1 / 116

Chapter 15: Applications of Aqueous Equilibria

Chapter 15: Applications of Aqueous Equilibria. AP Chemsitry. Titrations and pH Curves. How does the pH of 50.0 ml 0.100 M HCl change as 0.200 M NaOH is added? STRONG ACID—STRONG BASE pH before any NaOH is added? major species? HCl  H + + Cl - [0.100 M] [0.100 M] [0.100 M].

miriam-gallagher
Download Presentation

Chapter 15: Applications of Aqueous Equilibria

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 15: Applications of Aqueous Equilibria AP Chemsitry

  2. Titrations and pH Curves How does the pH of 50.0 ml 0.100 M HCl change as 0.200 M NaOH is added? STRONG ACID—STRONG BASE • pH before any NaOH is added? major species? HCl  H+ + Cl- [0.100 M] [0.100 M] [0.100 M] H+ Cl- H2O pH = -log (0.100) = 1.000

  3. 50.0 ml 0.100 M HCl 10.0 ml 0.200 M NaOH 2.pH after some NaOH is added • Add 10.0 ml 0.200 M NaOH major species? H+ Cl- Na+ OH- H2O MoVo = MTVT Step 1: DILUTION For H+ (0.1)(50) = MT(60) MT = 0.0833 M For OH- (0.2)(10) = MT(60) MT = 0.0333 M 60.0 ml total

  4. Step 2: Neutralization Banana Chart: Before, Neutralization, After H+ + OH- H2O B N A 0.08330.0333 --- 0.0333 0.0333 --- 0.0500≈0 --- major species? H+ H2O pH = -log (0.0500) = 1.300

  5. 50.0 ml 0.100 M HCl 20.0 ml 0.200 M NaOH pH after some NaOH is added ii. Add 20.0 ml 0.200 M NaOH major species? H+ Cl- Na+ OH- H2O MoVo = MTVT Step 1: DILUTION For H+ (0.1)(50) = MT(70) MT = 0.0714 M For OH- (0.2)(20) = MT(70) MT = 0.0571 M 70.0 ml total

  6. Step 2: Neutralization Banana Chart: Before, Neutralization, After H+ + OH- H2O B N A 0.07140.0571 --- 0.0571 0.0571 --- 0.0143≈0 --- major species? H+ H2O pH = -log (0.00143) = 1.845

  7. 50.0 ml 0.100 M HCl 24.0 ml 0.200 M NaOH pH after some NaOH is added iii. Add 24.0 ml 0.200 M NaOH major species? H+ Cl- Na+ OH- H2O MoVo = MTVT Step 1: DILUTION For H+ (0.1)(50) = MT(74) MT = 0.0676 M For OH- (0.2)(25) = MT(74) MT = 0.0649 M 74.0 ml total

  8. Step 2: Neutralization Banana Chart: Before, Neutralization, After H+ + OH- H2O B N A 0.06760.0649 --- 0.0649 0.0649 --- 0.0027≈0 --- major species? H+ H2O pH = -log (0.0027) = 2.569

  9. 50.0 ml 0.100 M HCl 24.9 ml 0.200 M NaOH pH after some NaOH is added iv. Add 24.9 ml 0.200 M NaOH major species? H+ Cl- Na+ OH- H2O MoVo = MTVT Step 1: DILUTION For H+ (0.1)(50) = MT(74.9) MT = 0.0668 M For OH- (0.2)(25) = MT(74.9) MT = 0.0665 M 74.9 ml total

  10. Step 2: Neutralization Banana Chart: Before, Neutralization, After H+ + OH- H2O B N A 0.06680.0665 --- 0.0665 0.0665 --- 0.0003≈0 --- major species? H+ H2O pH = -log (0.0003) = 3.523

  11. 50.0 ml 0.100 M HCl 25.0 ml 0.200 M NaOH pH after some NaOH is added vi. Add 25.0 ml 0.200 M NaOH major species? H+ Cl- Na+ OH- H2O MoVo = MTVT Step 1: DILUTION For H+ (0.1)(50) = MT(75) MT = 0.0667 M For OH- (0.2)(25) = MT(75) MT = 0.0667 M 75.0 ml total

  12. Step 2: Neutralization Banana Chart: Before, Neutralization, After H+ + OH- H2O B N A 0.06670.0667 --- 0.0667 0.0667 --- ≈0≈0 --- major species? H2O pH = -log(1 X 10-7) = 7

  13. equivalence point (mol H+ = mol OH-)

  14. 50.0 ml 0.100 M HCl 30.0 ml 0.200 M NaOH 3. pH after 30 ml NaOH added major species? H+ Cl- Na+ OH- H2O MoVo = MTVT Step 1: DILUTION For H+ (0.1)(50) = MT(80) MT = 0.0625 M For OH- (0.2)(25) = MT(80) MT = 0.075 M 80.0 ml total

  15. Step 2: Neutralization Banana Chart: Before, Neutralization, After H+ + OH- H2O B N A 0.06250.0750 --- 0.0625 0.0625 --- ≈00.0125 --- major species? OH- H2O pOH = -log (0.0125) = 1.903 pH = 14 – pOH = 14 – 1.903 = 12.097

  16. Figure 15.1The pH Curve for the Titration of 50.0 mL of 0.200 M HNO3 with 0.100 M NaOH

  17. WEAK ACID—STRONG BASE How does the pH of 50.0 ml 0.100 M acetic acid sol’n change as 0.200 M NaOH is added? • pH before any NaOH is added? major species? HC2H3O2 H2O

