gooby
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Understanding the rate of change in titration curves
I asked this over at the Chemistry StackExchange, which is usually pretty good, but no one has answered so far:
http://chemistry.stackexchange.com/questions/1220/understand...
Maybe it's because it's a weekend. In any case I think I should have answers to this before I try to wrap up this general chemistry text I have here:
| Quote: | As I am looking at titration curves, a few things stand out. For now, I have a two part question:
When titrating an acid with a base, for instance, the pH rises more or less abruptly around the equivalence point. Does this have to do with the
logarithmic nature of pH? In other words, is this caused by the fact that the number of particles involved in the jump from e.g. a pH of 2 to one of 3
is several orders of magnitude than the number of particles involved in the jump from 6 to 7?
When a weak acid is titrated, the rise to the equivalence point is a lot more shallow. Does this happen because of Le Chatelier's principle?
That is to say, as the base is titrating the acid, some of the acid that has not dissociated will dissociate in an attempt to restore equilibrium.
This should retard the titration to some degree. Is that the case?
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blogfast25
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Quote: Originally posted by gooby  | I asked this over at the Chemistry StackExchange, which is usually pretty good, but no one has answered so far:
http://chemistry.stackexchange.com/questions/1220/understand...
Maybe it's because it's a weekend. In any case I think I should have answers to this before I try to wrap up this general chemistry text I have here:
| Quote: | As I am looking at titration curves, a few things stand out. For now, I have a two part question:
When titrating an acid with a base, for instance, the pH rises more or less abruptly around the equivalence point. Does this have to do with the
logarithmic nature of pH? In other words, is this caused by the fact that the number of particles involved in the jump from e.g. a pH of 2 to one of 3
is several orders of magnitude than the number of particles involved in the jump from 6 to 7?
When a weak acid is titrated, the rise to the equivalence point is a lot more shallow. Does this happen because of Le Chatelier's principle?
That is to say, as the base is titrating the acid, some of the acid that has not dissociated will dissociate in an attempt to restore equilibrium.
This should retard the titration to some degree. Is that the case?
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Point one: the shape of the curve is really due to the (simple) mathematics of the way the oxonium ions
(H<sub>3</sub>O<sup>+</sup> are reacted away. Try and work
this out, assuming the acid is strong and completely deprotonated and the shape of the S-curve will become self-evident.
Point two: you’re almost there. Weak acids dissociate only by a small fraction of the total acid present. The initial pH is also higher (for the
same concentration) than for a strong acid and the jump (at equivalence point) smaller. Here the math is slightly more complicated.
There are plenty of excellent internet resources on these problems, forums usually aren’t optimal.
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gooby
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| Quote: Originally posted by blogfast25 | | Quote: Originally posted by gooby | I asked this over at the Chemistry StackExchange, which is usually pretty good, but no one has answered so far:
http://chemistry.stackexchange.com/questions/1220/understand...
Maybe it's because it's a weekend. In any case I think I should have answers to this before I try to wrap up this general chemistry text I have here:
| Quote: | As I am looking at titration curves, a few things stand out. For now, I have a two part question:
When titrating an acid with a base, for instance, the pH rises more or less abruptly around the equivalence point. Does this have to do with the
logarithmic nature of pH? In other words, is this caused by the fact that the number of particles involved in the jump from e.g. a pH of 2 to one of 3
is several orders of magnitude than the number of particles involved in the jump from 6 to 7?
When a weak acid is titrated, the rise to the equivalence point is a lot more shallow. Does this happen because of Le Chatelier's principle?
That is to say, as the base is titrating the acid, some of the acid that has not dissociated will dissociate in an attempt to restore equilibrium.
This should retard the titration to some degree. Is that the case?
|
|
Point one: the shape of the curve is really due to the (simple) mathematics of the way the oxonium ions
(H<sub>3</sub>O<sup>+</sup> are reacted away. Try and work
this out, assuming the acid is strong and completely deprotonated and the shape of the S-curve will become self-evident.
Point two: you’re almost there. Weak acids dissociate only by a small fraction of the total acid present. The initial pH is also higher (for the
same concentration) than for a strong acid and the jump (at equivalence point) smaller. Here the math is slightly more complicated.
There are plenty of excellent internet resources on these problems, forums usually aren’t optimal.
|
I did a lot of searching both on Google web search and Google Books for an answer to this question. None of the answers were particularly helpful
until I found this:
http://www.chemicalforums.com/index.php?topic=6140.0
...which was incidentally a forum post.
| Quote: | | don't forget that the y-axis is a logarithmic scale. |
So that would appear to confirm my suspicion. It's steep because pH is a logarithmic value.
That settles my question, I guess.
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Mildronate
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Its beter to use titration curve derivative.
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blogfast25
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Yes, but in most real lab situations not very practical.
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Mildronate
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its easy to graph derivative with exel.
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blogfast25
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Yes, but most don't draw the graph because that's time consuming. Most use indicators or look for the largest pH jump nearthe end point (using a
meter). Robotitrators use the derivative, of course.
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AJKOER
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To say that the relation is actually linear in the log is a most likely an approximation. It could be more complex like quadratic in the log, for
example.
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blogfast25
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| Quote: Originally posted by AJKOER | | To say that the relation is actually linear in the log is a most likely an approximation. It could be more complex like quadratic in the log, for
example. |
Where did anyone say it was linear in the log? But ‘quadratic in the log’? Not likely IMHO.
The simple case of a dilute strong acid titrated with a dilute strong alkali (we assume both are completely dissociated):
Initial amount of H<sub>3</sub>O<sup>+</sup> (mol): V x c, with V the volume of solution to be titrated and c the
concentration of the acid to be titrated.
After addition of V<sub>titrant</sub> ml of base titrant solution (at concentration c<sub>titrant</sub> the remaining amount of H<sub>3</sub>O<sup>+</sup> (mol) is:
V x c - V<sub>titrant</sub> x c<sub>titrant</sub>, and thus:
pH ≈ - log [(V x c - V<sub>titrant</sub> x c<sub>titrant</sub> / (V + V<sub>titrant</sub>]
(End point: V x c = V<sub>titrant</sub> x c<sub>titrant</sub>
So the function pH = f[g(V<sub>titrant</sub>] is a linear function of
the logarithm g(V<sub>titrant</sub> of but the function
g(V<sub>titrant</sub> isn’t linear itself.
If c ≈ c<sub>titrant</sub> then the function can be re-written as:
pH ≈ - log c – log [(V - V<sub>titrant</sub>/ (V+
V<sub>titrant</sub>]
This is an approximation which doesn’t work very near the end point because there Kw kicks in.
[Edited on 10-10-2012 by blogfast25]
Edit:
By introducing a new variable, α = V<sub>titrant</sub>/V and setting c = c<sub>titrant</sub> the equation can be
re-written to:
pH ≈ - log c – log [(1 – α ) / (1 + α )], a formula you find in some textbooks. α can be considered the ‘degree of
completion of the titration’.
And a bit of further digging around the end-point shows that this equation is accurate up to values of α very, very close to 1, because Kw is so
very small.
[Edited on 10-10-2012 by blogfast25]
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