Silvermask ([info]silvermask) wrote in [info]chemistryhelp,
@ 2008-12-18 23:25:00
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Current mood: confused

I should know acid/base titrations by now, I really should.
Why is my biochem texbook (Lehninger Principles of Biochem, 5th ed) showing titrations of weak acid behaving in a completely opposite fashion compared with everything else I've ever learned?

It's weirding me out. =\


My understanding from Gen Chem and spending way too much time on acid/base in Quantitative Analysis:

S-shaped titration curve (x: base added, y: pH)
pH changes most quickly nearest the equivalence point
buffer region before the equivalence point


Biochem book:
rotated-S-shaped titration curve (x: base added, y: pH)
pH changes most *slowly* nearest the equivalence point, rapidly at beginning and end
buffer region around and at the equivalence point

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I hate to think that there could be something that wrong in the textbook, but then again I also hate to think that I have no understanding of such a simple and fundamental concept.
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Edit: Thanks guys! Now, in the light of day, it's perfectly obvious that the Biochem book was graphing half the titration curve I was used to seeing.


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[info]nybiara
2008-12-19 03:28 pm UTC (link)
(Disclaimer: I'm a spectroscopist who hasn't done a real titration for several years.)

When you say "rotated S-shape", I assume you mean that the pH is initially low, rises rapidly with increasing base concentration, then levels out and finally rises rapidly again. Is that right? Perhaps they only included the first part of the titration curve?

Taking this curve (stolen from here) as an example:



If you took the graph above and only plotted the region from 0 to 25 cm3, it would look similar to how you described the graph in the textbook. It would be a confusing thing to do, even if the only point of the graph is to show where the buffer region is. Labelling that region as the equivalence point would definitely be wrong. :/

A quick Googling suggests that the publishers haven't yet released any corrections to the 5th edition, so that doesn't help.

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[info]cowbert
2008-12-19 04:29 pm UTC (link)
Perhaps it's just a typo or semantic error. The midpoint of the buffering area (12 cm^3) would be the *half*-equiv point, which can be calculated with Henderson-Hasselbach (i.e. when pH = pKa). Since in studying buffers you don't care about anything beyond the equiv point (because at the equiv point, the buffering capacity of the buffer is exhausted so it's more or less useless), the context sort of makes sense.

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[info]uberjason
2008-12-19 05:49 pm UTC (link)
Could this be a typo? The titration curves we drew in biochemistry always focused on the half-equivalence point (eq OH- = 0.5), where the slope flattens out, and we stopped drawing at right about the equivalence point (unless it was polyprotic). Part of the reason is for its use in buffers, and another part of the reason is that when you're roughly sketching a titration curve, the points useful to know are the half-equivalence and equivalence points - the former being the pKa's, and the latter by averaging between the half-equivalence points. Then you sketch a near flat slope around the half-equivalence points and a steep slope around the equivalence points.

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[info]silvermask
2008-12-19 09:10 pm UTC (link)
Yes, that's exactly it--the graph ends at the equivalence point. Thanks!

Maybe I would have figured that out if I had actually gone back to my Quant notes and looked for the ph=pKa at half-EP.

Like many things, it all makes sense in the light of day, haha.

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