1. C. Y. Yeung (CHW, 2009) p.01 Acid-Base Eqm (3): Explain the Strength of Organic Acids & Bases Review Questions Q.1:pK b of CH 3 NH 2 (an organic weak base) at 298K is 4.75. Find the pH of 0.05M CH 3 NH 2 at 298K. Kb =Kb =Kb =Kb = x2x2x2x2 0.05 – x = 10 -4.75 x2x2x2x20.05 = 1.778  10 -5  pOH = -log(9.43  10 -4 ) = 3.03 pH = 14 – pOH = 11.0 pH = 14 – pOH = 11.0 x = 9.43  10 -4

2. p.02 Q.2:25.0 cm 3 of a monobasic organic weak base B(aq) requires 20.25 cm 3 of 0.50M HCl(aq) for complete neutralization. Given that the pH of the B(aq) at 298K is 11.25, what is K b of B at 298K? As pH = 11.25, i.e. pOH = 14 – 11.25 = 2.75 conc. of B(aq) = (20.25/1000)  0.50 = 0.405 M (25.0/1000) B + H 2 O HB + + OH - at start at eqm 0.405 0 0 0.405 – x xx  x = [OH - ] = 10 -2.75 = 1.778  10 -3 Kb =Kb =Kb =Kb = (1.778  10 -3 ) 2 0.405 – 1.778  10 -3 = 7.84  10 -6 M

3. p.03 Stability of Conjugate Base … e.g. CCl 3 COO - is the conjugate base of CCl 3 COOH i.e. CCl 3 COO - is stable, thus it is less likely to react with H 3 O + to regenerate CCl 3 COOH. CCl 3 COO - is stabilized by “negative inductive effect”. electronegative atom !! The -ve charge is dispersed /delocalized over the anion.  The -ve charge is less concentrated on the O atom. CCl 3 COOH + H 2 O CCl 3 COO - + H 3 O + Cl C Cl Cl C O O -

4. p.04 Conjugate Base may be stabilized by “resonance effect” + H 2 O + H 3 O + phenol phenoxide O-O-O-O- O-O-O-  The -ve charge is delocalized over the aromatic ring. (resonance effect) i.e. phenoxide ion is stabilized by resonance effect. (resonance structures)

5. p.05 pK a of Phenol = 9.95 pKa of Ethanoic acid = 4.76 stronger acid! How to explain …? Although CH 3 COO - is destabilized by positive inductive effect, it is stabilized by resonance effect! CH 3 COO - C O O- both are electronegative! i.e. –ve charge is shared equally. In the case of phenoxide, delocalization of –ve charge by resonance is less effective, because the O atom is more electronegative than the carbon atoms in aromatic ring. The –ve charge is more delocalized in CH 3 COO -, and thus it is more stable.  CH 3 COOH is the stronger acid.

6. (resonance structures) p.06 Strength of Base … (depends on the availability of lone pair e -.) CH 3 NH H push e - availability of lone pair e - is increased by +I effect  stronger base (than NH 3 )! NH H NH H availability of lone pair e - is decreased by resonance.  weaker base (than NH 3 )! NH H -+ HN H -+ HN H-+ Methylamine (pK b = 3.38) phenylamine / aniline (pK b = 9.40) Ref: pK b of NH 3 = 4.80

7. a.1 Advanced Understanding on the Strength of Organic Acid (1) --- 4-nitrophenol (p-nitrophenol) (p-nitrophenol)(p-nitrophenol) O O O N - +- How can 4-nitrophenoxide ion be stabilized by resonance effect? O OON - + - O OON - + - O OON - + - O OON + - - O OON + - - The -ve charge on the p-nitro phenoxide ion is more dispersed than that on the phenoxide ion. Therefore, it is stronger than phenol. pK a of phenol = 9.95 pK a of p-nitrophenol = 7.14

8. a.2 Advanced Understanding on the Strength of Organic Acid (2) --- 3-nitrophenol (m-nitrophenol) Is the pK a of m-nitrophenol smaller than p-nitrophenol? O O O N - +- O OON - + - O OON - + - O OON - + - The -ve charge cannot be dispersed over the nitro group. Therefore, it is weaker than p-nitrophenol, and its pK a is larger. (But it is still stronger than phenol due to negative inductive effect.) pK a of phenol = 9.95 pK a of p-nitrophenol = 7.15 pK a of m-nitrophenol = 8.35

