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Final Exam Wednesday, December 17 7:00 – 10:00 pm

Final Exam Wednesday, December 17 7:00 – 10:00 pm. 100 Noyes Lab AQB,AQ1 Davis AQ2,AQ3 Roberts AQ5,AQ6 Tsai. 150 Animal Science AQ7,AQ8 Vuong AQJ Livingston. 192 Lincoln AQ4,AQ9 Patel. 217 Noyes Lab AQA,AQC Chang. Free Tutoring! Chem 101, Chem 102, Chem 103, Chem 104

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Final Exam Wednesday, December 17 7:00 – 10:00 pm

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  1. Final Exam Wednesday, December 17 7:00 – 10:00 pm 100 Noyes Lab AQB,AQ1 Davis AQ2,AQ3 Roberts AQ5,AQ6 Tsai 150 Animal Science AQ7,AQ8 Vuong AQJ Livingston 192 Lincoln AQ4,AQ9 Patel 217 Noyes Lab AQA,AQC Chang

  2. Free Tutoring! Chem 101, Chem 102, Chem 103, Chem 104 Econ 102, Econ 103 Math 221, Math 231, Math 234 PHYS 211 Stop by GREGORY HALL ROOM 327 AND 329 FROM 2-4PM on December 11 for free tutoring from Phi Eta Sigma Questions? PEStutoring2@gmail.com

  3. Final Exam Friday, December 19 1:30 – 4:30 pm 100 Noyes Lab EQ2,EQ3,EQC,EQ8 Steele EQ4 Ip 141 Wohler EQ1,EQ6,EQ7,EQB Small EQ5 Ip

  4. Titration Curves I. Strong Acid + Strong Base 0.1 M HCl 0.1 M NaOH 25.0 mL 25.0 mL 2.5 x 10-3mol 2.5 x 10-3mol 1. Initial pH HCl  H+ + Cl- 0.1 M 0.1 M [H+] = 0.1 M pH = - log H+ = 1.00 .

  5. Strong Acid + Strong Base 0.1 M HCl 0.1 M NaOH 25.0 mL 10.0 mL 2.5 x 10-3mol - 1.0x 10-3 mol = 1.5 x 10-3 mol 25 + 10 mL V = [H+] = 1.5 x 10-3 mol 35 x 10-3 L 4.28 x 10-2 M [H+] = . . pH = 1.37

  6. Strong Acid + Strong Base 0.1 M HCl 0.1 M NaOH 25.0 mL 20.0 mL 2.5 x 10-3mol - 2.0 x 10-3 mol = 0.5 x 10-3 mol 25 + 20 mL V = [H+] = 0.5 x 10-3 mol 45 x 10-3 L 1.11 x 10-2 M [H+] = . . . pH = 1.95

  7. Strong Acid + Strong Base 0.1 M HCl 0.1 M NaOH 25.0 mL 25.0 mL 2.5 x 10-3mol - 2.5 x 10-3 mol = 0.0 mol [H+] = 1.00 x 10-7 M . pH = 7.00 . . .

  8. Strong Acid + Strong Base 0.1 M HCl 0.1 M NaOH 25.0 mL 35.0 mL 2.5 x 10-3mol 3.5 x 10-3 mol OH- 25 + 35 mL V = [OH-] = 1.0 x 10-3 mol . 60 x 10-3 L . 1.67 x 10-2 M [OH-] = . . . pOH = 1.78 pH = 12.22

  9. Titration Curves Weak Acid + Strong Base 0.1 M CH3COOH 0.1 M NaOH 25.0 mL 25.0 mL Initial weak acid pH = pKa + log [CH3COO-] [CH3COOH] Ka = 1.8 x 10-5 = [H+] [CH3COO-] [CH3COOH] half-way point pH = pKa = 4.74 equivalence point CH3COO- + H2O  CH3COOH + OH- Kb = 5.6 x 10-10 = [OH-] [CH3COOH] [CH3COO-] strong base

  10. Titration Curves Weak Base + Strong Acid 0.1 M NH3 0.1 M HCl 25.0 mL 25.0 mL 1. Initial pH NH3 + H2O NH4+ + OH-  Kb= 1.8 x 10-5 = [OH-] [NH4+] [NH3] [OH-] = 1.34 x 10-3 M pH = 11.12 pOH = 2.87

