University:
University of Houston
Course:
PHYS 4356 | Particle and Nuclear Physics
Academic year:

2022
Views:

30

Pages:

16
Author:

Rebekah Woods

Observation If you add .01 ml i.e 1/100 ml of 1M HCl to 1000 ml of water, the pH of the water drops from 7 to 5!! i.e 100 fold increase in H+ concentration: Log = 2 change. Problem: Biological properties change with small changes in pH, usually less than 1 pH unit. How does a system prevent fluctuations in pH? Buffers A buffer can resist pH changes if the pH is at or near a weak acid pK value. Buffer range: the pH range where maximum resistance to pH change occurs when adding acid or base. It is = +1 pH from the weak acid pK If pK is 4.8 the buffering range is 3.8 Why? 5.8 Henderson - Hasselbalch equation From Rearrange Take (-)Log of each   [ H ][ A ] K [ HA] [ HA] [H ]  K  [A ]  [ A ] pH   log K  log [ HA]  [A ] pH  pK  log [ HA] - [A ] pH  pK  log [HA]  [A ] 1 10 ratio varies from  [HA ] 10 1 Above and below this range there is insufficient amount of conjugate acid or base to combine with the base or acid to prevent the change in pH. For weak acids A- + H+ HA This equilibrium depends on concentrations of each component . If [HA] = [A-] or 1/2 dissociated  Then [A ] log  log 1  0 : pH = pK [ HA] By definition the pK is the pH where [HA] = [A-] : 50% dissociated The buffer effect can be seen in a titration curve. To a weak acid salt, CH3C00-, add HC1 while monitoring pH vs. the number of equivalents of acid added. or do the opposite with base. Buffer capacity: the molar amount of acid which the buffer can handle without significant changes in pH. i.e 1 liter of a .01 M buffer can not buffer 1 liter of a 1 M solution of HCl but 1 liter of a 1 M buffer can buffer 1 liter of a .01 M solution of HCl Distribution curves for acetate and acetic acid Titration curve for phosphate Table 2-3 Dissociation constants and pK’s of Acids & buffers Acid K pK Oxalic H3PO4 Succinic Acid Succinate H2PO4NH4+ Glycine 5.37x10-2 7.08x10-3 6.17x10-5 2.29x10-6 1.51x10-7 5.62x10-10 1.66x10-10 1.27 2.15 4.21 (pK1) 5.65 (pK2) 6.82 9.25 9.78 Exams Problems You will be asked to solve Henderson Hasselbalch type problems: - [A ] pH  pK  log [HA] You may be asked the pH, pK, the ratio of acid or base or solve for the final concentrations of each. The 6 step approach 1. Write the Henderson + Hasselbalch equation. 2. Write the acid base equation 3. Make sure either an H+ or OH- is in the equation. 3. Find out what you are solving for 4. Write down all given values. 5. Set up equilibrium conditions. 6. Plug in H + H equation and solve. What is the pH of a solution of that contains 0.1M CH3C00and 0.9 M CH3C00H? 1) pH = pK + Log [A-] [HA] 2) CH3C00H CH3C00- + H+ 3) Find pH 4) pK = 4.76 A- = 0.1 M HA = 0.9 M 5) Already at equilibrium 6) X = 4.76 + Log 0.1 0.9 Log 0.111 = -.95 X = 4.76 + (-.95) X = 3.81 What would the concentration of CH3C00- be at pH 5.3 if 0.1M CH3C00H was adjusted to that pH. 1) 2) 3) 4) 5) pH = pK + Log [A-] [HA] CH3C00H CH3C00- + H+ Find equilibrium value of [A-] i.e [CH3C00-] pH = 5.3; pK = 4.76 Let X = amount of CH3C00H dissociated at equilibrium [A-] = [X] [HA] = [0.1 - X] 6) 5.3 = 4.76 + Log [X] [0.1 - X] Now solve. Blood Buffering System • Bicarbonate most significant buffer • Formed from gaseous CO2 CO2 + H2O <-> H2CO3 H2CO3 <-> H+ + HCO3• Normal value blood pH 7.4 • Deviations from normal pH value lead to acidosis

Acids And Bases 2

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