[glossolalia]

Hello - first post here, slogging through some practice exams.

Part A is to determine the pH of 11.7g of EDTA (MW=292) in 1L of solution, and I have (I hope successfully) determined that to be 1.656.  Feel free to double-check me.

Part B is where I am getting stuck:  titrating the solution in part A with 1L of .08M NaOH.  The polyprotic-ness (polyproticity) is what's throwing me for a loop here.  Thanks for any help in walking me through finding the pH of this solution.

Part C is the addition of 1L .04M Na3EDTA, which I can't see having any appreciative effect aside from doubling the volume.  Am I crazy here?  Thanks!

Cheers,
g

[Borek]

Part B is where I am getting stuck:  titrating the solution in part A with 1L of .08M NaOH.  The polyprotic-ness (polyproticity) is what's throwing me for a loop here.  Thanks for any help in walking me through finding the pH of this solution.

Please elaborate - what pH? Initial? End point? Final? At 37.24% neutralization?

Part C is the addition of 1L .04M Na3EDTA, which I can't see having any appreciative effect aside from doubling the volume.  Am I crazy here?  Thanks!

Buffers come to mind, but it is not even clear to me to what you are going to add the salt. If to the initial solution, you should expect reaction between acids and their conjugate bases.

[glossolalia]

Here is the question in its original text.  Thanks for your reply!

3.   a. A solution of the chelating agent ethylenediamine tetraacetic acid (EDTA) is made by
dissolving 11.7 grams of the polyprotic acid in sufficient water to make a total volume of 1
liter. What is the pH of the EDTA solution? (For EDTA: Mr = 292, pKa1 = 1.7, pKa2 = 2.6,
pKa3 = 6.3, pKa4 = 10.6)
b. If you add an equal volume of 80 mM NaOH to the EDTA solution in part a, what is the
pH?
c. If you add an equal volume of 40 mM Na3EDTA to the EDTA solution in part a, what is
the pH?

[Borek]

a. A solution of the chelating agent ethylenediamine tetraacetic acid (EDTA) is made by
dissolving 11.7 grams of the polyprotic acid in sufficient water to make a total volume of 1
liter. What is the pH of the EDTA solution? (For EDTA: Mr = 292, pKa1 = 1.7, pKa2 = 2.6,
pKa3 = 6.3, pKa4 = 10.6)

pKa1/2 values are so close, you have to take both steps of dissociation into account.

http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-simplified

b. If you add an equal volume of 80 mM NaOH to the EDTA solution in part a, what is the
pH?

What will be present in the solution? (hint: neutralization) There is a nice formula to calculate pH of this type of the substance.

c. If you add an equal volume of 40 mM Na3EDTA to the EDTA solution in part a, what is
the pH?

Think about stoichiometry of the neutralization reactions. First step will be something like

2H₄EDTA + Na₃HEDTA -> 3NaH₃EDTA

Question is, how far neutralization will go. Assuming these reactions are the only ones responsible for the final concentrations of EDTA forms in the solution, you can calculate these concentrations, then use Henderson-Hasselbalch equation to find pH. You will be slightly off, but I don't see how to find an approximate answer otherwise. Forget about finding an exact answer on paper, system is too complicated for that.

[glossolalia]

Well, for part B, I've set up an ice chart, halving the initial concentrations as we are doubling the volume:

H₄EDTA:   i=.02M  c=-.02M  e=0
NaOH:      i=.04M  c=-.02M  e=.02M
H₃EDTA^-:  i=0      c=+.02M  e=.02M
Na⁺:        i=0      c=+.02M  e=.02M
H₂O:        i=0      c=+.02M  e=.02M

Is this a right first step?  Then repeat with the reaction of H₃EDTA- and the remaining NaOH?  Doesn't feel right...

[Borek]

No need for ICE table, just assume neutralization went to completion - and use stoichiometry to find out what you have.

[glossolalia]

OK. okokokok O-KAY.

I left this alone to focus on OChem.  Tell me what concept I am misunderstanding, please.

The way I see it (on Part B), you mix the two solutions giving you .02M H4EDTA and .04M NaOH, in effect.  The NaOH will neutralize all .02M of the H4EDTA, generating .02M NaH3EDTA in the process and leaving us with .02M NaOH.  This, in turn, will neutralize all .02M of the H3EDTA, generating .02M of Na2H2EDTA. 

From there, it's an easy weak monoprotic acid dissociation problem with a pKa of 6.3 giving a pH of 4.0.  This obviously isn't right.  Ugh.  Where am I screwing up?

[Borek]

generating .02M of Na2H2EDTA.

OK 

From there, it's an easy weak monoprotic acid dissociation problem

It is not. You have an amphiprotic substance, that tries to both dissociate and hydrolyze.

http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt

[glossolalia]

Beautiful - that was the basic concept I was having trouble with:  seeing the difference between neutralizing the first two hydrogens off the H4EDTA vs just dumping .02M of H2EDTA2- into some water.

Roundabout way I wound up arriving at the answer is that after neutralization, the solution is at the 2nd equivalence point, which is going to lie midway between pka2/3.

I appreciate it, Borek - I knew we'd get to the core concept I wasn't graspin eventually