1 / 49

Chapter 14

Chapter 14. Applications of Equilibrium Principles (acid/base systems). Chapter 14. Buffers - applies stoichiometry and equilibria Acid-Base Titrations. Chapter 14. Two system types: a weak acid, HB, and its conjugate base, B - (buffer system). Chapter 14.

tansy
Download Presentation

Chapter 14

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. Chapter 14 • Applications of Equilibrium Principles (acid/base systems)

  2. Chapter 14 • Buffers - applies stoichiometry and equilibria • Acid-Base Titrations

  3. Chapter 14 • Two system types: • a weak acid, HB, and its conjugate base, B- (buffer system)

  4. Chapter 14 • An acid and a base used in an acid-base titration (especially a polyprotic acid)

  5. Buffer System • Buffers contain roughly equal amounts of a weak acid and its conjugate base.

  6. Buffer System Highly resistant to changes in pH brought about by the addition of a strong acid or strong base. (why?)

  7. Buffer System • The buffer can react with either H3O+ (H+) ions or OH- ions. • HB(aq) + OH-(aq) --> B-(aq) + H2O • B-(aq) + H+(aq) --> HB(aq)

  8. Buffer System • The reactions have large Keq values. They go almost to completion, so the H+ ions or OH- ions are consumed. • The pH is not directly affected.

  9. Buffer System Has a pH close to the pKa of the weak acid. Why?

  10. Buffer System • HB(aq) <--> H+(aq) + B-(aq) • Ka = [H+][B-]/[HB] • Solve for [H+]

  11. Buffer System • 1.) Assume equilibrium is established without appreciably changing the concentrations of either HB or B-.

  12. Buffer System • 1.) [HB] = [HB]o • [B-] = [B-]o

  13. Buffer System • 2.) Since HB and B-are present in the same solution, the ratio of their concentrations equals their mole ratio. • [HB]/[B-] = nHB/nB-

  14. Buffer System • 2.) [H+] = Ka(nHB/nB-) • Or… Henderson-Hasselbach equation: pH=pKa+log([B-]/[HB])

  15. Buffer System Calculate [H+] and pH in a solution prepared by adding 0.200 mol of acetic acid and 0.200 mol of sodium acetate to one liter. [H+] = (1.8 x 10-5) x (.200/.200) = 1.8 x 10-5 M, pH = 4.74

  16. Choosing a Buffer • To make a buffer in the lab at a required pH: note that HB and B- are present in roughly equal amounts and that pH is roughly equal to pKa of the weak acid.

  17. Choosing a Buffer • Most important factor: choose a buffer system with a Ka close to the pH desired (see table 14.1, p. 382 - weak acids and their salts).

  18. Choosing a Buffer • Ex. To establish a buffer of pH 7, choose a system such as H2PO4-, HPO42-, where Ka = 6.2 x 10-8.

  19. Choosing a Buffer • Ex. At pH 7, [H+] = 1.0 x 10-7 M. • So…[H2PO4-] / [HPO42-] = [H+] / Ka = (1.0 x 10-7)/(6.2 x 10-8)=1.6

  20. Choosing a Buffer • Ex. So… add 160 ml of 0.100 M H2PO4- to 100 ml of 0.100 M HPO42-.

  21. Choosing a Buffer • Another method: partial neutralization of a weak acid and weak base gives a buffer.

  22. Choosing a Buffer • Ex. Reacting 0.10 mol of HCl with 0.20 mol of NH3, consumes the H+, producing 0.10 mol of NH4+ and 0.10 mol NH3

  23. Distribution Curves • See figure 14.3, p. 385. • Buffers are only effective over small ranges: there must be appreciable amounts of both HB and B- for the system to work.

  24. Distribution Curves • Ex. P. 385, 389, 390 • At a pH of 6.4, [HB] = [B-], pH = pKa. • At pH of 5.4 - 7.4 there is enough of either species to consume the addition of a strong acid or base.

  25. Added H+ or OH- • The pH of a buffer does change slightly as a strong acid or base is added. • Addition of H+ converts B- to HB. Addition of OH- converts HB to B- and H2O.

