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

2. Quiz #5 3. 2 NO (g) + O 2(g)  2NO 2(g) 1.0 1.0 1.0 (init)  x  x +2x (change) 2  2x 1.5  x 2x (equil.) K eq = (2x) 2 (2) 2 (1  ) K is small  1.0   1.0 2.0 1.5 0.0 (init) neglect

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 + H 2 O  H 3 O + + Cl  acidbaseConjugate acid Conjugate base

4. Acids & Bases: Some chemicals can do both: they are called “amphoteric”  HSO 3  + H 2 O  H 3 O + + SO 3  acidbaseConjugate acid Conjugate base  HSO 3  + H 2 O  H 2 SO 3 + OH  acidbaseConjugate acid Conjugate base

5. Calculation Examples: Calculate the pH of a solution of 2.5  10  2 M HClO 4 ? (perchloric acid) n Step #1: identify the acid If strong acid  complete dissociation Step #2: pH =  log [H + ]  log (2.5  10  2 ) = 1.60  HClO 4  H + + ClO 4   2.5  10  2 M 0 0 init  0 equil  2.5  10  2 M 2.5  10  2 M

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

7. Weak Base Example: Calculate the pH of a solution of 2.5  10  2 M trimethylamine (CH 3 ) 3 N ? n Step #1: identify it as a weak base  look up K b = 6.5  10  5  (CH 3 ) 3 N  H 2 O  OH  + (CH 3 ) 3 N H +  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 K eq 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. Acids & Bases: What are K a & K b ?   HA  H + + A   H 2 A  H + + HA   HA   H + + HA  K eq  BOH  B + + OH   A  + H 2 O  HA + OH  B + + H 2 O  H + + BOH  H 2 O  H + + OH  KaKa K a1 K a2

10. Acids & Bases: What are K a & K b ?  H 2 A  H + + HA   HA   H + + HA  K eq K a1 K a2 “polyprotic acids”

11. Acids & Bases: What are K a & K b ?   HA  H + + A   H 2 A  H + + HA   HA   H + + HA  K eq  BOH  B + + OH   A  + H 2 O  HA + OH  B + + H 2 O  H + + BOH  H 2 O  H + + OH  KaKa K a1 K a2 KaKa

12. Acids & Bases: What are K a & K b ? K eq “Hydrolysis” (often metal salts….) B + + H 2 O  H + + BOH KaKa

13. Acids & Bases: What are K a & K b ?   HA  H + + A   H 2 A  H + + HA   HA   H + + HA  K eq  BOH  B + + OH   A  + H 2 O  HA + OH  B + + H 2 O  H + + BOH  H 2 O  H + + OH  KaKa K a1 K a2 KaKa KbKb KbKb

14. Acids & Bases: What are K a & K b ? K eq  A  + H 2 O  HA + OH  “hydrolysis” (conjugate base salts) KbKb

15. Acids & Bases: What are K a & K b ?   HA  H + + A   H 2 A  H + + HA   HA   H + + HA  K eq  BOH  B + + OH   A  + H 2 O  HA + OH  B + + H 2 O  H + + BOH  H 2 O  H + + OH  KaKa K a1 K a2 KaKa KbKb KbKb KwKw

16. Acids & Bases: What are K a & K b ?   HA  H + + A  K eq  A  + H 2 O  HA + OH  KaKa KbKb  H 2 O  H + + OH  KwKw Or: K a  K b = K w Remember: Hess’ Law

17. Example: n Determine K eq for:  HCO 3  + OH   CO 3  + H 2 O Step #1: identify the acid H 2 CO 3 (carbonic acid) Step #2: what needs to be added?  H + + OH   H 2 O  HCO 3   H + + CO 3  K a2  4.8  10  11  1/K w 1/1.0  10  14 K eq = 4.8  10  11 / 1.0  10  14

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

19. Worksheet 7 Try some more…