A 50mL solution of Histidine-HCl (Histidine: pKa1=1.8, pKa2=9.2, pKa3=6.0)is titrated with 0.500M NaOH. What is the expected pH of the histidine solution at the points in the titration when 14.0mL, 26mL, and 38mL of the NaOH titrant is added?
Here is a good tutorial on histidine HCl.
http://www.chem.uiuc.edu/CLCtutorials/104/AminoAcidTitrations/SeeIt.html
To determine the expected pH of the histidine solution at various points in the titration, we need to consider the acid-base properties of histidine and the stoichiometry of the reaction between histidine and NaOH.
Histidine is an amino acid with three ionizable groups: the amino group (pKa1), the alpha-carboxyl group (pKa2), and the imidazole group (pKa3).
At the start of the titration, before any NaOH is added, the histidine is in its fully protonated form as Histidine-HCl. The pH of this solution can be calculated by considering the ionization of the first proton from the amino group:
pKa1 = 1.8
When no NaOH is added, the concentration of Histidine-HCl is equal to the initial concentration of the histidine solution. Let's assume it's 0.500M.
Using the Henderson-Hasselbalch equation:
pH = pKa1 + log([A-]/[HA])
where [A-] is the concentration of the conjugate base and [HA] is the concentration of the acid, we can calculate the pH. Since no NaOH has been added, the concentration of the conjugate base [A-] is 0 and the concentration of the acid [HA] is 0.500M.
pH = 1.8 + log(0/0.500) = 1.8
Therefore, at the beginning of the titration, the expected pH of the histidine solution is 1.8.
Now, let's determine the pH at the points in the titration when 14.0mL, 26.0mL, and 38.0mL of the NaOH titrant is added.
To calculate the pH at these points, we need to determine the amount of Histidine-HCl that has been neutralized by the NaOH. We can use the stoichiometry of the reaction between histidine and NaOH to do this.
The balanced reaction is as follows:
Histidine-HCl + NaOH → Histidine + NaCl + H2O
From the balanced equation, we can see that one mole of Histidine-HCl is neutralized by one mole of NaOH. Therefore, the amount of Histidine-HCl neutralized is equal to the amount of NaOH added.
Now, let's calculate the amount of Histidine-HCl neutralized for each volume of NaOH added.
For 14.0mL of NaOH:
Amount of Histidine-HCl neutralized = (14.0mL) x (0.500mol/L) = 7.00mmol
For 26.0mL of NaOH:
Amount of Histidine-HCl neutralized = (26.0mL) x (0.500mol/L) = 13.0mmol
For 38.0mL of NaOH:
Amount of Histidine-HCl neutralized = (38.0mL) x (0.500mol/L) = 19.0mmol
To find the remaining concentration of Histidine-HCl, subtract the amount neutralized from the initial concentration:
For 14.0mL of NaOH:
Remaining concentration of Histidine-HCl = (0.500mol/L) - (7.00mmol / 0.050L) = 0.36M
For 26.0mL of NaOH:
Remaining concentration of Histidine-HCl = (0.500mol/L) - (13.0mmol / 0.050L) = 0.26M
For 38.0mL of NaOH:
Remaining concentration of Histidine-HCl = (0.500mol/L) - (19.0mmol / 0.050L) = 0.12M
Once we have the remaining concentration of Histidine-HCl, we can calculate the pH using the Henderson-Hasselbalch equation.
For 14.0mL of NaOH:
pH = pKa1 + log([A-]/[HA]) = 1.8 + log(0.0 / 0.36) = 1.8 + log(0) = 1.8
For 26.0mL of NaOH:
pH = pKa1 + log([A-]/[HA]) = 1.8 + log(0.0 / 0.26) = 1.8 + log(0) = 1.8
For 38.0mL of NaOH:
pH = pKa1 + log([A-]/[HA]) = 1.8 + log(0.0 / 0.12) = 1.8 + log(0) = 1.8
Therefore, at the points in the titration when 14.0mL, 26.0mL, and 38.0mL of the NaOH titrant is added, the expected pH of the histidine solution remains at 1.8.