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Outline: 2/28/07

Outline: 2/28/07. Seminar – Friday @ 4pm Pick up Quiz #5 – from me Mid-term grades due today…. Today: Chapter 17 (cont’d). Acid-base Equilibria : More weak acid/base/salt pH calcs The relationship K a  K b = K w Hydrolysis & Neutralization. + 1.0 +0.5 - 1.0 .

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Outline: 2/28/07

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  1. Outline: 2/28/07 • Seminar – Friday @ 4pm • Pick up Quiz #5 – from me • Mid-term grades due today… • Today: Chapter 17 (cont’d) • Acid-base Equilibria: • More weak acid/base/salt pH calcs • The relationship Ka Kb = Kw • Hydrolysis & Neutralization

  2. +1.0 +0.5-1.0 -2x -x +2x (change) neglect (2x)2 Keq = (2)2(1.5) Quiz #5 3. 2 NO(g) + O2(g) 2NO2(g) K is small 1.0 1.0 1.0 (init) 2.0 1.5 0.0 (init) 2 - 2x 1.5-x 2x (equil.)

  3. Acids & Bases: Chapter 17 • Any chemical that donates a proton is a Bronsted-Lowry acid… any chemical that accepts a proton is a Bronsted-Lowry base. • HCl  H+ + Cl- • HCl + H2O  H3O+ + Cl- acid base Conjugate acid Conjugate base

  4. Acids & Bases: • Some chemicals can do both: they are called “amphoteric” • HSO3- + H2O  H3O+ + SO32- acid base Conjugate acid Conjugate base • HSO3- + H2O  H2SO3 + OH- base acid Conjugate acid Conjugate base

  5. Calculation Examples: • Calculate the pH of a solution of 2.510-2 M HClO4? (perchloric acid) • Step #1:identify the acid • If strong acid  complete dissociation • HClO4H++ ClO4- • 2.510-2 M 0 0 init • 0 equil 2.510-2M 2.510-2 M • Step #2: pH = - log [H+]- log (2.510-2) = 1.60

  6. Examples: • Calculate the pH of a solution of 2.510-2 M HClO? (hypochlorous acid) • Step #1:identify the acid • if weak acid  look up Ka = 3.5  10-8 • HClOH++ ClO- • 2.510-2 M 0 0 init • (2.510-2-x)M x M x M equil • Step #2: pH = - log [H+]- log (2.710-5) = 4.56

  7. Weak Base Example: • Calculate the pH of a solution of 2.510-2 M trimethylamine (CH3)3N ? • Step #1:identify it as a weak base • look up Kb = 6.5  10-5 • (CH3)3N + H2OOH-+ (CH3)3NH+ • 2.510-2 M 0 0 init • (2.510-2-x)M x M x M equil • Step #2: pOH = - log [OH-]- log (1.310-3) = 2.89 • Step #3: 14 -pOH = pH = 11.11

  8. Problem solving overview: • Skills needed to solve these problems: • 1) Identify strong/weak acid/bases • 2) Identify conjugate acid/bases & salts • 3) Solving equilibrium problems • 4) Identifying & making the correct Keq Know the 6 strong acids / 4 strong bases… Know solubility rules & acid/base definitions Solve for x…forward or backwards... Add equations to make new one & K’s multiply

  9. Keq Ka1 Ka2 Acids & Bases: What are Ka & Kb? Ka • HA  H+ + A- • H2A  H+ + HA- • HA- H+ + HA- • B+ + H2O  H+ + BOH • BOH  B+ + OH- • A-+ H2O HA + OH- • H2O  H+ + OH-

  10. Keq Ka1 Ka2 • “polyprotic acids” Acids & Bases: What are Ka & Kb? • H2A  H+ + HA- • HA- H+ + HA-

  11. Keq Ka1 Ka2 Acids & Bases: What are Ka & Kb? Ka • HA  H+ + A- • H2A  H+ + HA- • HA- H+ + HA- • B+ + H2O  H+ + BOH Ka • BOH  B+ + OH- • A-+ H2O HA + OH- • H2O  H+ + OH-

  12. Keq Acids & Bases: What are Ka & Kb? • B+ + H2O  H+ + BOH Ka • “Hydrolysis” • (often metal salts….)

  13. Keq Ka1 Ka2 Acids & Bases: What are Ka & Kb? Ka • HA  H+ + A- • H2A  H+ + HA- • HA- H+ + HA- • B+ + H2O  H+ + BOH Ka Kb • BOH  B+ + OH- • A-+ H2O HA + OH- Kb • H2O  H+ + OH-

  14. Keq Acids & Bases: What are Ka & Kb? • A-+ H2O HA + OH- Kb • “hydrolysis” (conjugate base salts)

  15. Keq Ka1 Ka2 Acids & Bases: What are Ka & Kb? Ka • HA  H+ + A- • H2A  H+ + HA- • HA- H+ + HA- • B+ + H2O  H+ + BOH Ka Kb • BOH  B+ + OH- • A-+ H2O HA + OH- Kb Kw • H2O  H+ + OH-

  16. Keq • H2O  H+ + OH- Kw • Or: • KaKb =Kw Acids & Bases: What are Ka & Kb? Ka • HA  H+ + A- • A-+ H2O HA + OH- Kb Remember: Hess’ Law

  17. Example: • Determine Keq for: • HCO3- + OH-  CO32-+ H2O • Step #1:identify the acid • H2CO3 (carbonic acid) • HCO3- H++ CO32- Ka2 • 4.810-11 • Step #2: what needs to be added? • H++ OH-  H2O • 1/Kw 1/1.010-14 Keq = 4.810-11 / 1.010-14

  18. Let’s try some more examples • The pH of an 0.0222 M solution of a weak acid is 5.43 Which acid is it? • HAH++ A- • (0.0222 -x) M x M x M Equil • You knowx = 10-5.43 = 3.71  10-6 • Solve forKeq • K = x2/ (0.0222 - x) = (3.71e-6)2/(0.0222) • = 6.210-10 App. E:HCN

  19. Worksheet 7 Try some more…

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