Dang titrations!!

TripleDegree

Joker Doctor
15+ Year Member
Nov 6, 2003
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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
     

    Cerberus

    Heroic Necromancer
    15+ Year Member
    Dec 13, 2001
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    1. Attending Physician
      TripleDegree said:
      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*

      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
       

      TripleDegree

      Joker Doctor
      15+ Year Member
      Nov 6, 2003
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        Cerberus said:
        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

        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.
         
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        daffy

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        Apr 17, 2002
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          TripleDegree said:
          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*

          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
           

          daffy

          Member
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            TripleDegree said:
            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.

            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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