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Equilibrium / buffer question

  1. Jan 8, 2005 #1
    Hiya, im just having abit of an issue understanding the whole concept of this particular chapter.

    If I have a mixture that is at equilibrium, it implies that the mixture can undergo a forward and reverse reaction ?? So then a buffered solution is also at equilibirum but is composed of an acid with its conjugate base / or a base with its conjugate acid and is able to resist large pH change ???
  2. jcsd
  3. Jan 8, 2005 #2


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    Equilibrium is a shifting process, in which the system tries to neutralize the external altering made by you. If a system is at equilibrium, you can imply that the system succeeds the compensation.

    If a buffered solution is at equilibrium, this means that the added acid or base has been consumed and the change compensated. This will be valid until one of the buffer constitutients finish.

    If they don't make sense, ask more questions, so that I can try rephrasing these.
  4. Jan 8, 2005 #3
    ok thanks that makes sense. Would you (or anyone else?) mind helping me out with some questions ??? ive got a test this monday -_- ...

    Calculate the pH after 0.020 mol of HCL is added to 1.0L of each of the following solutions:

    a) 0.10M propanoic acid (HC3H5O2, Ka = 1.3 x 10^-5)
    b) 0.10M sodium propanoate (NaC3H5O2)
    c) pure H20

    For b) I had to use the ICE chart to find my final [H] concentration (which was 3.25 x 10^-6 M = pH 5.48) and I matched the correct answer given. My solution sheet says the answer for a) and c) is pH = 1.70, which is the result of simply doing -LOG[0.02], but why is that ??? Why is the pH of a) and c) simply found by -LOG[0.02] while b) required me to do the ICE chart ?? Or is a) and c) solved in a different manner ???
  5. Jan 8, 2005 #4


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    I am lil bit rust on my chem knowledge, but as i remember HCL is a strong acid and it diassociates completely in H+ ions, so if you mix it with pure water, the ph will be -log[concentration] of HCL.

    Also, what chem_tr is refering to is Le Ch√Ętelier principle, which is used equilibrium reactions, such as the weak acids and weak bases reactions, and the common ion effect.
  6. Jan 8, 2005 #5


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    Start from the definitions. Let me do (a), and you do the rest. There should be no need to look up ICE charts for (a), (c) (you figure out why).

    0.02M of [itex]HCl[/itex] has 0.02M of [itex]H^+[/itex] and 0.02M of [itex]Cl^-[/itex].
    So, from this alone, [tex] [H^+]_{HCl} = 0.02M [/tex]

    Now, consider the propanoic acid, [itex] C_2H_5COOH [/itex]

    You are given that [tex] K_a = 1.3*10^{-5} = \frac{[H^+][C_2H_5COO^-]}{[C_2H_5COOH]} [/tex]

    But, obviously, [tex] [H^+]=[C_2H_5COO^-] [/tex] (neglecting contribution from water = [itex]10^{-7}M [/itex])

    and you are given that [tex] [C_2H_5COOH] = (about)~0.1M [/tex]

    So substituting, you get :

    [tex] \frac{[H^+]^2}{0.1} = 1.3*10^{-5}M [/tex]

    Or, [tex] [H^+]_{prop~ac} = \sqrt {1.3*10^{-6}} = 0.00114M [/tex]

    Adding the contribution from [itex]HCl[/itex] gives :

    [tex] [H^+]_{total} = 0.02114 [/tex]

    And [tex] pH = -log[H^+]_{total} = -log 0.02114 = 1.675 [/tex], which is about 1.7

    For (c), ask yourself how much [itex]H^+[/itex] is contributed by water, and whether or not this number is significant in comparison to that from the acid.
    Last edited: Jan 8, 2005
  7. Jan 8, 2005 #6


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    I think I did too much this time.:grumpy:

    Please make sure you understand everything rather than just copying the approach.
  8. Jan 8, 2005 #7
    ohh ok ok that really really helps... i had a pile of questions similar to a) so i can go get them done now... AHrg! I should have been able to solve c), my chem teacher would kill me ! thank you all very very very much! :biggrin: !!!!
  9. Jan 8, 2005 #8
    oh will do, i think my issue was I neglected to consider the [H] contributed by the prop. acid ; in my rough work I think i kept assuming it was zero and only considered the 0.02 mol contributed by the HCL.
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