1. A Different Look At Buffers
Write the equation for the dissociation of hydrogen fluoride. (B-L perspective)
HF(aq) + H2O(l) ↔ H3O+(aq) + F-(aq)
Write the Ka expression:
Ka = [H3O+] [F-] / [HF]
= 7.2 x 10-4
What is the Ka value?
Small
What does that mean about the [F-]?
Small
What about the [H3O+]?
Small
Can you calculate those concentrations with the info given?
NO!
If you added a quantity of strong acid to the solution would LCP kick in?
Yes, but not too much!
Would the pH of the solution change?
Yes, pH ↓

2. A Different Look At Buffers (cont’d)
HF(aq) + H2O(l) ↔ H3O+(aq) + F-(aq)
How would you describe the substance NaF in terms of this system?
A salt of the conjugate base!
If I added some NaF, would LCP kick in?
Yes, but not too much!!
What would happen to the pH of the solution as a result of the addition?
pH ↑
Rearrange the mass action expression to solve for [H3O+]?
[H3O+] = Ka ([HF] / [F-])
If I added enough NaF to get to the [F-] to be the same as the [HF] that would be significant…how come?
[H3O+] = Ka
Another look… (take –log of both sides of this):
-log [H3O+] = -log (Ka ([HF] / [F-]))
pH = -log Ka + -log ([HF] / [F-])
pH = pKa + -log ([HF] / [F-])
pH = pKa + log [F-]
Henderson-Hasselbalch Eq.
[HF]

3. A Different Look At Buffers (cont’d)
HF(aq) + H2O(l) ↔ H3O+(aq) + F-(aq)
What is the pH of a solution of 1.5 M HF?
7.2 x 10-4 = [H3O+] [F-] / [HF]
7.2 x 10-4 = (x) (x) / (1.5-x)
x = [H3O+] = 0.033
pH = 1.48
What would the pH of a solution described as 1.5 M HF and 1.3 M NaF?
Use Henderson-Hasselbalch:
neglect these
at eq: [F-] = x
pH = pKa + log [F-]
[HF]
[HF] = x
= -log 7.2 x log ([1.3] / [1.5])
= 3.1 – = 3.0

4. A Different Look At Buffers (cont’d)
HF(aq) + H2O(l) ↔ H3O+(aq) + F-(aq)
What if 3.00 mL of KOH (0.90 M) was added to 75.0 mL of the buffer (1.50 M HF; 1.30 M NaF). What would happen to the pH of the solution?
First do the stoichiometry…
Moles of OH- added = mol
Moles of HF initially = mol
Moles of F- initially = mol
Moles of HFafter reaction with OH-= mol
Moles of F-after reaction with OH-= mol
[HF] before equilibrium = mol/0.078 L = 1.41 M
[F-] before equilibrium = mol/0.078 L = 1.28 M
Use Henderson-Hasselbalch:
at eq: [F-] = x
pH = pKa + log [F-]
[HF]
[HF] = x
neglect these
= -log 7.2 x log ([1.28] / [1.41])
= 3.1 – = 3.1

5. A Different Look At Buffers (cont’d)
Summary
pH of Weak Acid =
1.48
pH of Buffer =
3.0
pH of Buffer =
3.1
(after addition of strong base)
pH = pKa + log [B-]
[HB]
Buffers work best when:
The [ ]’s are large for both acid and the conjugate base
The ratio of base to acid is close to 1.
Teacher note: follow this up with video on buffering found in chapter 14 folder