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AP Chapter 17

AP Chapter 17. Ionic Equilibria of Weak Electrolytes. pH and pOH scales. acidic. basic. pH + pOH = 14.00. H + = H 3 O +. pH = -log[H + ]. The pH scale is a logarithmic scale. This means that in order to change the pH by one unit there must be a tenfold change in the [H + ].

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AP Chapter 17

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  1. AP Chapter 17 Ionic Equilibria of Weak Electrolytes

  2. pH and pOH scales acidic basic pH + pOH = 14.00

  3. H+ = H3O+

  4. pH = -log[H+] • The pH scale is a logarithmic scale. • This means that in order to change the pH by one unit there must be a tenfold change in the [H+]. • Find the pH of a solution with [H+] = 0.010M • Find the pH of a solution with [H+] = 0.10M

  5. Example 17.1 Page 521 • What is the pH of a solution of HCl with a concentration of 1.2 x 10-3M? pH = 2.92

  6. Example 17.2 Page 521 • Calculate the pH of: • 0.10 M solution of HNO3 • 0.10 M solution of CH3CO2H (1.3% ionized) pH = 1.00 pH = 2.89

  7. Example 17.4 Page 522 • Calculate the [H3O+] of a solution with a pH of 9.0 [H3O+] = 1.0 x 10-9

  8. Example 17.6 Page 523 • Calculate the pH and pOH of a 0.0125 M KOH solution. pOH = 1.903 pH = 12.097

  9. Calculate the [OH-] of a solution with a pH of 5.56. [OH-] = 3.63 x 10-9

  10. What are the ion concentrations in a 0.10M HCl solution? 0.10M HCl 0.10M H+ 0.10M Cl- Strong electrolytes dissociate completely

  11. What are the ion concentrations in a 0.15 M K2SO4 solution? 0.15M K2SO4 0.30M K+ 0.15M SO42-

  12. At equilibrium, a solution of acetic acid, [ CH3CO2H ] = 0.0788M and [H3O+] = [CH3CO2-] = 0.0012M. What is the Ka of acetic acid? Example 17.8 page 527

  13. Example 17.9 page 527 The pH of a 0.0516 M solution of nitrous acid, HNO2, is 2.34. What is the Ka?

  14. Additional Ka values Table contains more Ka values than are pictured here

  15. Calculate the [H3O+], [CH3CO2-], and [ CH3CO2H ] in a 0.100 M solution of acetic acid. What is the Ka = 1.8 x 10-5. Example 17.10 page 528

  16. What else can you determine in the previous example? • pH, pOH, [OH-] • percent ionization Calculate this value. Percent ionization = 1.3%

  17. What is the percent ionization of a 0.25 M solution of trimethylamine, (CH3)3N, a weak base with a Kb = 7.4 x 10-5 1.7%

  18. What is the pH of the trimethylamine solution? [OH-] = 4.3 x 10-3 pOH = -log[4.3 x 10-3] = 2.37 pH = 14 – 2.37 = 11.63 [OH-] = 4.3 x 10-3 [H+] = 1 x 10-14 ÷ 4.3 x 10-3 =2.3 x 10-12 pH = -log[2.3 x 10-12] = 11.63

  19. Appendix G has additional values

  20. Diprotic and Triprotic Acids • A diprotic acid ionizes in two steps because it has two ionizable hydrogens. • A triprotic acid ionizes in three steps because it has three ionizable hydrogens.

  21. The Stepwise Dissociation of Phosphoric Acid. A triprotic acid. H3PO4 (aq) + H2O(l) H2PO4-(aq) + H3O+(aq) H2PO4-(aq) + H2O(l) HPO42-(aq) + H3O+(aq) HPO42-(aq) + H2O(l) PO43-(aq) + H3O+(aq) H3PO4 (aq) + 3 H2O(l) PO43-(aq) + 3 H3O+(aq)

  22. Each ionization step occurs to a lesser extent than one preceding it. H3PO4 (aq) + H2O(l) H2PO4-(aq) + H3O+(aq) Ka = 7.5 x 10-3 H2PO4-(aq) + H2O(l) HPO42-(aq) + H3O+(aq) Ka = 6.3 x 10-8 HPO42-(aq) + H2O(l) PO43-(aq) + H3O+(aq) Ka = 3.6 x 10-13 H3PO4 (aq) + 3 H2O(l) PO43-(aq) + 3 H3O+(aq) Ka = ?