  26. Added H+ or OH- • Use stoichiometry to determine how the [HB] to [B-] ratio has changed. Then use equilibrium principles to ultimately calculate the new pH.

  27. Added H+ or OH- Calculate the pH of a buffer containing 0.200 mol of acetic acid and 0.200 mol of sodium acetate after the addition of 0.020 mol NaOH. [H+] = (1.8 x 10-5) x (.180/.220) = 1.5 x 10-5 M, pH = 4.82

  28. Indicators • Useful in determining the equivalence point of a titration.

  29. Indicators • It is a weak acid and its conjugate base. When one dominates, it is a different color then when the other dominates.

  30. Indicators • HIn (aq) <--> H+ (aq) + In- (aq). Ka = [H+][In-]/[HIn] • 1. If [HIn]/[In-] is >10, then the acid color dominates. • 2. If < 0.1, then the base color dominates.

  31. Indicators • 3. If [HIn]/[In-] approx. 1, then the color is an intermediate of the two.

  32. Indicators • Ka = [H+][In-]/[HIn], so…. • [HIn]/[In-] = [H+]/Ka • What two factors, then, effect [HIn]/[In-], and hence, the color of the indicator?

  33. Indicators • 1. [H+] or the pH of the solution. • High pH vs. low pH? • 2. Ka of the indicator. Indicators change color at different pHs (when pH = pKa).

  34. Distribution Curves • See table 14.2, p. 387. • Indicators are only effective over small ranges. There must be appreciable amounts of both the HB and B- for the system to work.

  35. Titrations • Titration: an analytical method in which a standard solution is used to determine the concentration of another solution.

  36. Titrations • A neutralization reaction occurs. Dependent on the indicator used, we can determine the volume at which the mixture reaches a certain pH.

  37. Titrations • Solving steps: 1.) balance the neutralization equation. • 2.) from the given determine the amount of moles used.

  38. Titrations • Solving steps: 3.) use the equation to determine the mole ratios. • 4.) Determine the molarity or the volume of the unknown.

  39. Titrations • If 26.4 cm3 of LiOH solution are required to neutralize 21.7 cm3 of 0.500 M HBr, what is the concentration of the basic solution?

  40. Acid/Base Titrations • Strong Acid-Strong Base • H+ + OH- --> H2O, K = 1/Kw = 1.0 x 1014. • pH = 7 at the end point, changes rapidly near end point.

  41. Acid/Base Titrations • Weak Acid-Strong Base • HA + OH- --> H2O + A-, K = 1/Kb. • Ex. If Ka = 1.0 x 10-5, Kb = 1.0 x 10-9, K = 1.0 x 109.

  42. Acid Base Titrations 0.10 M HA (Ka = 1 x 10-5) is titrated with a strong base. Determine the initial pH, the pH when the titration is half finished, and the pH at the end point. pH = 3.0, 5.0, and 9.0

  43. Acid/Base Titrations • In general, pH > 7 at the end point, changes slowly near the end point (buffer system).

  44. Acid/Base Titrations • Strong Acid-Weak Base • H+ + A- --> HA, K = 1/(Ka of HA). • pH < 7 at the end point, changes slowly near the end point.

  45. Choice of Indicator • 1. Weak acid - strong base: use an indicator such as (table 14.2 p. 387) phenolphthalein, which changes color above pH 7.

  46. Choice of Indicator • 2. Strong acid - weak base: use an indicator such as methyl red, which changes color below pH 7.

  47. Choice of Indicator • 3. Strong acid - strong base: use any indicator because the pH changes rapidly near the end point.

  48. Polyprotic Acids • Virtually all the H+ comes from the first dissociation. • Ex. H2CO3 <--> H+ + HCO3-, K1 = 4x10-7. • HCO3- <--> H+ + CO32-, K2 = 4 x 10-11

  49. Polyprotic Acids • In 0.1 M H2CO3, [H+] = [HCO3-] = 2 x 10-4 M, [CO32-] = 4 x 10-11 M

More Related