Calculating pKa of X's Second Ionizable Group

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In summary, a solution of an unknown compound, X, with a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 and 8, had 75 mL of 0.1 M NaOH added to it. This caused the pH to increase from 2.0 to 6.72. Using the Henderson-Hasselbalch equation, the pKa of the second group of X was calculated to be 7.2, contradicting the answer given in the book of 7.3. The use of accurate nomenclature in biochemistry problems can lead to discrepancies between solutions and textbook answers.
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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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  • #2
7.2 seems OK. Approach seems OK. Book seems wrong.
 
  • #3
which text are you using? I'll look at this problem later, just busy at the moment.
 
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  • #4
I'm using lehninger, Principles of Biochemistry... Not enjoying it much, the problems that is.
 
  • #5
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.
 
  • #6
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".
 

What is pKa and why is it important for calculating the second ionizable group of a molecule?

PKa is a measure of the acidity of a molecule, specifically the dissociation constant of the molecule's second ionizable group. It is important because it helps determine the strength of the molecule's acid or base properties, which can impact its chemical reactivity and behavior.

How do you calculate pKa for a molecule's second ionizable group?

To calculate the pKa of a molecule's second ionizable group, you need to know the concentration of the molecule in its acidic and basic forms, as well as the corresponding pH values. Using the Henderson-Hasselbalch equation, you can then solve for the pKa value.

What factors can affect the accuracy of pKa calculations for a molecule's second ionizable group?

The accuracy of pKa calculations can be affected by various factors such as temperature, solvent, and ionic strength. These factors can alter the molecule's acidity and basicity, leading to discrepancies in the pKa value.

How does the pKa of a molecule's second ionizable group impact its chemical behavior?

The pKa value of a molecule's second ionizable group can greatly impact its chemical behavior. A lower pKa value indicates a stronger acid, while a higher pKa value indicates a weaker acid. This can affect the molecule's reactivity, solubility, and other properties.

Can pKa values be experimentally determined for a molecule's second ionizable group?

Yes, pKa values can be experimentally determined through methods such as titration, spectrophotometry, and NMR spectroscopy. These techniques involve measuring the concentration of the molecule in its acidic and basic forms and solving for the pKa value using mathematical equations.

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