How Much Acid Rain Can Limestone Neutralize?

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Discussion Overview

The discussion revolves around estimating the volume of acid rain that can be neutralized by a specific mass of limestone (calcium carbonate). It involves concepts from chemistry and meteorology, particularly focusing on the reaction between sulfuric acid and limestone, as well as calculations related to pH and molarity.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a homework problem involving the neutralization of acid rain by limestone, specifying the mass of limestone and the pH of the rainwater.
  • Another participant reminds the original poster of the definition of pH and suggests using it to find the molarity of hydrogen ions in the rainwater.
  • A participant calculates the molarity of H+ ions in the rainwater but expresses confusion about how to incorporate this value into the overall calculation.
  • Another participant emphasizes that the pH of acid rain should be lower than 7 and that the pH after neutralization should approach 7, suggesting a clarification on these points.
  • This participant also advises converting the mass of CaCO3 into moles and using stoichiometry to relate it to the moles of sulfuric acid, while noting the need for equilibrium considerations if the pH is influenced by factors beyond the first acidic proton of sulfuric acid.

Areas of Agreement / Disagreement

The discussion includes multiple viewpoints on how to approach the problem, with no consensus reached on the specific calculations or assumptions regarding pH and the reaction dynamics.

Contextual Notes

Participants express uncertainty about the relationship between pH, molarity, and the stoichiometry of the reaction. There are also assumptions regarding the contributions of sulfuric acid to the pH that remain unresolved.

Punchlinegirl
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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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