  23. 0.10 mol CH3CO2H and 0.50 mol NaCH3CO2 are dissolved in 1.0L of solution. Calculate [H3O+]. Example 17.19 page 548

  24. 0.10 mol CH3CO2H and 0.50 mol NaCH3CO2 are dissolved in 1.0L of solution. Calculate [H3O+]. We can work the problem using either Ka or Kb Example 17.19 page 548

  25. Compare this answer to [H3O+] from problem 17.10 pp. 528. How does this illustrates LeChatlier’s Principle.

  26. Example 17.10 page 528 Calculate the [H3O+], [CH3CO2-], and [ CH3CO2H ] in a 0.100 M solution of acetic acid. What is the Ka = 1.8 x 10-5. CH3CO2H + H2O ↔ CH3CO2- + H3O+ [H3O+] = 1.3 x 10-3 0.10 mol CH3CO2H and 0.50 mol NaCH3CO2 are dissolved in 1.0L of solution. Calculate [H3O+]. [H3O+] = 3.6 x 10-6

  27. 10.0ml of 0.10M HCl and 25.0ml of 0.10M NH3 are mixed. Calculate the [OH-].

  28. Example 17.20 page 549 10.0ml of 0.10M HCl and 25.0ml of 0.10M NH3 are mixed. Calculate the [OH-].

  29. Common Ion Problems.What is the “common ion” in each example? Example 17.20 page 549 • 10.0ml of 0.10M HCl and 25.0ml of 0.10M NH3 are mixed. Calculate the [OH-]. Example 17.19 page 548 • 0.10 mol CH3CO2H and 0.50 mol NaCH3CO2 are dissolved in 1.0L of solution. Calculate [H3O+]. Buffers are also examples of the common ion effect since they are mixtures where both substances produce the same ion. Both of the solution mixtures above are buffers.

  30. Buffers • A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. • Buffer solutions have the property that the pH of the solution changes very little when a small amount of acid or base is added to it.

  31. Demonstration: Buffered vs. Non-buffered solutions

  32. Why are these solutions buffers? Example 17.20 page 549 • 10.0ml of 0.10M HCl and 25.0ml of 0.10M NH3 are mixed. Calculate the [OH-]. Example 17.19 page 548 • 0.10 mol CH3CO2H and 0.50 mol NaCH3CO2 are dissolved in 1.0L of solution. Calculate [H3O+].

  33. How buffers work • Weak acids and weak bases tend to remain in high concentrations when added to water because by definition they do not ionize much in water since they are weak. • However, they are very likely to react with any added strong base or strong acid.

  34. Why are weak acids/bases used to create buffers?

  35. Remember this problem Calculate the pH of: 0.10 M solution of HNO3 0.10 M solution of CH3CO2H (1.3% ionized) What would the [CH3CO2H] have to be for it to have the same pH as the HNO3 assuming the 1.3% ionization factor does not change? 0.10M = [CH3CO2H](0.013) [CH3CO2H] = 7.7 M pH = 1.00 pH = 2.89

  36. Why are weak acids/bases used to create buffers? That’s right 7.7 M[CH3CO2H] vs. 0.10M have the same pH. In other words the [CH3CO2H] is 77 times greater [HNO3]. It takes much more base to change the pH of a weak acid solution because there is a large reservoir of undissociated weak acid.

  37. Weak Acid (HA) and its conjugate base (A-) buffer

  38. Adding a strong base to the buffer • If a strong base is added to a buffer, the weak acid will give up its H+ in order to transform the base (OH-) into water (H2O) and the conjugate base: • HA + OH- → A- + H2O. • Since the added OH- is consumed by this reaction, the pH will change only slightly.

  39. Adding a strong acid to the buffer • If a strong acid is added to a buffer, the weak base will react with the H+ from the strong acid to form the weak acid, HA: • H+ + A- → HA • The H+ gets absorbed by the A- instead of reacting with water to form H3O+ (H+), so the pH changes only slightly.

  40. An effective buffer requires relatively equal amounts of weak acid and conjugate base.

  41. Basic Buffers • Note that the same ideas hold true for weak bases, (B), and their conjugate acids (BH+).

  42. For the most effective buffers Ka = [H3O+] HA + H2O ↔ A- + H3O+ Consider acetic acid

  43. You want to make the most “effective” buffer you can using acetic acid. What would the [H3O+] be? Calculate the pH of this buffer.

  44. I want to make a buffer with a pH of 3.14. Which acid should I use?

  45. I want to make a buffer with a pH of 3.14. Which acid should I use? Ka = [H3O+] [H3O+] = antilog (-pH) Ka = antilog (-3.14) = 7.2 x 10-4 HF

  46. I want to make a buffer with a pH of 3.75. Which acid should I use?

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