1. C. Y. Yeung (CHW, 2009) p.01 Acid-Base Eqm (2): K a and K b Review Questions Calculate the pH : Q.110 -8 HCl(aq) [100% ionized] Q.20.01M NaOH(aq) [100% ionized] [H 3 O + ] = (10 -7 + 10 -8 ) M = 1.1  10 -7 M,  pH = -log(1.1  10 -7 ) = 6.96 [OH - ] = 0.01M, pOH = -log(0.01) = -2  pH = 14 – 2 = 12

2. weak acid p.02 Acid Dissociation Constant (K a ) [temperature dependent!!] HA + H 2 O A - + H 3 O + Kc =Kc =Kc =Kc = [H 3 O + (aq)] [A - (aq)] [HA(aq)] [H 2 O(l)] K c [H 2 O(l)] = [H 3 O + (aq)] [A - (aq)] [HA(aq)] Ka =Ka =Ka =Ka = [HA(aq)]

3. p.03 Ka =Ka =Ka =Ka = [H 3 O + (aq)] [A - (aq)] [HA(aq)] Q:K a of methanoic acid (HCOOH) at 298K is 1.78  10 -4 mol dm -3. Calculate the pH value of 1.00M HCOOH. HCOOH + H 2 O HCOO - + H 3 O + at start at eqm 1.0000 1.00 – x xx Ka =Ka =Ka =Ka = x2x2x2x2 1 – x = 1.78  10 -4 x 2 = 1.78  10 -4 (x << 1,  K a is small) x = 0.0133  pH = -log(0.0133) = 1.88 solved eqn without approximation, x = 0.01325, pH = 1.88

4. p.04 Ka =Ka =Ka =Ka = x2x2x2x2 0.01 – x = 1.78  10 -4 x 2 = 1.78  10 -6 (x << 1,  K a is small) x = 1.33  10 -3  pH = -log(1.33  10 -3 ) = 2.87 Note: Different conc. of HA has different “% of dissociation” For 1.00M HCOOH, % of dissociation = (0.01325/1)  100% = 1.33 % Q:K a of methanoic acid (HCOOH) at 298K is 1.78  10 -4 mol dm -3. Calculate the pH value of 0.01 M HCOOH. For 0.01M HCOOH, % of dissociation = (1.242  10 -3 /0.01)  100% = 12.4 % solved eqn without approximation, x = 1.242  10 -3, pH = 2.91 *** i.e. [HA] decreases, % of dissociation increases.

5. p.05 For Polyprotic Acid : K a1, K a2, ….. [Note: K a1 > K a2 > … ]  WHY? Note:It is more difficult to remove a H + from the negatively charged HSO 4 -. H 2 SO 4 + H 2 O HSO 4 - + H 3 O + HSO 4 - + H 2 O SO 4 2- + H 3 O + K a1 K a2 H 3 O + is already produced in the first dissociation, it would suppress the formation of H 3 O + from the second dissociation. [COMMON ION EFFECT (p.148)] Overall K a of Polyprotic acid = K a1  K a2  …

6. p.06 Q:Assuming that the first ionization of H 2 SO 4 is 100%, and the K a2 of sulphuric acid is 0.01M. Find the pH of 0.10M H 2 SO 4. HSO 4 - + H 2 O SO 4 2- + H 3 O + K a2 = 0.01M at start at eqm 0.100.10 0.10 – x x 0.1+x K a2 = x(0.1+x) 0.1 – x = 0.01 x = 8.44  10 -3  pH = -log (0.1 + 8.44  10 -3 ) = 0.965

7. p.07 Importance of K a ?? 1.Indicates the strength of acid: (a)larger K a  stronger acid (b)smaller pK a  stronger acid (ref.: p. 141) 2.Indicates the relative stability of conjugate bases: e.g.K a of ethanoic acid = 1.80  10 -5 mol dm -3, K a of butanoic acid = 1.48  10 -5 mol dm -3. i.e.CH 3 CH 2 CH 2 COO - is less stable than CH 3 COO -.

