Calculating pKa of X's Second Ionizable Group

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The discussion focuses on calculating the pKa of the second ionizable group of compound X, which is believed to have a carboxyl group with a pKa of 2.0 and another group with a pKa between 5 and 8. After adding 75 mL of 0.1 M NaOH to a 100 mL solution of X at pH 2.0, the pH increased to 6.72. The calculated pKa of the second group is 7.2, while a reference book suggests 7.3. The discrepancy arises from differing assumptions about the dissociation of the second group during the reaction.

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An unknown compound, X is thought to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 and 8. When 75 mL of 0.1 M NaOH was added to 100mL of a 0.1 M solution of X at pH 2.0, the pH increased to 6.72. Calculate the pKa of the second group of X.

ok, here's another problem. I got the answer of 7.2 while the answer in the back says 7.3.

I just found out that i was actually wrong in my other post of "Dissociation of glycine". This book is pretty accurate as far as the answers go and when we're dealing with logs, .1 is still quite a big difference.

Here's how i solved it:

i knew that since pKa = pH at the beginning there must be equal amounts of X carboxyl group and it's ionized form. Since there were .01 moles in total there must be .005 moles of X carboxyl group.

When .0075 moles of OH(-) were added the carboxyl group completely dissociated and there were still .0025 moles of OH(-) left in the solution.

there were .01 moles of X in the beginning, and I'm assuming that none of the second group has been dissociated. The remaining OH(-) will react with the .01 moles X-second group, producing .0025 moles of the ionized form of X-second group, while only 0.0075 moles of X-second group remain.

Thus we go back the henderson hasselbach equation of

6.72 = pKa + log(.0025/.0075) and solve for pKa

i get 7.2. However, I'm thinking that the book could be wrong, in that it assumed that the .01 moles of X-second group did not react, and thus

6.72 = pKa + log(0.0025/.01) will be roughly 7.3.

Can some one explain this. Thanks.
 
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7.2 seems OK. Approach seems OK. Book seems wrong.
 
which text are you using? I'll look at this problem later, just busy at the moment.
 
Last edited:
I'm using lehninger, Principles of Biochemistry... Not enjoying it much, the problems that is.
 
That's what I thought, biochemistry. A heed of warning, be very careful about nomenclature, most biochem teachers especially professors are quite careless, they'll expect you to understand their perspective and not necessairly what's the most accurate. For your other problem, "glycine" meant the molecule as a whole, the IUPAC nomenclature for the protonated forms would be to tedious.

At the moment, I'm getting 7.2 for the answer, I may be missing something, I'll get back to you later.
 
I also work with "Lehningers priciples of biochemistry" and frankly some of the problems are just a b*tch. Unfortunately the solutions arent always perfect either...

Suffice it to say: "you right, book wrong".
 

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