Using exponents and logarithms to calculute pH

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Homework Help Overview

The discussion revolves around calculating the change in pH when a strong acid solution is diluted with an equal volume of water. The subject area includes concepts of pH, logarithms, and acid-base chemistry.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between hydrogen ion concentration and pH, questioning how dilution affects these values. There is a focus on the logarithmic nature of the pH scale and the implications of doubling the volume of the solution.

Discussion Status

Some participants have pointed out potential errors in the original poster's reasoning regarding the pH calculation and the effect of dilution. There is an ongoing examination of the assumptions made about the concentration of hydrogen ions and the resulting pH change.

Contextual Notes

Participants note the importance of the negative sign in the pH definition and discuss the conditions under which significant changes in pH occur, particularly emphasizing that a change in concentration must be substantial to affect pH meaningfully.

interhacker
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Homework Statement



If a solution containing a heavy concentration of hydrogen ions(i.e., a strong
acid) is diluted with an equal volume of water, by approximately how much is its
pH changed? (Express (pH)diluted in terms of (pH)original.)

Homework Equations



I think the question requires the equation defining pH:
[itex]pH = \log[H+][/itex]

The Attempt at a Solution



According to the document, the answer is supposed to be:

[itex]pH_{dilute} = pH_{original} + \log 2[/itex]
 
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If the concentration of H+ ions in the original (undiluted) solution is [H+] and you add approximately an equal amount of H+ ions by diluting the original solution, what is the total concentration of H+ ions in the dilute solution?
 
interhacker,
pH means negative of the logarithm of hydrogen ion concentration; you were missing the negative sign.
pH = -log[H+]

If you just double the volume to make the dilution, this is only a fraction of a pH unit change. The concentration change must be up or down by a factor of 10 for a change of pH of one unit.
 
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He must have made another mistake that canceled out the first one and made the answer right! :biggrin:
 

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