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Outline: 2/26/07

Outline: 2/26/07. CAPA 10 & 11 – due today Pick up CAPA 12 & 13 – outside Turn in seminar reports – to me. Today: Start Chapter 17 . Acid-base Equilibria : Acid & Bases: definitions Defining/Calculating pH Simple pH Calculations from K a. Any questions on equilibria?.

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Outline: 2/26/07

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  1. Outline: 2/26/07 • CAPA 10 & 11 – due today • Pick up CAPA 12 & 13 – outside • Turn in seminar reports – to me • Today: Start Chapter 17 • Acid-base Equilibria: • Acid & Bases: definitions • Defining/Calculating pH • Simple pH Calculations from Ka

  2. Any questions on equilibria? Lots of suggested problems to do…. 16.1, 16.9, 16.11, 16.17, 16.19, 16.21, 16.23, 16.25, 16.27, 16.29, 16.31, 16.33, 16.35, 16.37, 16.43, 16.45, 16.47, 16.49, 16.51, 16.57, 16.59, 16.63. 16.65, 16.67, 16.69, 16.71, 16.73, 16.75, 16.79, 16.81

  3. Quiz #5 Please put books/papers away… Remember: lots of partial credit for setting up the problem correctly… Solving it just gets you the last points

  4. Quiz #5 Please turn your quizzes over and pass them to your right…

  5. Acids & Bases: • These are all examples of equilibria big Keq = strong acid (dissociation) • little Keq = weak acid • HCl + H2O  H3O+ + Cl- Keq = 1000+ • HSO3- + H2O  H3O+ + SO32- Keq = 6.4  10-8

  6. p. 140 H2O H3O+ NO3- Chapter 17: More Equilibrium! What exactly is in an aqueous solution? HNO3

  7. p. 141 • Does this differ from a weak acid? HOAc H2O

  8. A combination of previous chapters: = Strong Acids • Solubility rules: • Highly soluble: • ionic salts of Cl-, Br-, I-, • ionic salts of NO3-, SO4-, • ionic salts of ClO4-, etc. • Equilibrium rules: • Keq = [products]/[reactants] • Big Keqlarge extent of reaction = Strong Acids

  9. Just a few strong acids & bases, the rest are weak:

  10. The extent of dissociation:Ka and Kb • Define acid dissociation constant: • Ka= [H+][A-]/[ HA] • Define base dissociation constant: • Kb = [B+][OH-]/[BOH] • OK, what exactly is water anyway? • H2O + H2O H3O+ + OH- • Keq = [H3O+][OH-] = Kw = 1  10-14 (by definition) • Called water dissociation constant • or “autoionization”

  11. Define water as neutral…. • Kw = 10-14 (arbitrary scale definition) • then: [H3O+] = [OH-] = x • or: x2 = 10-14 or x = 10-7 M Define pH = - log [H+] • What’s the pH of neutral water? • = - log(10-7) = 7.0

  12. Same thing for bases…. • Define pOH = -log(OH-) • Since Kw = [H3O+][OH-] = 10-14 • log[H3O+] + log[OH-] = log (10-14) • or pH + pOH = 14 • None of this is really new….using acids and bases in lab all year! • e.g. pHydron paper, pH electrodes, • Concentration vs. strength!

  13. 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

  14. 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

  15. 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

  16. 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

  17. And Practice, Practice, Practice…

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