1 / 16

Buffers & Buffer Capacity

Buffers & Buffer Capacity. Brown & LeMay 17.2 Pages 664 - 671. Buffer Solutions (or Buffers). Can resist drastic changes in pH upon the addition of small amounts of strong acid or strong base Interesting example is Human Blood Contains both An acidic species to neutralize OH -

Télécharger la présentation

Buffers & Buffer Capacity

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Buffers & Buffer Capacity Brown & LeMay 17.2 Pages 664 - 671

  2. Buffer Solutions (or Buffers) • Can resist drastic changes in pH upon the addition of small amounts of strong acid or strong base • Interesting example is Human Blood • Contains both • An acidic species to neutralize OH- • A basic species to neutralize H+ • The acidic and basic species must not consume one another.

  3. How do you make a buffer solution? • Buffers are usually composed of a weak acid-base conjugate pair. • Example might be HC2H3O2 / C2H3O2- made by adding HC2H3O2 with NaC2H3O2 salt or NH4+ / NH3 made by adding NH3 and NH4Cl salt

  4. How do buffer solutions work? • Consider HX (aq)  H+ (aq) + X- (aq) • Ka = [H+] [X-] [HX] Rearranged to [H+] = Ka [HX] [X-] As long as the amounts of HX and X- in the buffer are large compared to the amount of OH- added, the ratio of [HX] / [X-] doesn’t change much. Therefore, [H+] and pH change is small.

  5. Consider HCN and NaCN • HCN = acidic component • CN - = basic component • When acid is added, then • H + + CN- HCN • When base is added, then • OH - + HCN  H2O + CN-

  6. Consider NH3 Mixed With NH4Cl • A. NH4+ = acidic component; • NH3 = basic component • When acid is added, then • H + + NH3 NH4+ • When base is added, then • OH - + NH4+ H2O + NH3

  7. Buffers are most effective when the following two conditions are met: • [weak acid] and [conj. base] are about equal • Try to select a buffer whose acid form has a pKa close to the desired pH.

  8. Calculate the pH of a buffer prepared by dissolving 0.25 mol of acetic acid and 0.40 mole of sodium acetate in water and diluting to one liter. • Use the ICE approach, where initial conc of HC2H3O2 and C2H3O2- are given.

  9. HC2H3O2 (aq)  H+(aq) + C2H3O2- (aq) • I 0.25 0 0.40 • C -x +x +x • E 0.25 –x x 0.40 + x Ka= [H+] [C2H3O2-] [H+] = Ka[HC2H3O2] [HC2H3O2] [C2H3O2] =1.8 x10-5 (0.40 / 0.25) = 1.1 x 10-5 Check to see if assumption is valid. pH = - log(1.1 x 10-5) = 4.96

  10. Buffer solutions (a) When OH-is added to a buffer solution, some of the weak acid is neutralized and thus converted to the conjugate base. (b) When H3O+is added to a buffer solution, some of the conjugate base is neutralized and thus converted to the weak acid. However, as long as the concentration ratio [weak acid]/[conjugate base] stays close to its original value, [H3O+] and the pH won’t change very much.

  11. Buffer Capacity and pH • The amount of acid or base the buffer can neutralize before the pH begins to change appreciably. • Depends on the amount of acid & base from which it was made. • Henderson-Hasselbalch Equation • pH = pKa + log [base] [acid]

  12. What is the pH of a buffer that is 0.12 M in lactic acid and 0.10 M in sodium lactate? • Ka for lactic acid = 1.4 x 10-4

  13. Can use the Henderson-Hasselbalch equation to calculate pH directly: • pH = pKa + log [base] [acid] = -log(1.4 x10-4) + log (0.10 / 0.12) = 3.85 + (-0.08) = 3.77

  14. Calculate the pH of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. Ka for benzoic acid = 6.3 x 10-5

  15. pH = -log (6.3 x 10-5) +log (0.20 / 0.12) • = 4.20 + 0.22 = 4.42 • Please read Page 669 – Chemistry and Life: Blood as a Buffered Solution

  16. Biological Acids And Their Salts that make Biological Buffers • Citric acid and citrate • Fumaric acid and fumarate • Pyruvic acid and pyruvate • Lactic acid and lactate • Glutamic acid and glutamate

More Related