1. Outline:2/28/07 è Hand in Seminar Reports – to me è Pick up Quiz #6 – from me è Pick up CAPA 14 - outside è 5 more lectures until Exam 2… Today: è End Chapter 17 è Start Chapter 18 Ù Polyprotic acids Ù Buffers

2. Factors that affect acid strength: 1.Polarity of H  X bond 2.Bond strength of H  X bond 3.Stability of conjugate base X  Acid-Base Behavior and Chemical Structure

3. Which is a stronger acid? Acid-Base Behavior and Chemical Structure HCl or HF HClO or HBrO or HIO HBrO 2 or HBrO 3 more electroneg weaker bond more O’s = electroneg

4. Carboxylic Acids Carboxylic acids all contain COOH. All carboxylic acids are weak acids. When the carboxylic acid loses a proton, it generate the carboxylate anion, COO . Acid-Base Behavior and Chemical Structure

5. Carboxylic Acids HCH 3 H   Formic Acid Formate Ion  Acetic Acid Benzoic Acid Acetate Ion Benzoate Ion

6. Representative Polyprotic Acids

7. Polyprotic Acid Problem n Determine the concentration of all ions in 0.050 M H 2 CO 3 n Step #1: Write down all reactions H 2 CO 3 + H 2 O  CO 3 - + H 3 O + K a1 =4.5 x 10 -7  HCO 3 - + H 2 O  CO 3 2- + H 3 O + K a2 =4.7 x10 -11  H 2 O + H 2 O  H 3 O + + OH  K w =1.0 x10 -14 Most of the H 3 O + will come from the first Dissociation step. Treat as a simple weak acid problem.

8. H 2 CO 3 + H 2 O  CO 3 - + H 3 O + K a1 =4.5 x 10 -7 0.050 M 0 0 init  (0.050  x )M x M x M equil Assume x is small and we obtain: x= (0.050  4.5x10 -7 ) 1/2 = 1.5  10 -4 M Hence:  [  CO 3 - ]= [H 3 O + ] = 1.5  10 -4 M  H 2 CO 3 = 0.050 M  1.5  10 -4 = 0.0498 M Step 2. Set up ICE Table

9. What is the concentration of CO 3 2- ? 1.5  10 -4 0 1.5  10 -4 init   x +x +x change (1.5  10  4  x) x 1.5  10 -4 + x equil x(1.5  10 -4 +x)/(1.5  10 -4 –x) = 4.7  10 -11 Notice K a2. Assume x is small. x = [CO 3 2- ] = 4.7  10 -11 HCO 3 - + H 2 O  CO 3 2- + H 3 O + K a2 =4.7 x10 -11

10. Chapter 17 : No Common Ions è A typical weak acid dissociation: HOAc  H + + OAc  0.25 0.0 0.0 K a 1.8 × 10  5 K a = x 2 0.25  x x = 2.1 × 10  3 pH = 2.67 0.25 M

11. Chapter 18 : Common Ions è Add ions to both sides of equation: HOAc  H + + OAc  0.25 0.0 0.25 K a 1.8 × 10  5 K a = x(0.25+x) 0.25  x x = 1.8 × 10  5 pH = 4.74 vs. 2.67 LeChâtelier again

12. Acid Base reactions: è Add 0.10 mol NaOH to 1L H 2 O : Strong Base = complete dissociation pOH =  log(OH  ) =  log(0.1) = 1.0 pH = 13.0 No common ions: pH goes from 7.0 to 13.0

13. Acid Base reactions: è Add 0.10 mol NaOH to 1L HOAc: HOAc + OH   H 2 O + OAc  What is the correct K eq to use? HOAc  H + + OAc  KaKa H + + OH   H 2 O 1/K w HOAc + OH   H 2 O + OAc  K a /K w K a /K w = 1.8 × 10  5 /1.0 × 10  = 1.8 × 10  Acid + Base reaction 0.25M buffer

14. Reaction with Common Ions è Add 0.10 moles of NaOH: HOAc + OH   H 2 O + OAc  0.25 0.10 0.25 K a 1.8 × 10  -0.10 -0.10 +0.10 0.15 0.0 0.35 +x +x -x K eq = (0.35  x) (0.15  x)x x = 1.3 × 10  9 pOH = 8.89 pH = 5.11

15. Acid Base reactions: è Add 0.10 mol NaOH to …. No common ions: pH goes from 7.00 to 13.00 With common ions: pH goes from 4.74 to 5.11 change = +0.37 change = +6.00 A Buffer!

16. Chapter 18 : Buffers  Can add lots of HA or A  and pH doesn’t change much mathematically. è Easier to use Henderson-Hasselbach equation (p. 811)…. pH = pK a + log [base] [acid]

17. pH = 4.74 + log [OAc  ]/[HOAc] pH = pK a + log ([conj base] / [acid]) If 50.0 g of sodium acetate is added to 100 mL of 0.100 M solution of acetic acid, what will be the pH of the resultant buffer? 50 g CH 3 COONa = 0.61 mol/0.1L = 6.1 M pH = 4.74 + log (6.1/0.1) = 4.74 + 1.78 = 6.53

18. Where does this come from?  K a = [H + ][A  ] / [HA]  pK a = pH + p[A  ] / [HA]  pK a = pH  log[A  ] / [HA] pH = pK a + log ([conj base] / [acid])

19. Buffer calculations…. pH = pK a + log ([conj base]/[acid]) This one also exists: pOH = pK b + log ([conj acid]/[base]) Practice!