Biochemistry: Henderson-Hasselbalch Problem

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Discussion Overview

The discussion centers around a homework problem involving the titration of a fully protonated peptide (V-A-Y-K-H) to a specific pH of 4.00 using the Henderson-Hasselbalch equation. Participants explore the necessary calculations to determine the moles of base required for this titration.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • One participant states the problem and provides the relevant equation, indicating the pKa of the alpha COOH group and its role in the calculation.
  • Another participant suggests that the initial concentration of the peptide is needed to calculate the amount of base required, linking it to the stoichiometry of neutralization.
  • A third participant interprets the ratio of A/HA, explaining that at pH 4.00, the carboxyl group will be nearly fully deprotonated, estimating that just under one mole of base will be needed.
  • A later reply acknowledges a previous oversight regarding the question's focus on moles rather than concentration.

Areas of Agreement / Disagreement

Participants express differing views on the necessary information for solving the problem, particularly regarding the need for initial concentration versus the moles of base required. The discussion remains unresolved on how to directly calculate the moles of base needed from the given ratio.

Contextual Notes

The discussion does not clarify the assumptions regarding the initial concentration of the peptide or the specific conditions under which the calculations are made. There is also an unresolved aspect regarding the transition from the ratio of A/HA to the actual moles of base required.

cookiebookie
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Homework Statement



How many moles of base would be needed to titrate one mole of this fully protonated peptide to a pH of 4.00?

This polypeptide is V-A-Y-K-H.

Homework Equations



Henderson-Hasselbalch: pH= pKa + log[A]/[HA]

pKa of alpha COOH = 2.4
The alpha COOH (at the C-Terminus) is the only portion of the molecule that is deprotonated (basic) at pH 4.00, so we use its pKa for the calculation.

The Attempt at a Solution



pH = pKa + log [A]/[HA]
4.0 = 2.4 + log [A]/[HA]
1.6= log [A]/[HA]
39.8 = A/HA

I'm not sure how to go from the ratio of A/HA to the moles of base needed.
 
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You need initial concentration of peptide. Then, knowing final ratio, you can easily calculate concentration of the neutralized peptide and amount of NaOH can be calculated from the neutralization stoichiometry.
 
What that
cookiebookie said:
39.8 = A/HA

is saying is a way of saying what should be evident from the fact that the pH 4 is substantially (by 1.6 units) above the pK, so the carboxyl will be almost all deprotonated (to within about 2.5% in fact) so the number of moles required is just under 1.

(Sorry Borek, the question was about moles so you do not need concentration.)
 
Yep, missed the "one mole" part. Sorry about that.
 

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