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Dang titrations!!

Discussion in 'MCAT Discussions' started by TripleDegree, Apr 11, 2004.

  1. TripleDegree

    TripleDegree Joker Doctor
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    I feel like I have a solid handle on most of Physical Chemistry except for these dang titration problems.

    Could someone with a large heart do me a huge favor and post a sample MCAT titration problem, and explain their approach for solving this problem. I've read EK, Kaplan and my Chem text till I was blue in the face, and I still don't get it.

    Hope it doesn't end up being my Achille's heel
     
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  3. TripleDegree

    TripleDegree Joker Doctor
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    I guess either

    (a) Titrations are a heartburn for most other folks

    OR

    (b) The size of the viewer's cardiac organ isn't quite enough :)

    Anyway - here goes...

    *BUMP*
     
  4. Cerberus

    Cerberus Heroic Necromancer
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    if you post one i'll try and answer it.

    key points to remember:

    At half equivalence point:
    [conj base]=[conj acid]
    so
    pH=pKa+Log[Conjugate Base]/[Conjugate Acid] = pKa+log1=pKa
    remember eq point of
    strong acid titrated with strong base = 7
    strong acid titrated with weak base<7
    weak acid titrated with strong base>7
     
  5. TripleDegree

    TripleDegree Joker Doctor
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    Thanks Cerb

    Ok, here's a question.

    In a classic weak acid - strong base reaction, several text books have mentioend that the reaction goes 100% to completion.

    CH3COOH + NaOH --> CH3COO- + H+ + Na+

    Therefore, since the resultant ion is weakly basic, the pH of the overall solution at the equivalence point is greater than 7.

    My question is this - I thought that weak acids do not completely dissociate. So how can we assume that the reaction proceeds 100% to completion? I think the text books calculate the Keq value for this reaction which turns out to be a large number, and hence they make the conclusion that the reaction is skewed far to the right.
     
  6. daffy

    daffy Member
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    Hi TripleDegree, here is what works best for me:
    1. Mb*Vb = Ma*Va - this works in all cases
    2. @1/2 Eq [A]=

    3. Str B + Str A
    @ 0ml B added pH = [H+] from A
    @ before Eq point use the: starting, change, eq [] -> pH=[H+] from leftover A
    @ Eq point: H2O equation pH=7
    @ after Eq point pOH=[OH-] from leftover B

    4. Wk B + Str A
    @ 0ml B added: starting, change, eq[] -> use Ka to find pH
    @ before Eq point: BUFFERs pH=pKa + Log /[A]
    @ Eq point: use Kb for conj B to find pH
    @ after Eq: starting, change, eq [], find pOH from [OH-] from leftover B

    Hope this helps, if you have more questions and I can help pls feel free.
    Best of luck to all
     
  7. daffy

    daffy Member
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    Hi there,

    at Eq point

    CH3COO- + H2O <-> CH3COOH + OH- this shifts the eq in your rxn to the R
    Calc Kb for this rxn gives you pH. B/c of hydrolysis [OH-] is higher then [H+]

    I am not sure I am helping (If I am confusing you pls ignore this)
    good luck
     

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