  11. Titration Curves Weak Base + Strong Acid 0.1 M NH3 0.1 M HCl 25.0 mL 10.0 mL 2.5 x 10-3mol - 1.0 x 10-3 mol = 1.5 x 10-3 mol 25 + 10 mL V = x = 2.67 x 10-5 pOH = 4.57 NH3 + H2O NH4+ + OH-  pH = 9.43 Kb= 1.8 x 10-5 = [NH4+] [OH-] [NH3] . [NH3] [NH4+] [OH-] 0.043 0.029 0.0 0.043 -x 0.029 + x x 1.8 x 10-5 = [x] [0.029 + x] [0.043 - x]

  12. Titration Curves Weak Base + Strong Acid 0.1 M NH3 0.1 M HCl 25.0 mL 20.0 mL 2.5 x 10-3mol - 2.0 x 10-3 mol = 5.0 x 10-4 mol 25 + 20 mL V = x = 4.5 x 10-6 pOH = 5.35 NH3 + H2O NH4+ + OH-  pH = 8.65 Kb= 1.8 x 10-5 = [NH4+] [OH-] [NH3] . . [NH3] [NH4+] [OH-] 0.011 0.044 0.0 0.011 -x 0.044 + x x 1.8 x 10-5 = [x] [0.044 + x] [0.011 - x]

  13. Titration Curves Weak Base + Strong Acid 0.1 M NH3 0.1 M HCl 25.0 mL 25.0 mL 2.5 x 10-3mol - 2.5 x 10-3 mol = 0.00 25 + 25 mL V = x = 5.9 x 10-6 pH = 5.27 NH4+ NH3 +H+  Ka= 5.6 x 10-10 = [NH3] [H+] [NH4+] . . [NH4] [NH3] [H+] 0.05 0.00 0.0 . 0.05 -x x x 5.6 x 10-10 = [x2] [0.05 - x]

  14. pOH = pKb + log [NH4+] [NH3] Titration Curves Weak Base + Strong Acid 0.1 M NH3 0.1 M HCl 25.0 mL 20.0 mL pH = 8.65 2.5 x 10-3mol Ka = 1.8 x 10-5 5.0 x 10-4 mol NH3 2.0 x 10-3 mol NH4+ V = 45 x 10-3 L pOH = 4.74 + log (0.44) (0.11) = 5.34

  15. Polyprotic Acid H2SO3 HSO3- Ka1 = 1.4 x 10-2  + H+ HSO3- SO32- Ka2 = 6.5 x 10-8  + H+ 2 equivalents of base 0.10 M H2SO3 0.10 M NaOH 40 mL 80 mL Initial pH 1.4 x 10-2 = [HSO3-] [H+] = x2 [H2SO3] 0.1 - x x = 0.03 pH = 1.51

  16. Polyprotic Acid H2SO3 HSO3- Ka1 = 1.4 x 10-2  + H+ HSO3- SO32- Ka2 = 6.5 x 10-8  + H+ 2 equivalents of base 0.10 M H2SO3 0.1 M NaOH half-way point pH = pKa - log 1.4 x 10-2 = 1.85 - log 6.5 x 10-8 = 7.19 1st equivalence point . . 1.84 + 7.19 = 4.52 . 2 2nd equivalence point conjugate base, SO3-

  17. Polyprotic Acid H2SO3 HSO3- Ka1 = 1.4 x 10-2  + H+ HSO3- SO32- Ka2 = 6.5 x 10-8  + H+ 2 equivalents of base Kb2 = 1 x 10-14 / 6.5 x 10-8 = 1.53 x 10-7 0.10 M H2SO3 [SO32-] = 1.53 x 10-7 = x2 / 0.83 0.1 mol = 0.83 .120 L x = 3.57 x 10-4 = [OH-] half-way point pH = pKa pH = 10.44 - log 1.4 x 10-2 = 1.85 buffering regions - log 6.5 x 10-8 = 7.19 . 1st equivalence point . . 1.84 + 7.19 = 4.52 . 2 2nd equivalence point conjugate base, SO3-

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