1. Buffers and Henderson-Hasselbalch
Chapter 15, ppt 1
Buffers and Henderson-Hasselbalch

2. Buffer - A Definition
Solutions that contain components that resist changes in pH
Imperative to living things which require specific pH ranges (ex: blood = 7.4 even during creation of lactic acid…)

3. Solutions of Acids/Bases Containing a Common Ion
Weak acid (HA) is present in solution along with a salt (NaA --> Na and I)
Salt will break up (strong electrolyte)
HA will determine the pH
HF(aq) <-> H+(aq) + F-(aq)
The common ion (in this case F-) will shift equilibrium LEFT (Le Chatelier’s principle)
Fewer H+ ions present, higher/more basic pH

4. NH3 + H2O <-> NH4+ + OH-
Example
Which way will the equilibrium shift if NH4Cl is added to a 1.0 M NH3 solution? How will this affect the pH?
NH4Cl -> NH4+ + Cl-
NH3 + H2O <-> NH4+ + OH-
The creation of NH4+ ions will shift the second reaction to the LEFT. This will lower the [OH-] and increase the [H+] and pH.

5. Example #2 Calculate the pH and the percent dissociation of the acid
0.200 M solution of HC2H3O2 (Ka = 1.8 X 10-5)
Answer: pH = 2.7, 0.95%
0.200 M HC2H3O2 in the presence of M NaC2H3O2
Answer: pH = 5.14, 3.6 X 10-3%

6. Change From Chapter 14 Problems
The initial concentration of the products will not be zero in the presence of the salt
**ALWAYS write the major species in solution!!
Example: The equilibrium [H+] in a 1.0 M HF solution is 2.7 X 10-2 M, and the percent dissociation of HF is 2.7%. Calculate the [H+] and the percent dissociation of HF in a solution containing 1.0 M HF (Ka = 7.2 X 10-4) and 1.0 M NaF.
Answer: [H+] = 7.2 X 10-4, 0.072% dissociation

7. Buffered Solutions
A buffered solution is one that resists a change in its pH with addition of OH- ions or protons
BLOOD (HCO3- and H2CO3)
Weak acid and its salt (ex HF and NaF)
Weak base and its salt (ex NH3 and NH4Cl)
Same calculation approach as chapter 14

9. Adding Acid to a Solution (HA and A-)
If acid (H+) is added instead of OH-, the reaction that will occur is
H+ + A- --> HA
The H+ ions are replaced by HA, which won’t change the pH very much.

10. But WHY?
When a weak acid (HA) dissociates, HA --> H+ + A- the corresponding equilibrium expression is Ka = [H+][A-]/[HA]. In order to determine pH, the equation should be rearranged: [H+] = Ka [HA]/[A-]

11. A buffered solution may contain weak acid (HA) and it’s conjugate base (A-). When a base (OH-) is added, the reaction that occurs is
OH- + HA --> A- + H2O
The added OH- ions react with HA to make A- ions. With less HA and more A-, the equilibrium expression from the last slide is affected. The [H+] is reduced which will change the pH. The reason why it doesn’t affect the pH much is because [HA] and [A-] are very large compared to the amount of OH- added. The change in [HA]/[A-] will be small and won’t affect the pH much.

12. Henderson-Hasselbalch Equation
[H+] = Ka [HA]/[A-] will be very helpful in determining [H+] and pH in buffered solutions…can be manipulated to make:
pH = pKa + log ([A-]/[HA]) or
pH = pKa + log ([base]/[acid])
NOTE: In a particular buffering system, all solutions that have the same ratio of [A-] to [HA] will have the same pH.
ICE table assumption for x is generally accepted when using this equation

13. Example
Calculate the pH of a solution containing 0.75 M lactic acid (Ka = 1.4 X 10-4) and 0.25 M sodium lactate. Lactic acid (HC3H5O3) is a common constituent of biologic systems. For example, it is found in milk and is present in human muscle tissue during exertion.
Answer: 3.38