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

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…)

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

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.

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 0.500 M NaC2H3O2 Answer: pH = 5.14, 3.6 X 10-3%

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

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

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.

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-]

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.

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

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