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Buffers

Buffers. Buffers are solutions that undergo a minimal pH change when small quantities of a strong acid or a strong base is added. A buffer can be made in either of two ways: A buffer can be prepared by mixing a weak acid such as benzoic acid,

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Buffers

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  1. Buffers Buffers are solutions that undergo a minimal pH change when small quantities of a strong acid or a strong base is added. A buffer can be made in either of two ways: • A buffer can be prepared by mixing a weak acid such as benzoic acid, HC7H5O2, and a salt containing the anion of that acid such as potassium benzoate, KC7H5O2.

  2. A buffer can be prepared by mixing a weak base such as, NH3, and a salt containing the cation of the base such as, NH4Br.

  3. HC7H5O2 - KC7H5O2 Buffer • Benzoic acid being a weak acid will ionize • slightly as shown below: • HC7H5O2(aq) H+(aq) + C7H5O2-(aq) • Potassium benzoate being a strong • electrolyte will completely dissociate as • shown below: • KC7H5O2(aq) K+(aq) + C7H5O2-(aq) →

  4. Benzoate, C7H5O2-, being the conjugate base • of a weak acid will hydrolyze as shown below: • C7H5O2-(aq) + H2O(l) • HC7H5O2(aq) + OH-(aq) • The key to this buffer as with all buffers lie • with the two simultaneous equilibria: • HC7H5O2(aq) H+(aq) + C7H5O2-(aq) • C7H5O2-(aq) + H2O(l) • HC7H5O2(aq) + OH-(aq)

  5. There are four reacting species to consider: • HC7H5O2, H+, C7H5O2-, and OH- • When a small amount of a strong acid such • as, HCl, is added to the buffer, it will • completely ionize. • As odd as it may sound, when you add small • amounts of a strong acid to a buffer, you • produce more weak acid as shown below: • H+(aq) + C7H5O2-(aq) → HC7H5O2(aq)

  6. The H+ mainly comes from the HCl that was • added to the buffer, so the original [H+] does • not change much. • When a small amount of a strong base such • as, NaOH, is added to the buffer, it will • completely dissociate. • As odd as it may sound, when you add small • amounts of a strong base to a buffer, you • produce more weak base as shown below: • HC7H5O2(aq) + OH-(aq) → C7H5O2-(aq) + H2O(l)

  7. The OH- mainly comes from the NaOH that • was added to the buffer, so the original [OH-] • does not change much. • As long as both the HC7H5O2 and the C7H5O2- • are present in the buffer to react with the • addition of OH- and H+, the pH of the buffer • only changes slightly.

  8. NH3 - NH4Br Buffer • Ammonia being a weak base will undergo • slight hydrolysis as shown below: • NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) • Ammonium bromide being a strong • electrolyte will completely dissociate as • shown below: • NH4Br(aq) → NH4+(aq) + C7H5O2-(aq)

  9. Ammonium, NH4+, being the conjugate acid • of a weak base will ionize as shown below: • NH4+(aq) H+(aq) + NH3(aq) • The key to this buffer as with all buffers lie • with the two simultaneous equilibria: • NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) • NH4+(aq) NH3(aq) + OH-(aq)

  10. There are four reacting species to consider: • NH4+, OH-, H+, NH3 • When a small amount of a strong acid such • as, HCl, is added to the buffer, it will • completely ionize. • As odd as it may sound, when you add small • amounts of a strong acid to a buffer, you • produce more weak acid as shown below: • H+(aq) + NH3(aq) → NH4+(aq)

  11. The H+ mainly comes from the HCl that was • added to the buffer, so the original [H+] does • not change much. • When a small amount of a strong base such • as, NaOH, is added to the buffer, it will • completely dissociate. • As odd as it may sound, when you add small • amounts of a strong base to a buffer, you • produce more weak base as shown below: • NH4+(aq) + OH-(aq) → NH3(aq) + H2O(l)

  12. The OH- mainly comes from the NaOH that • was added to the buffer, so the original [OH-] • does not change much. • As long as both the NH3 and the NH4+ are • present in the buffer to react with the • addition of OH- and H+, the pH of the buffer • only changes slightly.

  13. A Buffer Problem • A buffer solution contains 0.12 mol of acetic • acid, HC2H3O2, and 0.10 mol of sodium • acetate, NaC2H3O2, in 1.00 L. • What is the pH of this buffer? • Ka = 1.8 x 10-5V = 1.00 L • n = 0.12 mol HC2H3O2 n = 0.10 mol NaC2H3O2 • [HC2H3O2] = 0.12 M [NaC2H3O2] = 0.10 M

  14. [H+] [C2H3O2-] Ka = [HC2H3O2] • HC2H3O2(aq) H+(aq) + C2H3O2-(aq) • [ ]I 0.120 0.10 • [ ]c -x+x+x • [ ]e 0.12 - xx0.10 + x 0.12 - x [H+] 1.8 x 10-5 = × 0.10 + x

  15. 0.12 2.2 x 10-5 = [H+] 1.8 x 10-5 = × 0.10 • . pH = -log[H+] = -log(2.2 x 10-5) = 4.66 (b) What is the pH of the buffer after adding 0.010 mol of NaOH? Before looking at the equilibria of the problem, you have to determine what effect 0.010 mol of OH- has on the stoichiometry of the problem.

