NavyDDS1990

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Feb 14, 2014
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A compound has 2 ionizable groups with pka values of 6.20 and 9.50.
A 1.0M solution of this compound (100ml total) is found to have a pH of 6.8.
A biochemist adds 60ml of 1.0M HCl to this solution.

The solution is now pH:

the answer was 5.60, but I don't understand how

Can someone solve this with good explanation for me??

Thanks!!!
 
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Aug 5, 2015
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This is a rather complicated question but it's solvable.

Before we go and solve for the answer, I'll give a quick conceptual note. Think about an amino acid with a non ionizable R group (eg. glycine, where R= hydrogen atom). An amino acid contains an amine group (NH3) and a carboxylic acid (COO-). At a low pH (eg. pH =1), the amine group and the carboxylic acid are both in the protonated form. As we raise the pH, we first pass the pKa of the carboxylic acid and then pass the pKa of the amine group. At the pKa point of each functional group, the amino acid exists in equilibrium between the conjugate acid and conjugate base. Heres a diagram to clarify:


Okay so here is the solution:

pKa1 = 6.20 and pKa2 = 9.50

We know that the pH of the solution is 6.8, a value is close to pKa1 of the first ionizable group. When the pH is equal to the pKa of one of the ionizable groups, the compound exists in equilibrium between two forms. The two forms merely differ by the presence of an additional proton. Given for example:

HA <-> A- + H+

We also know that a buffer (hence, the use of the H-H equation) works ideally in the pH range of the buffer's pKa +/- 1.0. Since the pH of 6.8 is closer to pKa1 (6.20), we can safely ignore pKa2 (9.50). I've posted the rest of the explanation as an attachment. Leave a reply if you still need any help.

Edit: It wasn't stated that you could use a calculator but the given values in the question can be approximated to get you close to exact answer.
 

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Graffix

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May 12, 2015
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Ah... I get it! We assume the pKa2 group wouldn't change the pH significantly enough to make a change. Do you guys think questions like this are fair game and something we might expect on the real dat... ?
 
Aug 5, 2015
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Pre-Dental
Ah... I get it! We assume the pKa2 group wouldn't change the pH significantly enough to make a change. Do you guys think questions like this are fair game and something we might expect on the real dat... ?
We can ignore the second pKa since the pH of the solution containing the compound isn't close to 9.50. Suppose the pH of the solution was 9.30, we would have surpassed the first pKa (6.20) and the first ionizable group would be fully de-protonated. However, the second ionizable group on the compound would exist as acid/conjugate base and the H-H equation can be used to solve for the appropriate pH.

I don't expect this to come up on the real DAT (much more of a biochem question) but understanding it would go a long way!
 
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NavyDDS1990

5+ Year Member
Feb 14, 2014
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Ah... I get it! We assume the pKa2 group wouldn't change the pH significantly enough to make a change. Do you guys think questions like this are fair game and something we might expect on the real dat... ?
This is never going to be on the DAT, this is actually for my graduate biochemistry course.... I'm currently in formal post-bacc to make up for my poor undergrad GPA
 
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NavyDDS1990

5+ Year Member
Feb 14, 2014
1,041
919
Status
Dental Student
This is a rather complicated question but it's solvable.

Before we go and solve for the answer, I'll give a quick conceptual note. Think about an amino acid with a non ionizable R group (eg. glycine, where R= hydrogen atom). An amino acid contains an amine group (NH3) and a carboxylic acid (COO-). At a low pH (eg. pH =1), the amine group and the carboxylic acid are both in the protonated form. As we raise the pH, we first pass the pKa of the carboxylic acid and then pass the pKa of the amine group. At the pKa point of each functional group, the amino acid exists in equilibrium between the conjugate acid and conjugate base. Heres a diagram to clarify:


Okay so here is the solution:

pKa1 = 6.20 and pKa2 = 9.50

We know that the pH of the solution is 6.8, a value is close to pKa1 of the first ionizable group. When the pH is equal to the pKa of one of the ionizable groups, the compound exists in equilibrium between two forms. The two forms merely differ by the presence of an additional proton. Given for example:

HA <-> A- + H+

We also know that a buffer (hence, the use of the H-H equation) works ideally in the pH range of the buffer's pKa +/- 1.0. Since the pH of 6.8 is closer to pKa1 (6.20), we can safely ignore pKa2 (9.50). I've posted the rest of the explanation as an attachment. Leave a reply if you still need any help.

Edit: It wasn't stated that you could use a calculator but the given values in the question can be approximated to get you close to exact answer.
WOW Thank you so much for such a detailed explanation!!!!! I really appreciate your help.
I've already finished my DAT as you see under my signature.
This was actually for my graduate biochemistry course because I'm currently in my post-bacc program to make up for poor undergrad GPA.
But Thank you again very much!!!!! Makes total sense now
 
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