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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 as compared to a strong acid solution of the same pH because there are many more moles of acid available to neutralize base in the weak acid as compared to the strong 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. Did you choose well? Or will you be flung into the “Gorge of Eternal Peril”.

  46. The Bridge of Death ≈ 3:10

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