1 / 75

AP Chemistry Chapter 11

AP Chemistry Chapter 11. Ionic Equilibria. Turn Test Corrections in to the Box You will need a calculator for these notes!. When salts, acids, or bases are added to water they ionize. For example: NaCl (s) → Na + ( aq ) + Cl - ( aq )

caia
Download Presentation

AP Chemistry Chapter 11

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. AP Chemistry Chapter 11 Ionic Equilibria Turn Test Corrections in to the Box You will need a calculator for these notes!

  2. When salts, acids, or bases are added to water they ionize. For example: NaCl(s) → Na+(aq) + Cl-(aq) Could an equilibrium be established for this reaction? What happens if you add a lot of salt to a small glass of water? What would you be able to see if it was at equilibrium? What would you not see if it was at equilibrium? How would this be written at equilibrium? NaCl(s)  Na+(aq) + Cl-(aq)

  3. Are there any salts that would never come to equilibrium? What about acids or bases? Why? For acids and bases to set up an equilibrium they must be weak by definition. When they come to equilibrium (in water) we can write equilibrium expressions for them.

  4. Ionization (Equilibrium) Constants for Monoprotic acids and bases

  5. In general for a weak, monoprotic acid: HA  H+ + A- so Ka = Why “Ka”? But acids must be in water, so it can look like this: HA + H2O  H3O+ + A- Ka = [H2O] usually considered = 1

  6. A solution of nicotinic acid was found at equilibrium to contain [HA] = 0.049M, [H3O+] = [A-] = 8.4 x 10-4 M. What is the Ka and pKa?

  7. Starting with a 0.0100 M solution of acetic acid, when added to water 4.2% ionizes. What is the Ka and pKa?

  8. The pH of a 0.15 M solution of chloroscetic acid (ClCH2COOH) is 1.92. What is the value of Ka?

  9. What are the equilibrium concentrations if the Ka of 0.10 M HOCl is 3.5 x 10-8?

  10. Calculate the percent ionization of a 0.10 M acetic acid solution.

  11. If you haven’t caught it by now: The higher the Ka value (less negative exponent) the lower the pKa value (closer to 0). Also, the “rule of thumb” for the simplification of the quadratic only works if Ka is “x 10-4” or a larger negative exponent. When compared to the [HA]!

  12. Calculate the [OH-] for a 0.20 M solution of NH3. Kb is 1.8 x 10-5.

  13. Calculate the pH for a 0.20 M solution of NH3. Kb is 1.8 x 10-5.

  14. Calculate the % ionization for a 0.20 M solution of NH3. Kb is 1.8 x 10-5.

  15. Ionization (Equilibrium) Constants for Polyprotic acids and bases

  16. Polyprotic acids can give off more than one H+, so it is considered to occur in steps, each with their own Ka value. For example: H3PO4 H+ + H2PO4- Ka1 = 7.5 x 10-3 H2PO4-  H+ + HPO42- Ka2 = 6.2 x 10-8 HPO42-  H+ + PO43- Ka3 = 3.6 x 10-13 So if the concentrations were wanted, we would need to solve all three reactions and then add up the results. (Not hard, but definitely time consuming.)

  17. Would anyone like a shortcut? Ka3 = 3.6 x 10-13 and (remember) Kw = 1 x10-14, so the contribution of the third equation to the total of the acid is only a little larger than water’s own contribution. Specifically, if we started with 0.10 M H3PO4, the third ionization would only add 0.00000000000000000093 M H+ to the total. Rule of Thumb: “x 10-9” and larger negative exponents indicate that reaction does not contribute in any noticeable way to the total (the additional amount will be rounded away).

  18. Calculate the concentrations in 0.10 M H2SO4. Ka2 = 1.2 x 10-2

  19. Salts and Hydrolysis

  20. When salts ionize it is possible one of the ions formed would react with water in a side reaction, but this reaction could change the pH. Bronstead Lowry definitions of acids and bases give us conjugate acid-base pairs. Strong acids/bases give weak conjugate bases/acids as products, and weak acids/bases give strong conjugate bases/acids.

  21. It is the conjugates of weak acids/bases that are the ones that typically react with water. For nonexample: NaCl → Na+ + Cl- Na+ would be the cation of NaOH(a strong base) so it would be a weak acid, and would not react with water. Cl- would be the anion of HCl(a strong acid) so it would be a weak base, and would not react with water.

  22. Strong base - weak acid example: NaC2H3O2 → Na+ + C2H3O2- Na+ would be the cation of NaOH(a strong base) so it would be a weak acid, and would not react with water. C2H3O2- would be the anion of HC2H3O2 (a weak acid) so it would be a strong base, and would react with water, like this: C2H3O2-+ H2O  HC2H3O2 + OH- What type of solution is the result? What would be the pH? Why would we use Kb?

  23. Strong acid - weak base example: NH4Cl → NH4+ + Cl- NH4+ would be the cation of NH4OH (a weak base) so it would be a strong acid, and would react with water, like this: NH4+ + H2O  NH3 + H3O+ Cl- would be the anion of HCl (a strong acid) so it would be a weak base, and would not react with water. What type of solution is the result? What would be the pH? Ka or Kb?

