You have a solution containing 1 mol each of two unknown amino acids in a mixture at pH6.5 (pKa=6.0). What would be the pH if you added 0.2 mol of NaOH?
I've used the Henderson-Hasselbalch to get this:
6.5 = 6.0 + log [B/A]
10^.5 = [B/A]
3.16/1 = [B/A]
What do I do next?
After finding the ratio of [B/A] to be 3.16/1, you can use this ratio to calculate the concentrations of the two amino acids in the mixture.
Let's denote the initial concentration of the two amino acids as [A] and [B] respectively. Since we have 1 mole of each amino acid initially, [A] = 1M and [B] = 1M.
Using the ratio 3.16/1, we can calculate the new concentrations of the amino acids after adding 0.2 mol of NaOH. Let's assume x represents the change in concentration for both amino acids.
So, the new concentrations of [A] and [B] after adding NaOH would be:
[A] = 1 - x
[B] = 1 + 3.16x
Now, let's consider the reaction that occurs when NaOH is added. NaOH is a strong base that will react with the acidic amino acids, causing the deprotonation of the carboxyl group (-COOH). This results in an increase in the concentration of the base form (B) and a decrease in the concentration of the acidic form (A) for both amino acids.
Since only NaOH is added, the total concentration of the amino acids remains the same. Therefore, we can write the equation:
[A] + [B] = 1 - x + 1 + 3.16x = 2
Simplify the equation:
4.16x = 2 - 2
4.16x = 0
x = 0
This means that the concentrations of the amino acids do not change after the addition of NaOH. Therefore, the pH of the solution will remain the same at pH 6.5.
A+B=2 moles
Based upon your calculations, 2*0.316=0.632 moles of A
2-0.632=1.37 moles of B
pH=pka+log[moles of B/moles of A]=6.0+log[moles of B/moles of A]
****Check my thinking. Not sure about the problem since you have to take into account the isolectric points of the amino acids