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Amino Acid Titration

  • #1

Homework Statement


How many moles of OH–
are required to raise the pH of 4.00 liters of 25 mM glycine, pH 1.90, to a final pH of
9.70? Assume that the addition of OH–
does not change the volume of the solution.



Homework Equations



I know at some point I will need to use the henderson hasselbalch equation, which is:

pH = pKa + log [Base]/[Acid].

The Attempt at a Solution



I started by using the final pH of 9.70, the average pKa of glycine (the pKa of the isoelectric point) which is 5.97, and converting 25mM into 0.025M. This gives the equation of

9.70 = 5.97 + log [Base]/[0.025-Base]

If we set the Base equal to x. we get:

9.70 = 5.97 + log [X]/[0.025-X]

Which becomes

3.73 = log [X]/[0.025]

I am stuck here. My teacher gave us the answer as 0.12 mol of OH is needed, but there was no work shown on how to get there. Any help would be greatly appreciated. Thank you.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Borek
Mentor
28,448
2,844
IMHO you need to treat glycine as a diprotic acid. Are you given two pKa values? Isoelectric point is of no use here.

Problem is, I am unable to reproduce 1.9 as pH of the 25 mM solution. Several sources I consulted are quite consistent, listing 2.34 or 2.35 for pKa1, and these values yield pH of 2.06 or 2.07 - but not 1.9.

Using Henderson-Hasselbalch equation when you are over 3 pH units from pKa is asking for troubles.
 
  • #3
epenguin
Homework Helper
Gold Member
3,725
769

Homework Statement


How many moles of OH–
are required to raise the pH of 4.00 liters of 25 mM glycine, pH 1.90, to a final pH of
9.70? Assume that the addition of OH–
does not change the volume of the solution.



Homework Equations



I know at some point I will need to use the henderson hasselbalch equation, which is:

pH = pKa + log [Base]/[Acid].

The Attempt at a Solution



I started by using the final pH of 9.70, the average pKa of glycine (the pKa of the isoelectric point) which is 5.97, and converting 25mM into 0.025M. This gives the equation of

9.70 = 5.97 + log [Base]/[0.025-Base]

If we set the Base equal to x. we get:

9.70 = 5.97 + log [X]/[0.025-X]

Which becomes

3.73 = log [X]/[0.025]

I am stuck here. My teacher gave us the answer as 0.12 mol of OH is needed, but there was no work shown on how to get there. Any help would be greatly appreciated. Thank you.

Homework Statement





Homework Equations





The Attempt at a Solution

Can you jusfy using the average pK? At pH 5 or 4 what is the charge state of the glycine overwhelmingly, and at pH 7 or 8 what is it? Is it reasonable it take a mole or two of alkali per mole of glycine to get from one to the other?

You have 0.1 moles of glycine there I think; I find it better to think on a 'per mole' basis, but as you prefer.

Per mole glycine, at pH 1.9 how many moles of the carboxyl group (pK 2.3) are in the -COO- form? And how many at pH 9.7? For second question approximate answer gives more insight than precise! But calculate that if you want to be sure. Then how many moles OH- needed to get from one to the other pH?

Per mole glycine at pH 1.9 how many moles per mole glycine are in state -NH3+? Approximately! :wink: And how many at pH 9.7? (Check - this pH is close to the pK of 9.6. So your answer to 2nd question will have to be not too far from 1/2. Won't it? :wink: ) So how many moles OH- to get from one state to the other?
 
Last edited:
  • #4
So I after doing more research it sounds like I need to use two Henderson Hassalbach equations, one for each pH. However, I don't know how to proceed from there. Do I subtract the values of base to get the difference?

Borek: On the test we will be given a table with pKa values, but here I used google to find them.

Epenguin: We have 0.1 moles of glycine? I'm probably just being dumb but wouldn't it be 0.025M/4L = 0.00625? I realize i used the wrong value in the work I showed.
 
  • #5
Borek
Mentor
28,448
2,844
So I after doing more research it sounds like I need to use two Henderson Hassalbach equations, one for each pH. However, I don't know how to proceed from there. Do I subtract the values of base to get the difference?
Think in terms of diprotic acid. You start with a pH 1.9 mixture of H2A/HA- (calculate concentrations of both from the HH equation). You end with a pH 9.7 mixture of HA-/A2- (again, use HH equation to find their concentrations). Then, just assume you need a simple neutralization stoichiometry to move from initial solution to the final one.

