean is becoming more acidic due to the dissolution of excess $\ce{CO2}$ from anthropogenic activities to form carbonic acid, $\ce{H2CO3}$, which even though mildly acidic is decreasing the pH of the ocean as you rightly point out.

So if we're trying to reduce the effect of this $\ce{CO2}$ on the ocean, we might consider removing it via some sink. One of these theoretical methods could be to react a base, such as $\ce{CaCO3}$, with this carbonic acid in a conventional neutralisation. In symbols:

$$\ce{H2CO3 + CaCO3 -> Ca(HCO_3)2}$$

This could work, and we could derive everything we wanted to know about how much base to use from information such as the amount of residual $\ce{CO2}$ left over from emissions after photosynthesis has been taken into account (and other sinks), and then assuming that all other $\ce{CO2}$ will be taken up by the sea. From here we can calculate the amount of $\ce{H2CO3}$ present in seawater from the equilibrium constant between $\ce{CO2}$ and $\ce{H2CO3}$.

But we won't do that!

Notwithstanding these nonsense assumptions which could grossly overpredict/underpredict the amount of base required, neutralisation reactions don't deal with the excess carbonate effectively. What you really want is a permanent sink (a sink so stable its contents are practically constant through time) at the bottom of the sea. See the long-term carbon cycle for more information on this.

So your question becomes: how do we get the $\ce{CO2}$ dissolved in the ocean to the deep-ocean sediments? And how do we do it 'permanently'?

An interesting paper came out in the Journal of Global Biogeochemical Cycles in 1990 trying to address the question about why phytoplankton in the Antarctic don't use the myriad nutrients in the Southern Ocean to photosynthesise. One might expect that it could be temperature or sunlight which limit photosynthesis in the Antarctic, but this actually isn't the case. The limiting factor in the Antarctic ocean is actually a lack of iron.

It has since be proposed to increase the amount of iron in the ocean (simply by adding it ourselves) which would promote the photosynthesis of phytoplankton in such locations. When these organisms die, they sink and take their biomass with them which contributes to the ocean sediments.

Both neutralisation and iron supplementation may work to offset ocean acidification, however we really don't know the full consequences of these bioremediation efforts—there are simply too many variables involved.

  • $\begingroup$ I corrected your equation and improved the coding. You can enclose the whole reaction equation in $$\ce{...}$$. This way the spacing is better and you safe a lot of typing... $\endgroup$ Commented Aug 25, 2014 at 8:53
  • $\begingroup$ Cheers Martin, still pretty new to the formatting. Will need to write my thesis in this so I'm bound to improve. $\endgroup$
    – user7232
    Commented Aug 25, 2014 at 12:05
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    $\begingroup$ Have a look at the documentation of the mhchem package on CTAN. $\endgroup$ Commented Aug 30, 2014 at 7:17

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