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!