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Buffer Solutions and pH Resistance

Learn about buffered solutions that resist changes in pH and how to calculate the pH of a buffer. Understand the Henderson-Hasselbach equation and the concept of buffer capacity.

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Buffer Solutions and pH Resistance

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  1. Chapter 17 Buffers

  2. Buffered solutions • A solution that resists a change in pH. • Buffers are: • A solution that contains a weak acid-weak base conjugate pair. • Often prepared by mixing a weak acid, or a weak base, with a salt of that acid or base.

  3. Buffered solutions • The solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. • We can make a buffer of any pH by varying the concentrations of these solutions.

  4. What is the pH of a buffer that is 0.12M in lactic acid, HC3H5O3 , and 0.10M in sodium lactate? For lactic acid, K = 1.4 x 10 -4 HC3H5O3↔ H+ + C3H5O3-1

  5. [ H +1] [C3H5O3-1] Ka = [HC3H5O3] 1.4 x 10-4 = (x) (0.10 – x) (0.12 – x) Because of the small Ka and the presence of the common ion, it is expected that x will be small relative to 0.12 or 0.10M. x = 1.7 x 10 -4 M therefore [H+1] = 1.7 x 10 -4 M pH = -log (1.7 x 10 -4 M) = 3.77

  6. Use Henderson-Hasselbach Instead [base] pH = pKa + log [acid] [ 0.10] pH = -log (1.4 x 10-4) + log [0.12] pH = 3.85 + (-0.08) = 3.77

  7. Practice Problem Calculate the pH of a buffer composed of 0.12M benzoic acid (HC7H5O2)and 0.20M sodium benzoate. Ka = 6.3 x 10-5 ans: 4.42 Remember the Aqueous Equilibrium Constants are located in Appendix D of your textbook.

  8. Buffered Solutions • Buffers resist changes in pH because they contain both an acidic species to neutralize OH-1 ions and a basic one to neutralize H+1 ions. • It is important that the acidic and basic species of the buffer do not consume each other through a neutralization reaction.

  9. Buffered Solutions • HA H+ + A- • Ka = [H+] [A-] [HA] • Buffers most effectively resist a change in pH in either direction when the concentrations of HA and A- are about the same. When [HA] equals [A-] then [H+] equals Ka. • Scientists usually try to select a buffer whose acid form has a pKa close to the desired pH.

  10. General equation • Ka = [H+] [A-] [HA] • so [H+] = Ka [HA] [A-] • The [H+] depends on the ratio [HA]/[A-] • Take the negative log of both sides • pH = -log(Ka [HA]/[A-]) • pH = -log(Ka)-log([HA]/[A-]) • pH = pKa + log([A-]/[HA])

  11. This is called the Henderson-Hasselbach equation • pH = pKa + log([A-]/[HA]) • pH = pKa + log(base/acid)

  12. Try an Acid • Calculate the pH of the following mixture: • Prob. #1: 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4) • Answer: • pH = -log(1.4x10-4) + log[0.25/0.75] • pH = 3.38

  13. Now Try a Base • Calculate the pH of the following mixture: • 0.25 M NH3 and 0.40 M NH4Cl • (Kb = 1.8 x 10-5) • Answer: • NH3 NH4+1 + OH-1 • (NH3 is B, NH4+1 is HB+1) • pOH = pKb + log[ HB+1 / B ] • pOH = -log(1.8x10-5) + log[0.4/0.25] • pOH – 4.94 now convert to pH

  14. Buffering Capacity • This is the amount of acid or base the buffer can neutralize before the pH begins to change to an appreciable degree.

  15. Buffering Capacity • The buffering capacity, that is, the effectiveness of the buffer solution, depends on the amount of acid and conjugate base from which the buffer is made. • The larger the amount the greater the buffering capacity. • In general, a buffer system can be represented as salt-acid or conjugate base-acid.

  16. **Buffer capacity** • The pH of a buffered solution is determined by the ratio [A-]/[HA]. • As long as this doesn’t change much the pH won’t change much. • The more concentrated these two are the more H+ and OH- the solution will be able to absorb. • Larger concentrations means bigger buffer capacity.

  17. Buffer Capacity • Calculate the change in pH that occurs when 0.010 mol of HCl(g) is added to 1.0L of each of the following: • 5.00 M HAc and 5.00 M NaAc • 0.050 M HAc and 0.050 M NaAc • Ka= 1.8x10-5

  18. Buffer capacity • The best buffers have a ratio [A-]/[HA] = 1 • This is most resistant to change • True when [A-] = [HA] • Make pH = pKa (since log1=0)

  19. Addition of Strong Acids or Bases to Buffers • See pg. 648 B&L for explanation and examples.

  20. Adding a strong acid or base • Do the stoichiometry first. • A strong base will take protons from the weak acid reducing [HA]0 • A strong acid will add its proton to the anion of the salt reducing [A-]0 • Then do the equilibrium problem.

  21. B & L pg 649

  22. Try the Buffer worksheet from the “Chang” chemistry text.

  23. Prove they’re buffers • What would the pH be if 0.020 mol of HCl is added to 1.0 L of both of the preceding solutions. • What would the pH be if 0.050 mol of solid NaOH is added to each of the proceeding.

  24. Thanks to Mr. Green

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