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Using exponents and logarithms to calculute pH

  1. Jul 28, 2013 #1

    interhacker

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    1. The problem statement, all variables and given/known data

    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.)

    2. Relevant equations

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

    3. The attempt at a solution

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

    [itex]pH_{dilute} = pH_{original} + \log 2[/itex]
     
  2. jcsd
  3. Jul 28, 2013 #2

    SteamKing

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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?
     
  4. Jul 28, 2013 #3

    symbolipoint

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    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.
     
  5. Aug 1, 2013 #4

    epenguin

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    He must have made another mistake that cancelled out the first one and made the answer right! :biggrin:
     
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