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Buffers

Buffers. Buffered Solutions. Solutions that resist changes in pH when either OH - or H + ions are added. Example: NH 3 /NH 4 + buffer system HC 2 H 3 O 2 / C 2 H 3 O 2 - buffer system. Usually contain a weak acid and its

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Buffers

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  1. Buffers

  2. Buffered Solutions Solutions that resist changes in pH when either OH- or H+ ions are added. Example: NH3/NH4+ buffer system HC2H3O2 / C2H3O2- buffer system

  3. Usually contain a weak acid and its salt or a weak base and its salt. Pure water has no buffering capacity---acids and bases added to water directly affect the pH of the solution.

  4. Systems that work • weak acid + salt of weak acid • weak base + salt of weak base • weak acid + ½ # of moles of strong base • weak base + ½ # of moles of strong acid • weak acid + weak base

  5. How does it work? • Since a buffer consists of both an acid or base and its conjugate….an acid and a base are present in all buffer solutions. • If a small amount of strong acid is added to the buffer, there is a base component ready and waiting to neutralize the ‘invader’.

  6. Buffer capacity The amount of acid or base that can be absorbed by a buffer system without a significant change in pH.

  7. The pH of a Buffered Solution A buffered solution contains 0.50M acetic acid (HC2H3O2, Ka = 1.8 X 10-5) and 0.50 M sodium acetate(NaC2H3O2). Calculate the pH. pH = 4.74

  8. Calculate the change in pH that occurs when 0.010 mol solid NaOH is added to 1.0 L of the buffered solution in the example before. (pH = 4.74)

  9. Compare this pH change with that which occurs when 0.010 mol solid NaOH is added to 1.0 L of water.

  10. One way to calculate the pH of a buffer system is with the Henderson- Hasselbach equation. pH = pKa + log [base] [acid] pH = pKa + log [A-] [HA] Remember conjugate acid/base pairs!

  11. Or you can use the McCormick equation…. [H+] = Ka [acid] [base]

  12. If acid is added to the buffer, simply add acid to the numerator AND subtract the same quantity from the base since it was self-sacrificing and neutralized the acid. If base is added, simply add the base to the denominator and subtract from the numerator. When equal concentrations of Acid and Base are present [which occurs at the ½ equivalence point of a titration] the ratio of acid to base equals ONE and the pH = pKa.

  13. Now, let’s try it. Same problem as before. Calculate the pH of 1.0 L of a buffer solution composed of 0.50 M acetic acid and 0.50 M sodium acetate before and after adding 0.010 mole solid NaOH. Ka = 1.8 X 10-5 before: [H+] = Ka [acid] = 1.8 X 10-5 [.50] = 1.8 X 10-5 [base] [.50] pH = 4.74 After: [H+] = 1.8 X 10-5 [.50-0.01] = 1.7 X 10-5 [.50 + 0.01] pH = 4.76

  14. Preparing buffer solutions • Use 0.10 M to 1.0 M solutions of reagents & choose an acid whose Ka is near the [H3O+] concentration. • pKa should be as close as possible to the pH

  15. A chemist needs a solution buffered at pH 4.30 and can choose from the following list of acids and their soluble salt: • chloroacetic acid Ka = 1.35 X 10-3 • propanoic acid Ka = 1.3 X 10-5 • benzoic acid Ka = 6.4 X 10-5 • hypochlorous acid Ka = 3.5 X 10-8

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