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Outline: 2/28/07

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

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Outline: 2/28/07

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

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

  3. Acid-Base Behavior and Chemical Structure weaker bond more electroneg more O’s = electroneg Which is a stronger acid? HCl or HF HClO or HBrO or HIO HBrO2 or HBrO3

  4. Acid-Base Behavior and Chemical Structure • 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-.

  5. H H CH3 CH3 Carboxylic Acids Formic Acid Formate Ion - Acetic Acid Acetate Ion - Benzoic Acid Benzoate Ion -

  6. Representative Polyprotic Acids

  7. Polyprotic Acid Problem • Determine the concentration of all ions in 0.050 M H2CO3 • Step #1:Write down all reactions • H2CO3 + H2O  HCO3- + H3O+ Ka1=4.5 x 10-7 • HCO3- + H2O  CO32- + H3O+ Ka2=4.7 x10-11 • H2O + H2O H3O+ + OH- Kw=1.0 x10-14 Most of the H3O+ will come from the first Dissociation step. Treat as a simple weak acid problem.

  8. Step 2. Set up ICE Table • H2CO3 + H2O  HCO3- + H3O+ Ka1=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: • [HCO3- ]= [H3O+] = 1.5  10-4 M • H2CO3 = 0.050 M -1.5  10-4 = 0.0498 M

  9. What is the concentration of CO32-? HCO3- + H2O  CO32- + H3O+ Ka2=4.7 x10-11 1.5  10-4 0 1.5  10-4 init • -x +x +xchange (1.510-4-x) x 1.5  10-4+ xequil x(1.5  10-4+x)/(1.5  10-4 –x) =4.7 10-11 Notice Ka2 . Assume x is small. x = [CO32- ] = 4.7 10-11

  10. Chapter 17 : No Common Ions Ka 1.8 × 10-5 0.25 0.0 0.0 x2 0.25-x Ka = • A typical weak acid dissociation: HOAc  H+ + OAc- 0.25 M x= 2.1 × 10-3 pH = 2.67

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

  12. Acid Base reactions: • Add 0.10 mol NaOH to 1L H2O : 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: 0.25M buffer HOAc  H+ + OAc- Ka H+ + OH - H2O 1/Kw HOAc + OH - H2O + OAc- Ka/Kw • Add 0.10 mol NaOH to 1L HOAc: HOAc + OH - H2O + OAc- What is the correct Keq to use? Ka/Kw= 1.8 × 10-5 /1.0 × 10-14 = 1.8 × 109 Acid + Base reaction

  14. Reaction with Common Ions 0.15 0.0 0.35 (0.35-x) (0.15+x)x Keq = pOH = 8.89 • Add 0.10 moles of NaOH: HOAc + OH - H2O + OAc- 0.25 0.10 0.25 -0.10 -0.10 +0.10 +x +x -x pH = 5.11 Ka 1.8 × 109 x = 1.3 × 10-9

  15. Acid Base reactions: change = +6.00 change = +0.37 • 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 A Buffer!

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

  17. 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? • pH = pKa + log ([conj base]/[acid]) pH = 4.74 + log [OAc-]/[HOAc] 50 g CH3COONa = 0.61 mol/0.1L = 6.1 M pH = 4.74 + log (6.1/0.1) = 4.74 + 1.78 = 6.53

  18. Ka = [H+][A-]/[HA] • pKa = pH + p[A-]/[HA] • pKa = pH -log[A-]/[HA] • Where does this come from? • pH = pKa + log ([conj base]/[acid])

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

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