How Much Calcium Carbonate to Add for Optimal pH in a Fish Tank?

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Discussion Overview

The discussion revolves around determining the appropriate amount of calcium carbonate to add to a fish tank in order to adjust the pH from 5.0 to 6.5, which is necessary for breeding small fish. The scope includes mathematical reasoning and conceptual understanding of pH, pOH, and buffering systems in water.

Discussion Character

  • Mathematical reasoning
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant presents a multiple-choice question regarding the mass of calcium carbonate needed to achieve the desired pH level.
  • Another participant suggests that the relationship between pH and pOH must be considered, referencing the ion product of water (Kw) and the stoichiometry of the reaction involving calcium carbonate.
  • A different participant argues that the initial pH's cause is unknown and that the presence of buffers complicates the calculation of the required amount of calcium carbonate.
  • Concerns are raised about the clarity of the mathematical reasoning, with one participant suggesting that mixing concepts of pH and hydrogen ion concentration may lead to confusion.

Areas of Agreement / Disagreement

Participants express differing views on the complexity of the problem and the necessary calculations. There is no consensus on the correct amount of calcium carbonate to add, and the discussion remains unresolved regarding the impact of buffering agents in the water.

Contextual Notes

Limitations include the unknown factors affecting the initial pH, the presence of buffering agents in the water, and the potential misinterpretation of pH and related concepts.

marcelo
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Considering a fish breeder decided to breed small fishes which needs a pH between 6,0 to 7,0 to stay alive. He needs to adjust the water's pH that is 5,0 to a value of 6.5, having available only calcium carbonate. The mass in mg added to 5L of water is about:
A)2,5

B)5,5

C)6,5

D)7,5

E)9,5
 
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What do YOU think?
 
I think if I have 10-5 of pH value we need 10-9 of pOH value to make Kw=-14
the reaction of water with CaCo3 is 2 per 1
(H+) + (OH-) + CO3(2-) + Ca(2+)----> HCO3(-) + Ca2(+) OH(-)
(H+) + (OH-) +HCO3(-) + (Ca(2+) + OH(-)-----> H2CO3 + Ca(OH)2
since I need only 10^-2 concentration of OH- and we have only 5L of water:

5.10(-2) mols of OH-. But 1 mol of CaCO3 gives 2 mols OH, we need only 2.5x10(-2) but this answer is wrong,
 
As asked this question has no answer, as we don't know what is responsible for the initial pH - typically water contains some buffers and the amount of base/acid required to change the pH depends of their concentration. Then, adding carbonates creates another buffer set, making the calculation a bit more convoluted than just the neutralization would do.

marcelo said:
I think if I have 10-5 of pH value we need 10-9 of pOH value to make Kw=-14
This is way too cryptic for me to guess what you mean. Or, if I guess right, you are mixing pH with [H+] and so on as if they were the same thing - they are not.
 
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