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Acid-base equilibria

  1. Aug 10, 2011 #1
    While I am studying my Chemistry lecture notes, I find the following notes with which I am rather confused:
    For a particular acid and its conjugate base
    HA + H2O = H3O+ + A-
    Ka= [A-][H3O+] / [HA]
    A- + H2O = HA + OH-
    Kb=[HA][OH-]/[A-]

    KaKb = Kw
    pKw= pKa + pKb = 14 (at 298K)

    I am really stuck at these notes. As far as I comprehend, first of all, why there are two equilibria in the solution? With repect to the first equation, the concentration of the acid and conjugate base is already fixed. Then, how can the conjugate base react with H2O forming back HA?
    I think I have wrongly interpreted the information. Could anyone please kindly recitfy and guide me ? How to think about the equilibria? Why A- reacts with H2O rather than H3O (which is the right hand side of the first equation) ?
     
    Last edited: Aug 10, 2011
  2. jcsd
  3. Aug 10, 2011 #2
    Your notes look correct to me.

    You have the dissociation of a weak acid, and its corresponding equilibrium constant.

    You have the reaction of a (conjugate) base with water, and its corresponding equilibrium constant.

    You have the correct equations describing the relation of the two equilibrium constants.

    You know how much the conjugate base is going to react with water based on the relation of the base equilibrium constant to the acid equilibrium constant with respect to KW.
     
  4. Aug 10, 2011 #3

    Ygggdrasil

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    I think you may be confused because, for the most part, these two equilibria are not occuring at the same time. For some acids, HA is more likely to react with water to form H3O+ and A-. Other acids, will not react significantly with water and instead will be much more likely to react with OH-.

    So, which acid/base reaction you use depends on the exact identity of the acid you're examining. For something like acetic acid, acetic acid will react with water and the top equation will apply. For a much weaker acid like ammonium (NH4+), the bottom equation would be more appropriate.
     
  5. Aug 11, 2011 #4
    You mean there is either equilibrium in a solution but not both?
     
  6. Aug 11, 2011 #5
    Both reactions are occurring at the same time. When the system is in equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. Protons are still being donated and abstracted but on a macro-scale there is no net change (IE pH will not change).

    The reason you have two equilibria expressions is because when you deprotonate an acid, you get a proton and the conjugate base. That base can then go and deprotonate something else. Therefore you have these conjugate pairs (acetic acid/acetate for instance). From the expressions you can see that as an acid becomes stronger the conjugate base becomes weaker. Acetic acid is not a strong acid and acetate can act as a base under certain conditions. Hydrochloric acid is a strong acid and its conjugate base (chloride) will pretty much never act as a base.

    So in summary, the Ka expression is describing the acidity of HA (aka the forward reaction, as written) while the Kb is describing the basicity of the conjugate base A- or the reverse reaction, as written. You can use Kw to find Ka if you know Kb and you can use pKa's to make the numbers easier to write/work with (because logarithms have magical powers).
     
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