  24. Once the side reaction is known, the amount of acid or base formed can be calculated just like we did earlier. It might be helpful to know: Kw = 1 x10-14 = KaKb so if you know Ka and need Kb or vise-versa you can calculate what you need

  25. Weak acid - weak base: If the salt of a strong base with weak acid gave a basic solution, and the salt of a strong acid with a weak base gave an acidic solution, then What happens if they are both weak? if Ka = Kb, then neutral if Ka > Kb, then acidic (acid was stronger than base, but still weak) if Kb > Ka, then basic (base was stronger than acid, but still weak)

  26. Let’s recap: NaCl Na+ + H2O NaOH + H+ Cl- + H2O HCl + OH-

  27. Let’s recap: NaCl Na+ + H2O NaOH + H+ Cl- + H2O HCl + OH- Strong Base

  28. Let’s recap: NaCl Na+ + H2O NaOH + H+ Cl- + H2O HCl + OH- Strong Base = No Reaction

  29. Let’s recap: NaCl Na+ + H2O NaOH + H+ Cl- + H2O HCl + OH- Strong Base = No Reaction Strong Acid

  30. Let’s recap: NaCl Na+ + H2O NaOH + H+ Cl- + H2O HCl + OH- Strong Base = No Reaction Strong Acid = No Reaction

  31. Let’s recap: NaCl Na+ + H2O NaOH + H+ Cl- + H2O HCl + OH- Strong Base = No Reaction Strong Acid = No Reaction Conclusion: Neither Acidic or Basic = Neutral

  32. Let’s recap: NH4Cl NH4+ + H2O NH4OH + H+ Cl- + H2O HCl + OH-

  33. Let’s recap: NH4Cl NH4+ + H2O NH4OH + H+ Cl- + H2O HCl + OH- Weak Base

  34. Let’s recap: NH4Cl NH4+ + H2O NH4OH + H+ Cl- + H2O HCl + OH- Weak Base = Yes Reaction

  35. Let’s recap: NH4Cl NH4+ + H2O NH4OH + H+ Cl- + H2O HCl + OH- Weak Base = Yes Reaction Strong Acid

  36. Let’s recap: NH4Cl NH4+ + H2O NH4OH + H+ Cl- + H2O HCl + OH- Weak Base = Yes Reaction Strong Acid = No Reaction

  37. Let’s recap: NH4Cl NH4+ + H2O NH4OH + H+ Cl- + H2O HCl + OH- Weak Base = Yes Reaction Strong Acid = No Reaction Conclusion: as H+ is produced = acidic

  38. Let’s recap: NaC2H3O2 Na+ + H2O NaOH + H+ C2H3O2- + H2O HC2H3O2 + OH-

  39. Let’s recap: NaC2H3O2 Na+ + H2O NaOH + H+ C2H3O2- + H2O HC2H3O2 + OH- Strong Base

  40. Let’s recap: NaC2H3O2 Na+ + H2O NaOH + H+ C2H3O2- + H2O HC2H3O2 + OH- Strong Base = No Reaction

  41. Let’s recap: NaC2H3O2 Na+ + H2O NaOH + H+ C2H3O2- + H2O HC2H3O2 + OH- Strong Base = No Reaction Weak Acid

  42. Let’s recap: NaC2H3O2 Na+ + H2O NaOH + H+ C2H3O2- + H2O HC2H3O2 + OH- Strong Base = No Reaction Weak Acid = Yes Reaction

  43. Let’s recap: NaC2H3O2 Na+ + H2O NaOH + H+ C2H3O2- + H2O HC2H3O2 + OH- Strong Base = No Reaction Weak Acid = Yes Reaction Conclusion: as OH- is produced = basic

  44. Let’s recap: NH4C2H3O2 – Your Turn…

  45. Let’s recap: NH4C2H3O2 NH4+ + H2O NH4OH + H+ C2H3O2- + H2O HC2H3O2 + OH-

  46. Let’s recap: NH4C2H3O2 NH4+ + H2O NH4OH + H+ C2H3O2- + H2O HC2H3O2 + OH- Weak Base = Yes Reaction Weak Acid = Yes Reaction

  47. Let’s recap: NH4C2H3O2 NH4+ + H2O NH4OH + H+ C2H3O2- + H2O HC2H3O2 + OH- Weak Base = Yes Reaction Weak Acid = Yes Reaction Conclusion: we cannot yet say if it is acidic or basic! As Kb = 1.8 x 10-5 and Ka = 1.8 x 10-5 it is neutral, otherwise the larger K wins! (larger = small -exponent

  48. More Ionic Equilibria (Acid-Base Interactions)

  49. Much of life involves acids and bases, including your body. • What is in your stomach? • What must your pancreas produce to avoid dissolving your intestines? • When carbon dioxide is dissolved in water (blood) it forms carbonic acid. What must also be in our blood to avoid dissolving our vessels? • There are people who get sick after drinking a coke (coke contain acids). What must they be lacking?

  50. Much of life, then, involves buffers. A buffer contains a conjugate acid-base pair with both the acid and base in reasonable concentrations. (By Bronstead-Lowry definition.) The acidic part would react with added base. The basic part would react with added acid.

More Related