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Interpreting buffer solution reaction

  1. Jul 28, 2013 #1
    So it seems I've been seeing two different ways of interpreting buffer solution reactions. One to me seems to work with Le Chatelier's Principle and the other doesn't.

    I am dropping a strong base, NaOH, into a weak acid/base buffer solution.

    View A:

    HA + H2O <--> (H3O+) + (A-)

    Reasoning: The (OH-) of the NaOH will end up consuming the (H3O+) and turning into H2O thus the pH isn't really effected. In response to the declining concentration of (H3O+), the equilibrium will shift to the right to produce more of the products and less of the reactants.

    View B:

    HA + (OH-) <--> H2O + (A-)

    Reasoning: The (OH-) will start consuming HA on the reactant side thus the pH is not changed as (OH-) is not produced.

    My confusion stems in how these two are connected. View A makes complete sense to me however View B is used often as well. View A seems to deal with the (OH-) consuming the product versus View B where (OH-) is consuming the reactant.

    The shifts in equilibrium also appear to be different. In View A the equilibrium is from L -> R to make up for a decreasing (H3O+) concentration to produce more of (H3O+) and (A-). In View B, there would be less reactant so it would seem then that the equilibrium shift would be from R -> L to produce more HA.

    Thanks for the help, appreciate it
     
    Last edited: Jul 28, 2013
  2. jcsd
  3. Jul 28, 2013 #2
    Just wanted to clarify or add:

    For View B, it is a forward reaction of course, not an equilibrium.

    If (OH-) is implicitly interacting with the product (H3O+), the equilibrium shift is driven by the consumption of the product (H3O+) and the need to replenish those (H3O+).

    If (OH-) is interacting with the reactant, HA, the equilibrium shift is driven by the forward reaction.
     
  4. Jul 29, 2013 #3

    epenguin

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    I've written a related answer about this not long ago. Basically all those species are there in equilibrium and the equilibria are all you ever have to worry about.

    If you really want to worry improductively about where the proton of HA 'goes to', the concentration of H2O is, even in rather alkaline solution maybe a hundred, and usually orders of magnitude greater than that of OH- so most of it will go to the water. But everything that goes round comes round and in the end a bit of OH- gets protonated by indirect path and a smaller amount directly.

    These equilibria are the most rapid chemical reactions known, there is nearly zero enthalpy barrier. Nearly everyone who uses chemistry has to understand about the equilibria, not one in a million of those is involved with the reactions happening. And I am not one of them, but I think the only way to measure a rate of ordinary protonation and deprotonation is by nmr. That is you can pick up a nmr signal of HA if the molecule lasts long enough for a few cycles of the nmr radio wave. And these are typically in the hundreds of MHz so that is 10-8 s kind of times. And you still have something in hand for faster stuff than that because you may be able to see only a minority of molecules and you can cool the sample and can slow the protonation by controlling the pH. As you speed up the reaction e.g. by warming or acidifying the solution, you will see the separate signals of HA and A- merge. This is rough memory, there may be someone more expert here, but the point is you cannot in general mix and watch the reaction happen, stopwatch in hand! So we only consider equilibria and what happens to the molecules you could say in a way doesn't matter!
     
    Last edited: Jul 29, 2013
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