9. a.3 Advanced Understanding on the Strength of Organic Acid (3) --- p-hydroxybenzaldehyde How can the conjugate base be stabilized by resonance effect? H O O C- H OOC - H OOC - H OOC - H OOC - H OOC - The -ve charge is dispersed over the anion, p-hydroxy benzaldehyde is acidic. Therefore, it is stronger than phenol. pK a of phenol = 9.95 pK a of p-hydroxybenzaldhyde = 7.66

10. a.4 Look at their pK a values again …… H OOHC OH O O OH N - + O O OH N - + pK a (at 298K) 9.957.158.357.66 nitro group (– NO 2 ) at the p-position stabilizes the conjugate base by delocalizing the – ve charge through: negative inductive effect negative inductive effect resonance effect resonance effect

11. a.5 Compare the pK b values of m-nitroaniline, p-nitroaniline and aniline. NH 2 O O N - + O O N - + pK b (at 298K) 9.4012.911.6 The availability of lone pair e - on the –NH 2 group is reduced by : negative inductive effect negative inductive effect resonance effect resonance effect

12. p.07 In Conclusion …. Stronger Acid: conjugate base stabilized by - I effect / resonance effect. (delocalization of –ve charge) Stronger Base: availability of lone pair e - increased by + I effect. (localization of –ve charge) Weaker Acid: conjugate base destabilized by + I effect. (localization of –ve charge) Weaker Base: availability of lone pair e - decreased by - I effect / resonance. (delocalization of –ve charge)

13. p.08 More practice on the calculation of K a, K b and pH Q.1:At body temperature, human blood pH is 7.40. Lactic acid, CH 3 CH(OH)COOH, is produced in muscle tissues during physical exertion. Whether the lactic acid produced exists mainly in form of “undissociated molecule” or “dissociated lactate ions”? (Given: K a of lactic acid at body temperature = 8.50  10 -4 mol dm -3 ) Ka =Ka =Ka =Ka = [H 3 O + ][CH 3 CH(OH)COO - ] [CH 3 CH(OH)COOH] pK a = [CH 3 CH(OH)COO - ] [CH 3 CH(OH)COOH] pH – log 2.14  10 4 = [CH 3 CH(OH)COOH] [CH 3 CH(OH)COO - ]  Lactic acid formed is mainly in form of dissociated lactic ions. With a given K a, the ratio of [salt] and [acid] could be found from the pH value.

14. p.09 Q.2:Calculate the pH value of the resultant solution for 10cm 3 of 0.20M CH 3 COOH are mixed with 10cm 3 of 0.40M CH 3 COONa. (K a = 1.75  10 -5 mol dm -3 ) x(0.20+x) 0.10 – x 1.75  10 -5 = x = 8.75  10 -6  pH = 5.06 After mixing, new [CH 3 COOH] = 0.10M new [CH 3 COONa] = 0.20M An acid-salt mixture (acid and salt have comparable conc.)  ACIDIC BUFFER! pH of acidic buffer could be adjusted by [acid] and [salt].

15. p.10 Q.3:Calculate the pH value of the resultant solution for 10cm 3 of 1.00M NH 3 are mixed with 10cm 3 of 1.00M NH 4 Cl. (K b = 1.78  10 -5 mol dm -3 ) x(0.50+x) 0.50 – x 1.78  10 -5 = x = 1.78  10 -5  pOH = 4.75 After mixing, new [NH 3 ] = 0.50M new [NH 4 + ] = 0.50M A base-salt mixture (base and salt have comparable conc.)  BASIC BUFFER!  pH = 9.25 pH of basic buffer could be adjusted by [base] and [salt].

16. Assignment Lab report [due date: 6/4(Mon)] p.11 Next …. Acidic Buffers and Basic Buffers (Book 2 p. 153 – 163) Pre-Lab: Expt. 13 Analysis of Two Commercial Brands of Bleaching Solution Book 3A p.141 Q.12, 13, 14 [due date: 20/4 (Mon)]

17. xxx (o)(o)(o)(o) (m)(m)(m)(m) (p)(p)(p)(p)