8. p.08 Base Dissociation Constant (K b ) [temperature dependent!!] B + H 2 O HB + + OH - Kb =Kb =Kb =Kb = [HB + (aq)] [OH - (aq)] [B(aq)] Q:K b of NH 3 at 298 K is 1.78  10 -5 mol dm -3. What is the pH value of 1M NH 3 at 298K? Kb =Kb =Kb =Kb = x2x2x2x2 1 – x = 1.78  10 -5 x = 4.22  10 -3  pOH = -log(4.22  10 -3 ) = 2.37 pH = 14 – pOH = 11.6 pH = 14 – pOH = 11.6

9. p.09 Q:K b of NH 3 at 298 K is 1.78  10 -5 mol dm -3. What is the pH value of a solution which is both 1M NH 3 and 1M NH 4 Cl at 298K? NH 3 + H 2 O NH 4 + + OH - at start at eqm 1 0 1 1 – x 1+x x Kb =Kb =Kb =Kb =x(1+x) = 1.78  10 -5 x = 1.78  10 -5  pOH = -log(1.78  10 -5 ) = 4.75 pH = 14 - 4.75 = 9.25

10. p.10 Importance of K b ?? 1.Indicates the strength of base: (a)larger K b  stronger base (b)smaller pK b  stronger base (ref.: p. 142) 2.Indicates the relative stability of conjugate acid: e.g.K b of NH 3 = 1.74  10 -5 mol dm -3, K b of (CH 3 ) 2 NH = 9,55  10 -4 mol dm -3. i.e.(CH 3 ) 2 NH 2 + is more stable than NH 4 +.

11. p.11 Relationship between K w, K a and K b ?? HA + H 2 O A - + H 3 O + Ka =Ka =Ka =Ka = [H 3 O + (aq)] [A - (aq)] [HA(aq)] Kb =Kb =Kb =Kb = [HA(l)] [OH - (aq)] [A - (aq)] A - + H 2 O HA + OH - Ka  Kb =Ka  Kb =Ka  Kb =Ka  Kb = [H 3 O + (aq)] [A - (aq)] [HA(aq)]  [HA(l)] [OH - (aq)] [A - (aq)] [H 3 O + (aq)] [A - (aq)] [HA(aq)]  [HA(l)] [OH - (aq)] [A - (aq)] = K w

12. Assignment Study all examples in p.142 - 149 p.152 Q.3 –12 [due date: 26/3(Thur)] p.12 Next …. Explain the Strength of Organic Acids & Bases (Book 3A p. 113 – 133) Pre-Lab: Expt. 12 Determination of K a

13. p.13 Expt. 12Dissociation Constant of Weak Acids HA + H 2 O A - + H 3 O + Ka =Ka =Ka =Ka = [H 3 O + (aq)] [A - (aq)] [HA(aq)] log K a = [H 3 O + (aq)] [A - (aq)] [HA(aq)] log pK a = [A - (aq)] [HA(aq)] + log log [H 3 O + (aq)] pK a = [A - (aq)] [HA(aq)] + log pH

14. p.14 HA + H 2 O A - + H 3 O + pK a = [A - (aq)] [HA(aq)] + log pH Step 6 (10cm 3 NaOH added) 1  10 -3 mol 0 mol Step 5 1  10 -3 mol 0 mol Step 7 (another 10cm 3 HA added) 1  10 -3 mol at eqm 1  10 -3 mol – x 1  10 -3 mol + x in the same volume, i.e. same conc.! = pH + log (1) = pH measured by pH meter

15. p.15 Simple explanation on the Relative Acidity of Organic Acids The more stable is the conjugate base, the more acidic is the acid. 1. If the –ve charge of conjugate base is localized, it is more likely to recombine with the H 3 O + to form the acid again.  the acid is less acidic! 2. If the –ve charge of conjugate base is delocalized, it is less likely to recombine with the H 3 O + to form the acid again.  the acid is more acidic!

16. p.16 Alkyl group (e.g. CH 3 ) is e - releasing, halo group (e.g. Cl) is e - withdrawing. H C H CH 3 C O O -HC H Cl C O O - e- releasing group e- withdrawing group more stable conjugate base,  ClCH 2 COOH is more acidic! (i.e. larger K a, smaller pK a ) *** Longer alkyl group, more e - realeasing. ***More halogen atoms, more e- withdrawing.