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PH, Buffers, and Dilution question

  1. Oct 12, 2007 #1
    I understand the henderson-hasselbalch equation, and how the pH of a solution depends on the ratio of acid to its conjugate base. However, I dont understand how this concept can be related to solutions that are extremely diluted.

    For example, lets say you have 0.00001M acid+c.base. According to Henderson-Hasselbalch, the pH will still = pKa, since the log ratio is equal to 1. But does this make sense logically? I mean with that little acid, there are so little H+ ions dissociated, so how can the pH possibly equal the pKa (especially if the pKa is very low for that specific acid)??
  2. jcsd
  3. Oct 13, 2007 #2


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    The dissociation of water becomes much more important for dilute solutions of acids (or bases). The K value for the weak acid or base still holds (not sure how well it holds when solutions are so dilute, like less than 0.001 F). Try a forum search. The more general form for the equilibrium constant expression has been discussed. The topic occurs fairly frequently.
  4. Oct 28, 2007 #3
    Well, in fact, there exists equation for calculation.
    if the acid is in very low concentration, the approximation towards the water cannot be neglected.
    Hence, the pH of the solution would be -log(sqrt Kw) =7 around 298K
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