Understanding the rate of change in titration curves

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 (H₃O⁺ 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.

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.

Quote: Originally posted by Mildronate

its easy to graph derivative with exel.

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 H₃O⁺ (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_titrant ml of base titrant solution (at concentration c_titrant the remaining amount of H₃O⁺ (mol) is:

V x c - V_titrant x c_titrant, and thus:

pH ≈ - log [(V x c - V_titrant x c_titrant / (V + V_titrant]

(End point: V x c = V_titrant x c_titrant

So the function pH = f[g(V_titrant] is a linear function of the logarithm g(V_titrant of but the function g(V_titrant isn’t linear itself.


If c ≈ c_titrant then the function can be re-written as:

pH ≈ - log c – log [(V - V_titrant/ (V+ V_titrant]

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_titrant/V and setting c = c_titrant 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]