[jamesac]

Hi there! First time poster on this forum, and I was hoping someone here might be able to help me out.  In my biochemistry class, we recently did a quick review of acid-base and buffer calculations (specifically the Henderson-Hasselbalch equation) from gen chem and are now going over titrations of amino acids. In calculating the charge of each amino acid group, we were told the following rules for weak acid groups:

-if pH is 1 or more greater than the pKa, the charge is -1
-if pH is equal to pKa, the charge is -1/2
-if pH is 1 or more less than the pKa, the charge is zero

And the following rules for weak base groups:

-if pH is 1 or more greater than the pKa, the charge is about zero
-if pH is equal to pKa, the charge is 1/2
-if pH is 1 or more less than the pKa, the charge is 1

These are approximations that come from the HH equation (below).



I perfectly understand how these approximations were arrived at. However, while studying this amino acid material, I decided to go back over some buffer calculations and found something that has stumped me. Take this problem, for example, where we are told to calculate the pH when you add 100mL of 0.1M NaOH to 150mL of 0.2M acetic acid. I know that for these types of problems you typically just make an ICE table, find how much acid and base you end up with, and plug the values into the HH equation. The answer for this problem comes out to be 4.46.

Here's what's confusing me though. To make that 150mL solution of acetic acid, you would dilute acetic acid in water. But water has a pH of ~7, while acetic acid has a pKa of just 4.75. So according to the rules above, wouldn't an acid with a pKa of 4.75 almost entirely dissociate in a solution with a pH that is 2.25 higher, before you've even added the NaOH? I know that you can use Ka=(products/reactants) and solve for the amount of acid dissociation, which turns out to be a very small amount. But doesn't that formula then contradict the HH equation above? 

If someone could help explain where my thought process is failing, I would greatly appreciate it. I know how to do these types of problems, but I'd rather actually have a full understanding of what I'm doing so I can better remember it in the future! 

[Borek]

Here's what's confusing me though. To make that 150mL solution of acetic acid, you would dilute acetic acid in water. But water has a pH of ~7, while acetic acid has a pKa of just 4.75. So according to the rules above, wouldn't an acid with a pKa of 4.75 almost entirely dissociate in a solution with a pH that is 2.25 higher, before you've even added the NaOH?

7 is jus a starting pH of water, before adding acid. What happens to the pH of the solution when the acid dissociates?

[Babcock_Hall]

I am not entirely comfortable with your terminology description of charge.  What you call a weak acid I would prefer to call a weak, neutral acid, such as a carboxylic acid group within an amino acid (this is a situation in which the conjugate acid is neutral).  What you call a weak base, I think may refer to a weak, cationic acid, namely the ammonium group of an amino acid.  The reason to pay attention to the charges on the amino acid is better to understand the isoelectric point of amino acids and proteins, the pH at which they have no net charge.

[jamesac]

Here's what's confusing me though. To make that 150mL solution of acetic acid, you would dilute acetic acid in water. But water has a pH of ~7, while acetic acid has a pKa of just 4.75. So according to the rules above, wouldn't an acid with a pKa of 4.75 almost entirely dissociate in a solution with a pH that is 2.25 higher, before you've even added the NaOH?

7 is jus a starting pH of water, before adding acid. What happens to the pH of the solution when the acid dissociates?

I know the dissociated H+ ions from the acid would lower the pH of the water solution, but it would take quite a lot of H+ ions to decrease the pH down to 4.75, and wouldn't the acid keep dissociating until the pH reaches this point?

Edit: I've just realized I was wrong and it actually wouldn't take much H+ to lower the pH. I think I understand now.

So now, when we're titrating a weak acid with, say, NaOH, the -OH consumes the H+ as the acid goes to base. If it didn't, and acid just dissociated, would the pH actually go down since we'd have more H+ in solution? But as it works, -OH consumes H+, we get more base, which means pH slowly goes up (from HH equation). Then after all of the acid has been converted to base and we continue to titrate, pH will shoot up as we add more -OH to the solution. Have I got this understanding correct? I'm trying to tie everything together.

[jamesac]

I am not entirely comfortable with your terminology description of charge.  What you call a weak acid I would prefer to call a weak, neutral acid, such as a carboxylic acid group within an amino acid (this is a situation in which the conjugate acid is neutral).  What you call a weak base, I think may refer to a weak, cationic acid, namely the ammonium group of an amino acid.  The reason to pay attention to the charges on the amino acid is better to understand the isoelectric point of amino acids and proteins, the pH at which they have no net charge.

Yes, that's true, when I said weak base, I was really referring to the protonated conjugate acid of the base. We were just taught it this way so that we can use pKa values to solve for charge/isoelectric point since that's what we are given.

[Borek]

Edit: I've just realized I was wrong and it actually wouldn't take much H+ to lower the pH.

That's the point.

So now, when we're titrating a weak acid with, say, NaOH, the -OH consumes the H+ as the acid goes to base.

Convoluted (I am not sure what you mean by "acid goes to base" - neutralization?).

If it didn't, and acid just dissociated, would the pH actually go down since we'd have more H+ in solution?

Acid would not dissociate further than it already dissociated. Again, hard to say what you mean, but I would not bother with predicting hypothetical situations that are not following what we already know about the system, as they are - buy definition - unpredictable.

But as it works, -OH consumes H+, we get more base, which means pH slowly goes up (from HH equation). Then after all of the acid has been converted to base and we continue to titrate, pH will shoot up as we add more -OH to the solution. Have I got this understanding correct? I'm trying to tie everything together.

Yes, I think you got it correctly here.

[jamesac]

Thank you for the help, Borek. I think I've got a better understanding of how this works now. And yes when I mentioned the acid going to base I was just meaning that when we add more strong base we get dissociation of the weak acid and neutralization i.e. HA + -OH ---> H2O + A-