Analytical Chemistry question: Find the pH and the concentration of each species of lysine in a solution of 0.0010 M lysine*HCL.
I know that the pKAs of lysine are
(1.77)(CO2H)
9.07 (�-NH3)
10.82 (�-NH3)
but I am not sure where to go from there.
Note: I am unfamiliar with how the alha amino acids dissociate so look at what follows with a skeptical eye. Here is what I think.
With a concn of 0.001 the hydrochloride will consist of the fully protonated amino acid plus the chloride ion. You can assign 0.001 to (Cl^-) right off the bat.
Then you note that the pKa2 and pKa3 is so much higher than pKa1 so the ionization of the first group (the COOH) can be calculated and ignore that from pKa2 and pKa3. Thus, treat the first ionization just like you would a simple monoprotic acid, HA
........HA ==> H^+ + A^-
I....0.001.....0......0
C......-x......x......x
E....0.001-x...x.......x
Solve for x and convert to pH. Undoubtedly you will need to use the quadratic equation. Here is a web site that may prove useful to you. The information on this site does not ignore the other species as I have done.
http://www.cmu.edu/bio/education/courses/03310/HomeworkProblems/ProblemC1.pdf
To find the pH and concentration of each species of lysine in a solution, we need to first understand the acid-base properties of lysine and how it behaves in water. Lysine is an amino acid and has multiple ionizable functional groups: the carboxylic acid group (CO2H) and two amino groups (-NH3+).
Let's break down the steps to solve this problem:
Step 1: Write the dissociation reactions for each ionizable group in lysine.
The carboxylic acid group (CO2H) can dissociate into H+ and the carboxylate ion (CO2-):
CO2H ⇌ H+ + CO2-
The first amino group (-NH3+) can also lose a proton (H+) and become an ammonium ion (NH4+):
-NH3+ ⇌ H+ + NH3
The second amino group (-NH3+) can also lose a proton (H+) and become an ammonium ion (NH4+):
-NH3+ ⇌ H+ + NH3
Step 2: Determine the predominant species at a given pH.
Based on the given pKa values, we can determine the predominant species of each ionizable group at different pH values. The pH at which a species is half-protonated and half-deprotonated is equal to its pKa.
For the carboxylic acid group (CO2H):
pH < pKa (1.77) → CO2H is protonated (predominant form)
pH > pKa (1.77) → CO2- is deprotonated (predominant form)
For the first amino group (-NH3+):
pH < pKa (9.07) → -NH3+ is protonated (predominant form)
pH > pKa (9.07) → NH3 is deprotonated (predominant form)
For the second amino group (-NH3+):
pH < pKa (10.82) → -NH3+ is protonated (predominant form)
pH > pKa (10.82) → NH3 is deprotonated (predominant form)
Step 3: Calculate the concentration of each species.
Given that the initial concentration of lysine•HCl is 0.0010 M, we need to calculate the concentration of each species at equilibrium.
To simplify the calculations, let's assume the amount of lysine•HCl that actually dissociates is negligible compared to the initial concentration.
For the carboxylic acid group (CO2H):
At equilibrium, the concentration of CO2H is given by the initial concentration minus the concentration of H+:
[CO2H] = 0.0010 M - [H+]
For the first amino group (-NH3+):
At equilibrium, the concentration of -NH3+ is given by the initial concentration minus the concentration of H+:
[-NH3+] = 0.0010 M - [H+]
For the second amino group (-NH3+):
At equilibrium, the concentration of -NH3+ is given by the initial concentration minus the concentration of H+:
[-NH3+] = 0.0010 M - [H+]
Step 4: Find the pH.
To find the pH, we need to determine the concentration of H+ ion. Since lysine is a weak acid, we can assume that the change in concentration of H+ is negligible compared to its initial concentration (i.e., [H+] ≈ 0). Thus, the concentration of H+ can be considered negligible and can be ignored in the calculations.
Now, you can apply these steps to calculate the pH and concentrations of each species at specific pH values or experimentally determined pH values of the solution.