How Much Acid Rain Can Limestone Neutralize?

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SUMMARY

The discussion focuses on calculating the volume of acid rain that can be neutralized by 1.00 g of limestone (calcium carbonate, CaCO3) in the context of a meteorology lab assignment. The net reaction involves sulfuric acid (H2SO4) and calcium carbonate, leading to the formation of calcium sulfate and carbon dioxide. Participants clarify the use of pH to determine the molar concentration of hydrogen ions in acid rain, emphasizing the importance of stoichiometry in the neutralization process. The final pH after neutralization is noted to be 4.36, indicating the solution remains acidic but closer to neutral.

PREREQUISITES
  • Understanding of acid-base reactions, particularly involving sulfuric acid and calcium carbonate.
  • Familiarity with pH calculations and the relationship between pH and hydrogen ion concentration.
  • Knowledge of stoichiometry and mole conversions in chemical reactions.
  • Basic chemistry concepts, including molecular weight and chemical equations.
NEXT STEPS
  • Learn how to calculate molarity from pH values using the formula [H3O+] = 10^(-pH).
  • Study the stoichiometric relationships in acid-base neutralization reactions.
  • Explore the implications of acid rain on environmental chemistry and limestone weathering.
  • Investigate the effects of different pH levels on the solubility of calcium carbonate.
USEFUL FOR

Students in meteorology or environmental science, chemistry learners needing a refresher on acid-base reactions, and anyone interested in the environmental impact of acid rain on geological formations.

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Homework Statement


Estimate the volume of acid rain in L that can be neutralized by 1.00 g of limestone. Assume the pH of rainwater collected in your rain gauge is a representative value for acid rain. Assume that acid rain contains sulfuric acid as the source of hydrogen ions. The molecular weight of calcium carbonate is 100.09 g/mol.
This is a question from my meteorology lab and I haven't had chemistry in 4 years, so I'm a bit rusty. Any help would be gladly appreciated!



Homework Equations


The net reaction between the two is
H2SO4 + CaCO3 --> CaSO4H20 +CO2


The Attempt at a Solution



I know that we had 1.00 g CaCO3 and the molecular weight is 100.09. So I think I divide and plug this into the moles of CaCO3. Then our pH for the rainwater which would be 7.04, but I'm not really sure what to do with it since its not in mols?
I'm pretty lost.
 
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Do you remember the definition of pH?

It's -log(molar concentration of H+ ions). You can use this to find the mols/L of H+ ions in the rain water.
 
Ok so if I did this right, we have 9.12e-8 mol/L of H+ ions. Then we had 100.09 mols of CaCO3. I also know that the pH of the solution after neutralization is 4.36, but I don't really understand how to plug these into the equation since all of the units are different.
 
The pH of acid rain should be lower than 7 and that of the neutralized solution should be close to 7 please have this aspect clarified.

Once you know the pH of the acid rain solution find the [H3O+] by using the pH equation. Also it must be assumed that the pH is mainly due to the first acidic proton of Sulfuric Acid = H2SO4 if this is the case then the [H3O+]=[H2SO4]. Otherwise we need to employ some equilibrium theory.

Convert the amount of CaCO3 into moles then use the stoichiometry of the net equation between CaCO3 and H2SO4 to find the amount of H2SO4 in moles. Remember that [H2SO4]=[moles of H2SO4/L of solution]. Use this mole value that you've found and divide it by the value of [H2SO4] you've found.
 

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