Consider the titration of a 50.0 mL 0.500 M F- with 0.500 M HCl. The pKa of HF = 3.46 and the pKb of F- is 10.54.

Calculate the pH at the following volumes of HCl added.
1. Vsa= 0.00 mL of 0.500 M HCl
2. Vsa= 25.0 mL of 0.500 M HCl
3. Vsa= 49.5 mL of 0.500 M HCl
4. Vsa= 50.0 mL of 0.500 M HCl

millimols F^- initially = mL x M = 50 x 0.5 = 25

At zero mL HCl added, the pH is determined by the hydrolysis of F^-
......F^- + HOH ==> HF + OH^-
I....0.5M...........0.....0
C.....-x............x.....x
E.....0.5-x.........x.....x

Kb = approx E-11 = (x)(x)/(0.5-x)
Solvr for x = (OH^-) and convert to pH.

At 25 mL HCl you have added 12.5 mmols HCl. The reaction is
......F^- + H^+ ==> HF
I.....25.....0.......0
add.........12.5........
C....-12.5..-12.5....+12.5
E.....12.5...0.......12.5

Substitute this into the Henderson-Hasselbalch equation. The pH, when you get it worked out, will be 3.46

is done the same way but of course the answer is not 3.46.

4. You will have converted all of the F^- to HF so (HF) = mmols HF/total mL. You work that as a weak acid problem.
Post your work if you get stuck.

To calculate the pH at each volume of HCl added, we need to determine the concentrations of the remaining species in the solution, namely, the amount of unreacted HF and the concentration of the resulting H3O+ ions.

Here's how you can calculate the pH at each volume of HCl added:

1. Vsa = 0.00 mL of 0.500 M HCl:
When no HCl is added, the initial concentration of F- is the same as the concentration of the original solution, which is 0.500 M. Since the pKb of F- is given, we can use the equation: pKw = pKa + pKb. The pKw (water auto-ionization constant) is 14, so we can solve for pKb using the given pKa value: pKb = pKw - pKa. Once we have pKb, we can calculate Kb (base dissociation constant) using the equation: Kb = 10^(-pKb). Finally, we can calculate the concentration of OH- using the equation: OH- = sqrt(Kb * [F-]), where [F-] is the initial concentration of F-. Since the solution is initially neutral, we know that [H3O+] = [OH-]. Therefore, we can use the equation: pH = -log[H3O+] to calculate the pH.

2. Vsa = 25.0 mL of 0.500 M HCl:
At this point, 25.0 mL of the 0.500 M HCl has been added. This will react with an equal volume of 0.500 M F- solution. To determine the amount of unreacted HF, we can use the balanced equation: HF + HCl -> H2O + F- + Cl-. Since the stoichiometry is 1:1, the amount of unreacted HF will be equal to the initial concentration minus the amount reacted from the volume of HCl added. The volume of the remaining solution is now 50.0 mL - 25.0 mL = 25.0 mL. Using the concentration of the remaining HF, we can calculate the concentration of F- using the stoichiometry of the reaction. Once we know the concentration of F-, we can follow the same steps as in the first scenario to calculate the pH.

3. Vsa = 49.5 mL of 0.500 M HCl:
In this case, 49.5 mL of the 0.500 M HCl has been added. By using the same reasoning as in scenario 2, we can determine the amount of unreacted HF and the resulting concentrations of F- and H3O+. Since the reaction is nearing completion, the concentration of HF will be much smaller, and therefore, the concentration of F- and H3O+ will be higher compared to previous scenarios.

4. Vsa = 50.0 mL of 0.500 M HCl:
When the total volume of the added HCl matches the volume of the original F- solution (50.0 mL), the reaction is complete. In this case, all of the HF has reacted, and the resulting solution will consist entirely of the Cl- ions from the HCl. Therefore, the pH will be determined solely by the presence of the Cl- ions.

By following these steps and calculations, you can determine the pH at each volume of HCl added in the titration of F- with HCl.