  16. nb 0.12 mol 0.010 mol 0.10 mol • HC2H3O2(aq) + OH-(aq) → C2H3O2-(aq) + • H2O(l) • na 0.11 mol 0 0.11 mol • This is a limiting reactant problem where OH- • is the limiting reactant and is totally • consumed. • The mole ratio is 1:1:1 giving the values for • na which represents the number of moles • remaining after the reaction.

  17. The previous stoichiometry of the problem is • critical and can not be ignored. • [HC2H3O2] = [C2H3O2-] = 0.11 M • The next step is to recalculate the pH with the • new molarities.

  18. [H+] [C2H3O2-] Ka = [HC2H3O2] • HC2H3O2(aq) H+(aq) + C2H3O2-(aq) • [ ]I 0.110 0.11 • [ ]c -x+x+x • [ ]e 0.11 - xx0.11 + x 0.11 + x ≈ 1.8 x 10-5 M [H+] 1.8 x 10-5 = × 0.11 - x

  19. pH = -log[H+] = -log(1.8 x 10-5) = 4.74 • A sanity check is in order here! • According to the calculations the pH • increased from 4.66 to 4.74. • Does this even sound reasonable? • The answer is yes because a strong base was • added to the buffer which would cause the pH • to increase.

  20. The buffer should become more basic (less • acidic) as a result. • Should such a small increase (ΔpH = 0.080) • result? • The answer is yes because the solution is • “buffered” meaning that it will resist changes • in pH. • Compare this to the change in pH of a neutral • solution (pH = 7.00) after the addition of • 0.010 mol OH-.

  21. [NaOH] = 0.010 M • NaOH(aq) → Na+(aq) + OH-(aq) • pOH = -log[OH-] = -log(0.010) = 2.000 • pH + pOH = 14.00 • pH = 14.00 – 2.000 = 12.00 • This results in a ΔpH = 5.00. • These gyrations should clarify what is meant • by a solution being buffered.

  22. (c) What is the pH of the buffer after adding 0.010 mol of HI? Before looking at the equilibria of the problem, you have to determine what effect 0.010 mol of H+ has on the stoichiometry of the problem. • .

  23. nb 0.010 mol 0.10 mol 0.12 mol • H+(aq) + C2H3O2-(aq) → HC2H3O2(aq) • na 0 mol 0.09 mol 0.13 mol • This is a limiting reactant problem where • H+ is the limiting reactant and is totally • consumed. • The mole ratio is 1:1:1 giving the values for • na which represents the number of moles • remaining after the reaction.

  24. The previous stoichiometry of the problem is • critical and can not be ignored. • [HC2H3O2] = 0.13 M • [C2H3O2-] = 0.09 M • The next step is to recalculate the pH with the • new molarities.

  25. [H+] [C2H3O2-] Ka = [HC2H3O2] • HC2H3O2(aq) H+(aq) + C2H3O2-(aq) • [ ]I 0.130 0.09 • [ ]c -x+x+x • [ ]e 0.13 - xx0.09 + x 0.13 - x [H+] 1.8 x 10-5 = × 0.09 + x 0.13 ≈ 2.6 x 10-5 M ≈ [H+] 1.8 x 10-5 × 0.09

  26. pH = -log[H+] = -log(2.6 x 10-5) = 4.59 • A sanity check is in order here! • According to the calculations the pH • decreased from 4.66 to 4.59. • Does this even sound reasonable? • The answer is yes because a strong acid was • added to the buffer which would cause the pH • to decrease.

  27. The buffer should become less basic (more • acidic) as a result. • Should such a small increase (ΔpH = 0.070) • result? • The answer is yes because the solution is • “buffered” meaning that it will resist changes • in pH. • Compare this to the change in pH of a neutral • solution (pH = 7.00) after the addition of • 0.010 mol H+.

  28. [HI] = 0.010 M • HI(aq) → H+(aq) + I-(aq) • pH = -log[H+] = -log(0.010) = 2.000 • This results in a ΔpH = 5.00. • These gyrations should clarify what is meant • by a solution being buffered.

