1. C. Y. Yeung (CHW, 2009) p.01 Titration Curves Acid-Base Eqm (5): Titration Curves Titration Curves Acid-Base Eqm (6): Titration Curves Plotting the Titration Curves with your calculated results. new [HCl] = E.g. (1a)25 cm 3 HCl + 20 cm 3 NaOH 25/1000  0.1 – 20/1000  0.1 (25+20)/1000 = 0.011 M [H 3 O + ] = 0.011 M pH = 1.95

2. p.02 [NaOH] = E.g. (1b)25 cm 3 HCl + 30 cm 3 NaOH 30/1000  0.1 – 25/1000  0.1 (30+25)/1000 = 9.09  10 -3 M [OH - ] = 9.09  10 -3 M pOH = 2.04 pH = 11.96

3. p.03 new [CH 3 COOH] = E.g. (2a)25 cm 3 CH 3 COOH + 20 cm 3 NaOH 25/1000  0.1 – 20/1000  0.1 (25+20)/1000 = 0.011 M = 5.36 (0.044) (0.011) pH = - log(1.76  10 -5 ) + log [CH 3 COO - ] = 20/1000  0.1 (25+20)/1000 = 0.044 M [H 3 O + ] = 10 -5.36 = 4.37  10 -6 M

4. p.04 E.g. (2b)25 cm 3 CH 3 COOH + 25 cm 3 NaOH pH = 8,73, [H 3 O + ] = 10 -8.73 = 1.86  10 -9 M pOH = 5.27 [CH 3 COO - ] = 25/1000  0.1 (25+25)/1000 = 0.05 M 1.00  10 -14 CH 3 COO - + H 2 O CH 3 COOH + OH - 1.76  10 -5 = x2x2x2x2 (0.05 – x) x = 5.33  10 -6 = [OH - ] Kb =Kb =Kb =Kb =

5. p.05 E.g. (2c)25 cm 3 CH 3 COOH + 45 cm 3 NaOH [NaOH] = 45/1000  0.1 – 25/1000  0.1 (45+25)/1000 = 0.0286 M [OH - ] = 0.0286 M pOH = 1.54 pH = 12.5

6. p.06 new [HCl] = E.g. (3a)25 cm 3 HCl + 10 cm 3 NH 3 25/1000  0.1 – 10/1000  0.1 (25+10)/1000 = 0.043 M [H 3 O + ] = 0.043 M pH = 1.37

7. p.07 E.g. (3b)25 cm 3 HCl + 25 cm 3 NH 3 pOH = 8,73, [OH - ] = 10 -8.73 = 1.86  10 -9 M pH = 5.27 [NH 4 + ] = 25/1000  0.1 (25+25)/1000 = 0.05 M 1.00  10 -14 NH 4 + + H 2 O NH 3 + H 3 O + 1.74  10 -5 = x2x2x2x2 (0.05 – x) x = 5.36  10 -6 = [H 3 O + ] Ka =Ka =Ka =Ka =

8. p.08 E.g. (3c)25 cm 3 HCl + 40 cm 3 NH 3 [NH 3 ] = 40/1000  0.1 – 25/1000  0.1 (40+25)/1000 = 0.0231 M [NH 4 + ] = 25/1000  0.1 (40+25)/1000 = 0.0385 M pOH = 4.98, [OH - ] = 10 -4.98 = 1.04  10 -5 M (0.0385) (0.0231) pOH = - log(1.74  10 -5 ) + log pH = 9.02

9. p.09 Titration Curves: Strong Acid VS Strong Base abrupt change of pH

10. p.10 Titration Curves: Weak Acid VS Strong Base abrupt change of pH pH of salt > 7 Due to hydrolysis of conjugate base of weak acid: A - + H 2 O HA + OH -

11. p.11 Titration Curves: Strong Acid VS Weak Base abrupt change of pH pH of salt < 7 Due to hydrolysis of conjugate acid of weak base: BH + + H 2 O B + H 3 O +

12. p.12 Titration Curves: Comparison strong acid VS strong base weak acid VS strong base strong acid VS weak base

13. Choosing a Suitable Indicator (1)  the abrupt change on the pH curve must fall across the “working range” of the indicator. pK In ± 1 Phenolphthalein pH 7 8.15 9.15 10.15 colourless pale pink pink Methyl orange pH 7 2.70 3.70 4.70 red orange yellow p.13

14. Choosing a Suitable Indicator (2) p.14 End point: Equivalent point: The sudden change in colour seen in a titration. The mixture in which amount of acid and base are exactly balance. If the correct indicator has been chosen, the end point will be very close to the equivalent point.

15. p.15 14 12 10 8 6 4 2 0 pH methyl orange red orange yellow phenolphthalein colourless pale pink pink 25 vol. of alkali added / cm 3 Strong Acid VS Strong Base end pt. eqv. pt.

16. p.16 14 12 10 8 6 4 2 0 pH methyl orange red orange yellow phenolphthalein colourless pale pink pink 25 vol. of alkali added / cm 3 Weak Acid VS Strong Base

17. p.17 14 12 10 8 6 4 2 0 pH methyl orange red orange yellow phenolphthalein colourless pale pink pink 25 vol. of acid added / cm 3 Strong Acid VS Weak Base

18. p.18 Explain why phenolphthalein turns pink in a solution of sodium carbonate, but remains colourless in a solution of sodium hydrogencarbonate. [1990]

19. p.19 Explain why at 298K, in a solution of pH7.0, the indicator methyl orange shows its alkaline colour (yellow), while phenolphthalein shows its acidic colour (colourless). [1994] Acid-base indicators are weak acids or bases. The dissociation of which can be represented by HIn(aq) + H 2 O(l) H 3 O + (aq) + In - (aq) The colour of an indicator depends on the relative concentrations of HIn and In - which are of different colours. The dissociation constant K In of different indicators are different, thus they change colour over different pH range. The pH range of methyl orange is below 7, while that of phenolphthalein is above 7.

20. p.20 HKALE:p. 233 Q.15(a),(b)

21. p.21 HKALE: Q.20(a)

22. Double Indicator Titration p.22 For mixtures containing TWO BASES. [Phenolphthalein & Methyl Orange] E.g. 25cm 3 mixture containing NaHCO 3 & Na 2 CO 3 11.2 cm 3 0.1M HCl: Phenolphthalein changes colour. 28.8 cm 3 0.1M HCl: Methyl Orange changes colour. to be neutralized first 11.2 cm 3 0.1M HCl NaHCO 3 28.8 cm 3 0.1M HCl NaCl no. of mol of Na 2 CO 3 = 1.12  10 -3 mol total no. of mol of NaHCO 3 = 2.88  10 -3 mol  Original no. of mol of NaHCO 3 = 2.88  10 -3 – 1.12  10 -3 = 1.76  10 -3 mol  [Na 2 CO 3 ] = 0.0448 M [NaHCO 3 ] = 0.0704 M

23. Assignment p.23 Next …. Solubility Product (K sp ) [p.172-176] p.229 Q.3(c), 4, 11, 18, 26, 29 p.171 Check Point 18-4 [due date: 29/4(Wed)]