  18. ICE, ICE, BABY Eqn? HC2H3O2 H+ + C2H3O2- I C E 0.100 ≈0 ≈0 -x +x +x 0.100-x x x

  19. HC2H3O2 H+ + C2H3O2- I C E 0.100 ≈0 ≈0 -0.00134 +0.00134 +0.00134 ≈ 0.100 0.00134 0.00134 x = 0.00134 pH = -log (0.00134) = 2.872

  20. 50.0 ml 0.100 M HAc 5.0 ml 0.200 M NaOH 2.pH after some NaOH is added • Add 5.0 ml 0.200 M NaOH major species? HC2H3O2 Na+ OH- H2O MoVo = MTVT Step 1: DILUTION For H+ (0.1)(50) = MT(55) MT = 0.0909 M For OH- (0.2)(5) = MT(55) MT = 0.0182 M 55.0 ml total

  21. Step 2: Neutralization HC2H3O2 OH- H2O major species: Eqn? HC2H3O2 + OH- H2O + C2H3O2- B N A 0.09090.0182 --- ≈0 0.0182 0.0182 --- 0.0182 0.0727≈0 --- 0.0182 major species? HC2H3O2 C2H3O2- H2O Eqn? HC2H3O2 H+ + C2H3O2-

  22. ICE, ICE, BABY HC2H3O2 H+ + C2H3O2- I C E 0.0727 ≈0 0.0182 -x +x +x 0.0727-x x 0.0182+x x = 7.19 X 10-5

  23. HC2H3O2 H+ + C2H3O2- I C E 0.0727 ≈0 0.0182 7.19 X 10-5 +7.19 X 10-5 + 7.19 X 10-5 ≈0.0727 7.19 X 10-5≈0.0182 pH = -log (7.19 X 10-5) = 4.143

  24. Terms: Common Ion Effect: HC2H3O2 H+ + C2H3O2- Add NaC2H3O2? A response to the addition of an ion already involved in the equilibrium

  25. Buffered Solution: A sol’n that RESISTS a change in the pH Made up of: A WEAK ACID and its SALT OR A WEAK BASE and its SALT i.e. HA and NaA i.e. B and BHCl NH3 and NH4Cl

  26. Why WEAK acids and bases? do not 100% dissociate HA + H2O H3O+ + A- add H+? Shift makes more acid (HA) add OH-? Shift makes more base (A-) Acid Base C.Acid C. Base

  27. Why WEAK acids and bases? do not 100% dissociate B + H2O BH+ + OH- add H+? Shift makes more acid (BH+) add OH-? Shift makes more base (B) Acid Base C.Acid C. Base

  28. Henderson-Hasselbalch Equation

  29. 50.0 ml 0.100 M HAc 10.0 ml 0.200 M NaOH 2.pH after some NaOH is added ii. Add 10.0 ml 0.200 M NaOH major species? HC2H3O2 Na+ OH- H2O MoVo = MTVT Step 1: DILUTION For H+ (0.1)(50) = MT(60) MT = 0.0833 M For OH- (0.2)(10) = MT(60) MT = 0.0333 M 60.0 ml total

  30. HC2H3O2 H+ + C2H3O2- Step 2: Neutralization HC2H3O2 OH- H2O major species: Eqn? HC2H3O2 + OH- H2O + C2H3O2- B N A 0.08330.0333 --- ≈0 0.0333 0.0333 --- 0.0333 0.0500≈0 --- 0.0333 major species? HC2H3O3 C2H3O2- H2O Eqn?

  31. ICE, ICE, BABY Or……….. Henderson-Hasselbalch Eqn

  32. ICE, ICE, BABY HC2H3O2 H+ + C2H3O2- I C E 0.0500 ≈0 0.0333 -x +x +x 0.05-x x 0.0333+x x = 2.70 X 10-5

  33. HC2H3O2 H+ + C2H3O2- I C E 0.0500 ≈0 0.0333 2.70 X 10-5 +2.70 X 10-5 +2.70 X 10-5 ≈0.0500 2.70 X 10-5≈0.0333 pH = -log (2.70 X 10-5) = 4.568

  34. Or…H-H Eqn 4.74

  35. 50.0 ml 0.100 M HAc 12.5 ml 0.200 M NaOH 2.pH after some NaOH is added iii. Add 12.5 ml 0.200 M NaOH major species? HC2H3O2 Na+ OH- H2O MoVo = MTVT Step 1: DILUTION For H+ (0.1)(50) = MT(62.5) MT = 0.0800 M For OH- (0.2)(12.5) = MT(62.5) MT = 0.0400 M 62.5 ml total

  36. HC2H3O2 H+ + C2H3O2- Step 2: Neutralization HC2H3O2 OH- H2O major species: Eqn? HC2H3O2 + OH- H2O + C2H3O2- B N A 0.08000.0400 --- ≈0 0.0400 0.0400 --- 0.0400 0.0400≈0 --- 0.0400 major species? HC2H3O3 C2H3O2- H2O Eqn?

  37. ICE, ICE, BABY Or……….. Henderson-Hasselbalch Eqn

  38. ICE, ICE, BABY HC2H3O2 H+ + C2H3O2- I C E 0.0400 ≈0 0.0400 -x +x +x 0.04-x x 0.04+x x =1.80 X 10-5

  39. HC2H3O2 H+ + C2H3O2- I C E 0.0400 ≈0 0.0400 1.80 X 10-5 +1.80 X 10-5 +1.80 X 10-5 ≈0.0400 1.80 X 10-5≈0.0400 pH = -log (1.80 X 10-5) = 4.745

  40. Or…H-H Eqn 4.74 =1!!!

More Related