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 2/17 e CHEM1002 [Part 2] A/Prof Adam Bridgeman (Series 1) Dr Feike Dijkstra (Series 2) Weeks 8 – 13 Office Hours: Monday 2-3, Friday 1-2 Room: 543a e-mail: adam.bridgeman@sydney.edu.au e-mail: feike.dijkstra@sydney.edu.au

3. Slide 3/17 e Lecture 3: Salts of Acids and Bases Buffer systems Blackman Chapter 11, Sections 11.3-11.6 Acids & Bases Reproduced from ‘The Extraordinary Chemistry of Ordinary Things, C.H. Snyder, Wiley, 2002 (Page 245) Lecture 4: Titrations Blackman Chapter 11, Section 11.7

4. Slide 4/17 e Strong acids completed dissociates in solution: HA + H 2 O  A - (aq) + H 3 O + (aq) pH = -log 10 ([strong acid] initial ) If strong base is added, it reacts with the H 3 O + so [H 3 O + (aq)] is reduced: H 3 O + (aq) + OH-(aq)  H 2 O(l) pH = -log 10 ([strong acid] remaining ) Equivalence point: when the amount of added base = initial amount of acid: [H + (aq)] = 10 -7.0 M pH = 7.00 After the equivalence point, any added base increases [OH - (aq)]:  pH = 14.00 - pOH = 14.00 - (-log 10 ([excess base])) Strong Acid/Strong Base Titration

5. Slide 5/17 e Initial pH is  pH = 14 - pOH = 14 -(-log 10 ([strong base] initial ) At the equivalence point,  [H + (aq)] = 10 -7.0 M  pH = 7.00 (at 25 °C) After the equivalence point,  pH = -log 10 ([excess acid]) Strong Base/Strong Acid Titration

6. Slide 6/17 e Strong Base/Strong Acid Titration Add HCl(g) in 0.10 mol amounts to 1.0 L of 0.5 M NaOH(aq)

7. Slide 7/17 x Weak Acid/Strong Base Titration Add NaOH(s) in 0.10 mol amounts to 1.0 L of 0.5 M acetic acid Initial pH: initially the solution contains just acetic acid and its pH must be calculated using the procedure outlined in slide 7 of lecture 2. pK a = 4.76. CH 3 COOHH2OH2OH3O+H3O+ CH 3 COO – initial (I)0.5large00 change (C)negligible equilibrium (E)large K a =

8. Slide 8/17 x Weak Acid/Strong Base Titration Add NaOH(s) in 0.10 mol amounts to 1.0 L of 0.50 M acetic acid cf lecture 3 Acidic region: in the acidic region, OH - has reacted with some of the acetic acid to make acetate. The solution then contains both acetic acid and its conjugate base and the Henderson-Hasselbalch equation can be used to calculate the pH. CH 3 COOH(aq) + OH - (aq)  CH 3 COO - (aq) + H 2 O(aq)

9. Slide 9/17 x Weak Acid/Strong Base Titration Add NaOH(s) in 0.10 mol amounts to 1.0 L of 0.50 M acetic acid cf lecture 2 Equivalence point: all of the acetic acid has reacted with OH - to form CH 3 COO -. At this point, the solution contains a weak base so is basic. The pOH must be calculated. CH 3 COO - H2OH2OCH 3 COOHOH - initial (I)0.5large00 change (C)negligible equilibrium (E)large K b =

10. Slide 10/17 x Weak Acid/Strong Base Titration Add NaOH(s) in 0.10 mol amounts to 1.0 L of 0.50 M acetic acid Alkaline region: after the equivalence point, the major source of OH - is that from the excess NaOH. Very little is contributed from the acetate ion. After the equivalence point, the calculations are then exactly the same as for the strong acid/strong base titration. When 0.60 mol of NaOH(s) is added, 0.50 mol react with the CH 3 COOH leaving a 0.10 M solution of OH - :

11. Slide 11/17 e Weak Acid/Strong Base Titration Initial pH is higher than for strong acid / strong base  A weak acid is present initially In acidic region, weak acid and conjugate base present  Buffering region with slow change in pH  At the 1/2 equivalence point, pH = pK a At equivalence point, conjugate base present  pH > 7 In alkaline region, excess strong base present  Curve is the same as for strong acid/strong base titration

12. Slide 12/17 e Weak Acid/Strong Base Titrations

13. Slide 13/17 e Weak Base/Strong Acid Titration

14. Slide 14/17 e Titrations Equivalence Point:  When number of moles of added base = original number of moles of acid  Strong acid/strong base pH = 7  Weak acid/strong base pH > 7  Strong acid/weak base pH < 7 End Point:  When a colour change in the indicator is observed  Choose an indicator that changes colour close to the equivalence point

15. Slide 15/17 e Indicators The pH at which acid  base depends on the pK a of the indicator weak acid base –each form has a different colour pH 3.2pH 4.4pH 4.8pH 5.4 methyl orangemethyl purplebromothymol bluephenolphthalein pH 6.0pH 7.6pH 8.2pH 10.0

16. Slide 16/17 e 1. Which one of the following combinations does the titration curve to the right represent? A. Addition of a strong base to a weak acid B. Addition of a weak base to a strong acid C. Addition of a weak acid to a strong base D. Addition of a strong acid to a strong base E. Addition of a strong acid to a weak base 2. What is the value of the pK a that can be obtained from this titration curve? A. 11.3 B. 10.0 C. 9.3 D. 5.3 E. 1.8 Practice Examples pH 8 10 12 6 4 2 1030504020 Amount of solution added (mL)

17. Slide 17/17 e Learning Outcomes - you should now be able to: Complete the worksheet Understand strong acid/strong base, strong base/strong acid, weak acid/strong base and weak base/strong acid titrations Be able to extract pK a from the half equivalence point Be able to work at the pH at the equivalence point Answer Review Problems 11.38-11.44 in Blackman Summary: Acids & Bases 4 Next lecture: Periodic Trends