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 = M [H 3 O + ] = M pH = 1.95

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  M [OH - ] = 9.09  M pOH = 2.04 pH = 11.96

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 = M = 5.36 (0.044) (0.011) pH = - log(1.76  ) + log [CH 3 COO - ] = 20/1000  0.1 (25+20)/1000 = M [H 3 O + ] = = 4.37  M

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

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 = M [OH - ] = M pOH = 1.54 pH = 12.5

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 = M [H 3 O + ] = M pH = 1.37

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

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 = M [NH 4 + ] = 25/1000  0.1 (40+25)/1000 = M pOH = 4.98, [OH - ] = = 1.04  M (0.0385) (0.0231) pOH = - log(1.74  ) + log pH = 9.02

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

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 -

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 +

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

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 colourless pale pink pink Methyl orange pH red orange yellow p.13

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.

p 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.

p pH methyl orange red orange yellow phenolphthalein colourless pale pink pink 25 vol. of alkali added / cm 3 Weak Acid VS Strong Base

p pH methyl orange red orange yellow phenolphthalein colourless pale pink pink 25 vol. of acid added / cm 3 Strong Acid VS Weak Base

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

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.

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

p.21 HKALE: Q.20(a)

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

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