Liming Lakes: A Solution for Acidification?

AI Thread Summary
Lakes affected by acid rain can be neutralized through liming, which involves adding limestone (CaCO3). A calculation for neutralizing a 3.8e9 L lake with a pH of 5.5 suggests that approximately 610 kg of CaCO3 is needed. However, there is some debate about the accuracy of this figure, with one participant suggesting that 601 kg might be more accurate based on simplified stoichiometry. The discussion highlights that the neutralization process is not straightforward due to the complexities of carbonate chemistry, including the solubility of CaCO3 and the behavior of carbonic acid and bicarbonate ions in water. It is noted that while the initial calculations provide a basic estimate, a more thorough analysis is necessary to account for these factors. The potential ecological benefits of adding CaCO3 to lakes are acknowledged, as it poses minimal harm and can support aquatic life.
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TL;DR Summary
Neutralizing acidified lakes
Lakes that have been acidified by acid rain can be neutralized
by liming, the addition of limestone (CaCO3). How much
limestone in kilograms is required to completely neutralize
a 3.8e9 L lake with a pH of 5.5? (question credited to Introductory Chemistry Essentials book)
I calculated it and found that is approximately 610Kg needed of CaCO3 to bring the lake pH to a neutral level.

Is my calculation right 1st :)?

Then if this is right, 610kg of CaCO3 is not that much in a matter of cost or weight to help the bio-life of the lake. I'm wondering if we are applying actually this solution? As far as I know, there is no harm to add CaCO3 to water lakes, water is even drinkable with dissolved CaCO3 ions to some extent.

Cheers
 
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610 or 601kg?

Not that it is a right answer, but at least let's make sure what we deal with. 601 kg is what I get using highly simplified approach (so simplified, it is actually completely wrong - the only thing that is correct is a stoichiometry).
 
Yes I used stochiometry and 2 significant figures as the result on the calculator was 608kg.
 
Perhaps some discrepancy in molar masses used, or you rounded intermediate results.

Trick is, this is incorrect. This is not a simple stoichiometric neutralization. What will happen is that the salt will dissolve and carbonate ions will get protonated, then some will get protonated further into carbonic acid.

To find out how much calcium carbonate needs to be added one should take into account its solubility, water autodissociation and buffering effects of the carbonic acid/hydrogen carbonate ion. Probably further analysis will show of these can be safely ignored - but it is not obvious to tell which ones without doing at least some initial estimates (at pH 7 amount of CO32- is negligible, so Ksp is probably not needed). This is actually quite a nice problem, but IMHO a little bit beyond introductory chemistry.
 
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I have re-done this problem, and indeed your "most simplified" :smile: is more accurate:
1632295023034.png

I also found this old magazine article about liming lakes:
1632295731954.png
Cheers!
 
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