1. Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille, Chemistry, 2007 (John Wiley) ISBN: 9 78047081 0866

2. Slide 2/15 e CHEM1002 [Part 2] Dr Michela Simone Weeks 8 – 13 Office Hours: Monday 3-5, Friday 4-5 Room: 412A (or 416) Phone: 93512830 e-mail: michela.simone@sydney.edu.au

3. Slide 3/15 e Strong Acids and Bases Completely ionise in water: Strong acids H 2 SO 4, HCl, HBr, HI, HNO 3, HClO 4 Strong bases All hydroxides of Groups 1 & 2 (except Be): NaOH, Ca(OH) 2, … Equilibrium lies completely to the right, K a  . e.g. HCl(aq) H + (aq) + Cl – (aq)

4. Slide 4/15 x Examples What is the pH of a 0.1 M HCl solution?  Thus [H + ] = 0.1 M and pH = – log 10 [H + ] = 1.0 What is the pH of a 0.002 M NaOH solution?  Completely ionised, so [OH – ] = 0.002 M,  pOH = – log 10 [OH – ] = – log 10 (0.002) = 2.7  pH = 14 – 2.7 = 11.3 HCl(aq H + (aq) + Cl - (aq) [H + ] = 0.1 M

5. Slide 5/15 e Weak Acids pK a = – log 10 K a HA(aq) H + (aq) + A – (aq) Acid dissociation constant: Most acids or bases are weak  they do not completely ionise in wat er

6. Slide 6/15 e Relationship between K a and pK a  K a = 1.02 x 10 -2 then pK a = -log 10 (1.02 x 10 -2 ) = 1.991  pK a = 1.991 then K a = 10 -1.991 = 1.02 x 10 -2 The larger the value of K a, the stronger the acid and the lower the value of pK a.

7. Slide 7/15 x Find the pH of 0.1 M acetic acid (CH 3 COOH (HAc)) HAc(aq) H + (aq) + Ac – (aq) Example DATA: pK a = 4.7, K a = 10 -4.7 Initial (I):0.100 Change (C): Equilibrium (E):

8. Slide 8/15 x Find % ionisation of 0.50 M HF (pK a = 3.1) HF(aq) H + (aq) + F – (aq) Example Initial (I):0.500 Change (C): Equilibrium (E): % ionisation = x / 0.50  100 =

9. Slide 9/15 e Weak Bases Ionisation of a weak base: Tactic: calculate pOH and then pH, given K b. Equilibrium constant is called base ionisation constant, K b : NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH – (aq)

10. Slide 10/15 e For conjugate systems (Brønsted-Lowry acid-base pairs)  We only need values of pK a, since pK b = 14 – pK a pK a + pK b = 14  Conjugate base (A – ): A – (aq) + H 2 O(l) HA(aq) + OH – (aq)  Acid (HA): HA(aq) H + (aq) + A – (aq) Relationship Between pK a and pK b

11. Slide 11/15 x Find the pH of 1.0 x 10 –2 M NaHCO 2 (pK a of formic acid (HCO 2 H) is 4.1) HCO 2 – (aq) + H 2 O(l) OH – (aq) + HCO 2 H(aq)  pK b = 14 – pK a = 14 – 4.1 = 9.9 Example Initial (I):0.01large00 Change (C): Equilibrium (E): -x-x +x+x +x+x 0.01 - x x x

12. Slide 12/15 x Example (Continued) x = [OH – ] so pOH = -log 10 (10 -5.95 ) = 5.95  pH = 14 – 5.95 = 8.05 = 8.1 (one significant figure)

13. Slide 13/15 e Polyprotic Acids removing more protons is harder: increasing pK a = decreasing K a : K a1 > K a2 > K a3  reason: harder to remove +ve charge against increasing -ve charge large difference in pK a values  only need to consider one equilibrium at a time (simplifies maths!) H 3 PO 4 (aq) H + (aq) + H 2 PO 4 – (aq)pK a1 = 2.2 H 2 PO 4 – (aq) H + (aq) + HPO 4 2– (aq)pK a2 = 7.2 HPO 4 2- (aq) H + (aq) + PO 4 3– (aq)pK a3 = 12.4

14. Slide 14/15 x Practice Examples 1.Which one of the following is NOT a conjugate acid-base pair? (a)HCO 3 – and CO 3 2– (b)H 3 O + and H 2 O (c)OH – and O 2– (d)SO 3 and HSO 3 – (e)NH 2 OH 2 + and NH 2 OH 2. What is the pH of a 0.20 M solution of boric acid? The pKa of boric acid is 9.24. (a) 0.70 (b) 2.73 (c) 4.97 (d) 5.12 (e) 5.87 3.Rank the following solutions (all 1.0 M) in DECREASING order of pH. NaCl, NaCN, KOH, HCl, CH 3 COOH (a)HCl > CH 3 COOH > NaCN > NaCl > KOH (b)KOH > NaCl > NaCN > CH 3 COOH > HCl (c)KOH > NaCN > NaCl > HCl > CH 3 COOH (d)KOH > NaCl > NaCN > HCl > CH 3 COOH (e) KOH > NaCN > NaCl > CH 3 COOH > HCl

15. Slide 15/15 e Learning Outcomes - you should now be able to: Complete the worksheet Complete acid/base calculations Use pH, pK w, pK a and pK b Define strong and weak acids & bases Answer Review Problems 11.12-11.35 in Blackman Summary: Acids & Bases 2 Next lecture: Buffer systems