  29. Variant of a Buffer Problem • A sample of 1.75 L of HBr gas at 17°C and • 0.960 atm is bubbled through 0.600 L of • 0.160 M NH3 solution. Assuming that all of the • HBr dissolves and the volume of the solution • remains the same, calculate the pH of the • final solution. • V = 1.75 L V = 0.600 L NH3 • T = 17°C = 290 K [NH3] = 0.150 M • P = 0.960 atm

  30. 1.75 L 0.960 atm × L•atm 0.0821 290 K × mol•K mol NH3 n = 0.160 0.600 L × L n = • PV = nRT • [NH3] = n/V 0.071 mol HBr n = n 0.096 mol NH3 =

  31. nb 0.071 mol 0.096 mol 0 mol • H+(aq) + NH3(aq) → NH4+(aq) • na 0 mol 0.025 mol 0.071 mol • [NH4+] = n/V 0.071 mol NH4+ [NH4+] = = 0.12 M 0.600 L 0.025 mol NH3 [NH3] = = 0.042 M 0.600 L

  32. The question which arises many times is, do I • use the NH4+ acting as an acid or the NH3 • acting as a base? • The best way to find out is to do both and then • compare the pH’s.

  33. NH4+(aq) H+(aq) + NH3(aq) • [ ]I 0.120 0.042 • [ ]c -x+x+x • [ ]e 0.12 - xx0.042 + x • At this point in the problem you may notice • something disturbing. You tried looking up • the Ka of NH4+ and couldn’t find it. [H+] [NH3] Ka = [NH4+]

  34. Or just as disturbing, you are taking an exam • and the problem gives the Kb for NH3. • At this point, you must calmly remember that • Ka × Kb = Kw = 1.00 × 10-14 • therefore, • Do not forget parenthesis in the denominator! 1.00 x 10-14 Ka = 5.6 x 10-10 = (1.8 x 10-5)

  35. x (0.042 + x) 0.042 x ≈ 0.12 - x 0.12 [H+] [NH3] Ka = [NH4+] • . 5.6 x 10-10 = [H+] = 1.6 × 10-9 M pH = -log[H+] = -log(1.6 × 10-9) = 8.80

  36. NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) • [ ]I 0.0420.12 0 • [ ]c -x+x+x • [ ]e 0.042 - x0.12 + xx [NH4+] [OH-] = Kb [NH3] (0.12 + x) x 0.12x ≈ 1.8 x 10-5 = 0.042 - x 0.042

  37. [OH-] = 6.5 x 10-6 M • pOH = -log[OH-] = -log(6.5 x 10-6) = 5.19 • pH + pOH = 14.00 • pH = 14.00 – 5.19 = 8.81 • So, in the final analysis, do you use the NH4+ • acting as an acid or the NH3 acting as a base?

  38. Another Variant of a Buffer Problem • Calculate the concentration of potassium • butanate that is needed in a 0.20 M solution of • butanoic acid to produce a pH of 4.00. • [KC4H7O2] = ? Ka = 1.5 x 10-5 • [HC4H7O2] = 0.20 M pH = 4.00 • [H+] = 10-pH = 10-4.00 = 1.0 x 10-4 M

  39. HC4H7O2(aq) H+(aq) + C4H7O2-(aq) • [ ]I 0.201.0 x 10-4 0 • [ ]c -x+x+x • [ ]e 0.20 - x1.0 x 10-4 + xx [H+] [C4H7O2-] = Ka [HC4H7O2] (1.0 x 10-4 + x ) x 1.0 × 10-4 x ≈ 1.5 x 10-5 = 0.20 - x 0.20

  40. [H+] = [C4H7O2-] = 0.030 M • [KC4H7O2] = [C4H7O2-] = 0.030 M

  41. More Buffer Variants • Calculate the mole ratio of acetic acid to • potassium acetate that is necessary to form a buffer solution with a pH = 5.28. The Ka for HC2H3O2 is 1.8 x 10-5. • pH = 5.28 Ka = 1.8 x 10-5 • [H+] = 10-pH = 10-5.28 = 5.2 x 10-6 M

  42. HC2H3O2(aq) H+(aq) + C2H3O2-(aq) • [ ]I y 0 0 • [ ]c -x+x+x • [ ]e y - x xx [H+] [C2H3O2-] = Ka [HC2H3O2] [HC2H3O2] [H+] 5.2 x 10-6 = = 0.29 = [C2H3O2-] Ka 1.8 x 10-5

  43. [HC2H3O2] [H+] = Ka × [C2H3O2-] n(HC2H3O2)/V 0.29 = n(C2H3O2-)/V • (b) How is the pH of this buffer system • changed upon adding 5.0 mL of distilled • water? • As shown above, the mole ratio is • independent of volume because within the • same solution, the volumes cancel out. n(HC2H3O2)/V = n(C2H3O2-)/V

  44. (c) Why should the pH of a buffer be close to • the pKa of the weak acid? • Ideally, [HC2H3O2] ≈[C2H3O2-]. This • means the concentrations in the above • equation cancel, leaving [H+] = Ka. • Taking the -log of both sides of the • equation yields, -log [H+] = -log Ka, which • simplifies to pH = pKa. [HC2H3O2] [H+] = Ka × [C2H3O2-]

  45. (d) What is buffer capacity? • Buffer capacity is the maximum amount • of strong acid or strong base that can be • added to a buffer before a significant pH • change results. • A buffer is destroyed when all the weak • acid or weak base is consumed.

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