Epenguin: We have 0.1 moles of glycine? I'm probably just being dumb but wouldn't it be 0.025M/4L = 0.00625? I realize i used the wrong value in the work I showed.
Huh?

[tex]C=\frac n V[/tex]

Solve for n.

I don't like this question. I don't like questions that don't clearly state what is present in the solution. In this particular case I was confused by the initial pH. "Solution of glycine" is not the same thing as "solution of glycine acidified with whatever to pH 1.9". And in fact if we don't know what was used to acidify the solution, we can't be sure our answer will be correct (it will only if a strong acid was used).
 
  • #6
epenguin
Homework Helper
Gold Member
3,725
769
So I after doing more research it sounds like I need to use two Henderson Hassalbach equations, one for each pH. However, I don't know how to proceed from there. Do I subtract the values of base to get the difference?

Borek: On the test we will be given a table with pKa values, but here I used google to find them.

Epenguin: We have 0.1 moles of glycine? I'm probably just being dumb but wouldn't it be 0.025M/4L = 0.00625? I realize i used the wrong value in the work I showed.
Borek does not like this slipshod laboratory slang and I agree. It would have been clear and cost little effort if she had just said "brought to pH 1.9 with strong acid" which we had to psych - unfortunately you will often meet this including in scientific papers.

I don't much like the Henderson-Hasselbalch equation either for beginners as it is like a magic formula or spell trusted to at the expense of better understanding of the situations in solutions. But whether you use [H+] and Ka or pH and pK, you need to know the moles of -COOH at the two pH's. The difference equals the moles of OH- you have to add to deprotonate that carboxyl group from what it is at the lower pH to what it is at the higher OK? (OK I said -COO- last time but if you know one you'll know the other won't you?)

Likewise you have to know the moles of -NH2 (or of -NH3+) at the two pH's to know the moles of deprotonating OH- so to speak. The moles of OH- to add will be the sum of those two differences I hope you understand. In fact you are possibly saying the same thing as me in different words.

So that is four numbers of moles you need to know and can calculate. Two of them you should be able to say off the top of your head to quite sufficient exactness, but if not calculate them and then you will see why it should have been easy.

For the last question don't take my word. :smile: Should you divide or multiply by 4? Give some reasoning.
 
  • #7
So after trying again, I used these equations:

1.90 = 2.34 + log [Base]/[0.1] --> 10^-.44 = [Base]/[0.1] --> 0.036 = Base

and

9.70 = 9.60 + log [Base]/[0.1] --> 10^0.1 = [Base]/[0.1] --> 0.126 = Base.

However, 0.126- 0.036 = 0.09.

I believe I am using the wrong [Base] value for the second HH equation, but I do not know how to find it.
 
  • #8
epenguin
Homework Helper
Gold Member
3,725
769
So after trying again, I used these equations:

1.90 = 2.34 + log [Base]/[0.1] --> 10^-.44 = [Base]/[0.1] --> 0.036 = Base

and

9.70 = 9.60 + log [Base]/[0.1] --> 10^0.1 = [Base]/[0.1] --> 0.126 = Base.

However, 0.126- 0.036 = 0.09.

I believe I am using the wrong [Base] value for the second HH equation, but I do not know how to find it.
You seem to agree now that there are 0.1 moles of glycine.(?)

Usually square [] brackets are used for molarity not for moles, but mole ratios and molarity ratios at least are the same so you might get away with something like your equations if you knew what you were doing and said it clearly.

However, confirming my reservations above about the HH equation you have misinterpreted the meaning of terms in it.

The base, whose molarity appears in the HH equation is the base form, the form that can accept a proton - in the case of the carboxyl group it is the molarity of -COO-, [-COO-] we might call it or [glyCOO-]. It is not the total molarity of glycine in all its forms. Unless these two things were practically equal. Which they are in one of the four cases I mentioned which you have not considered yet.

So try again, possibly using my previous post as guide.

The whole thing is really no different than if you had a mixture of 0.1 moles of an acid of pKa 2.3 at pH 1.9 that you had to bring to 9.7, and 0.1 moles of a base of pKa 9.6 for which you had to do the same thing. So the problem is a pretty minor extension of stuff with acids and bases you must have done in the very previous lesson or chapter.

May I say your work would benefit from setting out more clearly and explicitly what you are doing. Probably you would make fewer mistakes seeing better what your own reasoning is, but also if you make a mistake and got a thing half right you would get half a credit instead of zero if a Prof finds it too hard to follow; not least you would understand and get recall from your own notes or worked examples when you come to revise later.
 
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