Okay I have another question. I answered this one below B but the correct answer is A. Why?
860. The Ka1 for H2CO3 is 4.3x10^-7. The Ka2 for H2CO3 is 5.6x10^-11. Which of the following depends upon the concentration of acid and conjugate base?
A) The pH of the first equivalence point
B) The pH of the first half equivalence point
C) The pH of the second half equivalence point
D) The pKa1
I was thinking that the 1st half eq. point (B) would be 50% H2CO3 and 50% HCO3-; therefore, the pH=pKa at this location and indeed depends on the concentration of the acid and its conjugate. No? Hmm... I dunno...help please?
Where did you get this question? You might want to check the source of this question and see if they have corrected their error in a later edition, because there is no correct answer in the choices. What they have presented can actually lead to a
HUGE misunderstanding of the subject. Some typos are obvious, and all books have them. But this is a case where the author doesn't really know what they are presenting. I hate to toss out such a strong comment and mean to pick no argument here (afterall, we all make mistakes), but this can prove to be very damaging to not only acid-base chemistry but also protein chemistry and the isolectric point. What they have in essence said with their wrong answer implies that the isoelectric point depends on the concentration of the zwitterion, which it definitely does not. You might want to use a better set of materials for acid-base chemistry or at the very least consult a chemist whenever you are told something is wrong that you don't fully understand.
The pH of a buffer equals the pKa at the half equivalence point. It does not matter how much acid or base you have, the pH at the half equivalence points will always the same.
Very true... so keep that in mind for pKa1, pKa2, and pKa3 in your example, because that will help clear up a misconception.
Say I start out with two sets of solutions, one containing 1 M phosphoric acid in its fully protonated state, and one containing .1 M phosphoric acid, also fully protonated. Ok, I start adding NaOH to both solutions. At some point, I'll get to the first half equivalence point. When I get there, the pH will equal the first pKa, so it will be 2.16.
Right on the mark to this point.
Ok, I keep titrating, until I get to the first equivalence point. At this point, the first solution will have approximately 1 M monobasic phosphate (H2PO4 2-) and basically no other form of phosphate (this is assuming that the NaOH added no volume to the solution. Obviously this wouldn't be the case, but it works either way, this just simplifies the math). The second solution will have approximately 0.1 M H2PO4(2-). If you calculate the pH of the 1 M solution at this point, you should be somewhere around 3.60. The pH of the .1M solution should be closer to 4.10. Thus, this is where the pH actually depends on the actual concentration of the species, as opposed to the relative concentration of the species (pKa and thus the half equivalence points depend on the relative concentration of the acid and conjugate base, not the actual concentration). If you keep titrating till the second half-equivalence point, you're now at the second pKa, so your pH is 7.21 for both solutions.
Just want to toss a correction into your thought process here. With polyprotic acids, and most commonly with amino acids, you determine the middle equivalence points by averaging the adjacent pKa values. This is what we do to ascertain the isoelectric point for amino acids, because of the symmetry of the polyprotic titration curve. The technique for attaining isoelectric point is based on a generic aspect of a polyprotic titration curve, which shows that the middle equivalence points lie perfectly displaced between the adjacent pKa values.
So, in your example comparing 1.0 M H3PO4 to 0.10 M H3PO4, the two curves will share the same pKa1, first equiv pt., pKa2, second equiv pt. and pKa3. They will differ according to concetration only at the start and the last equivalence point.
If you find this counterintuitive based on what you have proposed above about concentration, then ask yourself what the pH will be after 1.5 equivalents of base have been added. As you stated above, in both curves you expect the pH = pKa2. That would mean that in both curves, the first equiv. pt. would lie halfway between pKa1 (about 2.1) and pKa2 (about 7.2), resulting in a pH of roughly 4.65. Drawing the curves overlayed onto one another should help to make this more apparent.
