[Cardinal]

I got AP Chem curriculum from a teacher who used to grade the actual test, through his protege, so to speak. I've been over this with a tutor a couple times, so I'm not just jumping over here to be lazy. It's only my 2nd year of AP, and some of this stuff hasn't come up since I took college chem 20 years ago.

Information:
Potassium sorbate, KC6H7O2 (molar mass 150. g/mol) is commonly added to diet soft drinks as a preservative. A stock solution of KC6H7O2(aq) of known concentration must be prepared. A student titrates 45.00 mL of the stock solution with 1.25 M HCl(aq) using both an indicator and a pH meter. The value of Ka for sorbic acid, HC6H7O2, is 1.7 × 10−5.

(e)   The initial pH and the equivalence point are plotted on the graph below. Accurately sketch the titration curve on the graph below. Mark the position of the half-equivalence point on the curve with an X.

Here's the answer given, but the line should be shifted a bit left to land on the dots. http://i86.photobucket.com/albums/k110/flavord/AP%20Chem%205-11%20graph%20practice%20test_zpsr6mdfbyq.jpg

Here's my current explanation of the graph, as far as I have one. Please poke holes in this at any point:

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The HCl releases the protons essentially 100%. At first, the molecule most available to take them is water, so the hydronium concentration spikes, which means the pH is falling fast. After a bit, there is so much hydronium that it meets the sorbate ion in quantities high enough to slow the pH fall very noticeably. After another bit of time/added HCl, enough of the sorbate is protonated so that some of it is even putting protons onto water, after all, it's a weak acid, and water is a better base than it is. The hydronium concentration grows rapidly, and the pH falls sharply.
==========================

If that all is correct above, I still don't understand why there's the tail that starts to flatten out to the right of the equivalence point. We're still adding HCl, but the sorbate ion is apparently taking the proton quite a bit there. Is it basically Le Chatelier's Principle, that if we load up the protonated side, the system shifts to the other side, and forms sorbic acid?

[chenbeier]

Think about buffer.

[Cardinal]

Yes, my main question is about the tail on the right that starts to flatten out. If I've overwhelmed the buffer, why is the pH going down slower at the last?

[Borek]

The HCl releases the protons essentially 100%. At first, the molecule most available to take them is water, so the hydronium concentration spikes, which means the pH is falling fast.

This is already a strange way of explaining the situation. You start with a solution in which HCl is already 100% dissociated, all protons are already present as hydronium, so its concentration doesn't "spike", actually initially most of hydronium added is consumed by the sorbate anion present in large quantities. Yes, pH goes down, but that's mostly because pH is a logarithmic scale and in solution close to neutral pH changes are quite fast despite concentration changes being quite small.

[Cardinal]

That's a good point: the hydronium is already formed from the HCl in the original acid solution. Your other point is that changing from 1E-7 to 1E-6 is a smaller actual numerical change than from 1E-6 to 1E-5. Is that right?

I still don't get the final tailout on the right, though. I can't understand my tutor's point, unless it's just Le Chatelier's Principle.

[Borek]

Are you asking why the pH stops going quickly down after the equivalence point? Once you are past the equivalence point there is no sorbate present, you are just approaching the pH of the titrant there, pH can't go down all the time. It is more or less just a dilution curve.

[Cardinal]

I guess I need it dumbed down farther. I assume you're right, but I don't understand why the curve flattens at the last.

[Borek]

What other behavior do you expect?

[Cardinal]

Talking with my paid tutor, I think I realize that you're right. After the sorbate ion is mostly used up, most of the hydronium is not being deprotonated, and the pH crashes again. The reason for the last "tail" flattening is that the move between each integer of pH is 10x harder than the last, and the natural graph of -log x makes the slope flatten toward zero.

Right?

[Borek]

The reason for the last "tail" flattening is that the move between each integer of pH is 10x harder than the last, and the natural graph of -log x makes the slope flatten toward zero.

Even without log the curve must be flattened out at the end. You are adding excess acid to some initial amount of solution (no reactions taking place, so you can assume this initial amount to be just water). That means concentration of the acid in the mixture asymptotically approaches concentration of the titrant, asymptotes always look "flattened out".

Note that initially pH goes down so fast, you will not observe the curve to be so prominently bent, just after adding single drop of the titrant pH will fall below 7.