TripleDegree

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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
 

TripleDegree

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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*
 

Cerberus

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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*
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
 
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TripleDegree

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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.
 

daffy

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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

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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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