1. C. Y. Yeung (CHW, 2009) p.01 Double Indicator Titration & Solubility Product (K sp ) Acid-Base Eqm (7): Double Indicator Titration & Solubility Product (K sp ) For mixtures containing TWO BASES. (e.g. NaOH / Na 2 CO 3 / NaHCO 3 ) Double Indicator Titration [Phenolphthalein & Methyl Orange] decreasing K b When titrate against standard HCl(aq): NaOH + HCl  NaCl + H 2 O Na 2 CO 3 + HCl  NaHCO 3 + NaCl NaHCO 3 + HCl  NaCl + CO 2 + H 2 O

2. p.02E.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 + CO 2 + H 2 O 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 All CO 3 2- converted to HCO 3 -. All HCO 3 - converted to CO 2 and H 2 O.

3. p. 171 Check Point 18-4 p.03 25cm 3 mixture containing NaOH & Na 2 CO 3 18.5 cm 3 0.05M HCl: Phenolphthalein changes colour. 10.0 cm 3 0.05M HCl: Methyl Orange changes colour. to be neutralized first 10.0 cm 3 0.05M HCl NaCl + CO 2 + H 2 O total no. of mol of NaHCO 3 = 5  10 -4 mol no. of mol of NaOH = 9.25  10 -4 – 5  10 -4 = 4.25  10 -4 mol  Original no. of mol of Na 2 CO 3 = 5  10 -4 mol  [NaOH] = 0.017 M [Na 2 CO 3 ] = 0.020 M 18.5 cm 3 0.05M HCl NaCl + NaHCO 3

4. p.04 Equilibrium between (s) and (aq) e.g. PbCl 2 (s) Pb 2+ (aq) + 2Cl - (aq) sparingly soluble salt (very low solubility in water!) “K” is very small! K sp = [Pb 2+ (aq)] [Cl - (aq)] 2 Key points : 1.When PbCl 2 is added to water, a very small amount of ions will be formed. (as K sp is very small) 2.If extra ions (e.g. Cl - ) are added to the solution, eqm will shift BW, and more PbCl 2 (s) will be formed. 3.If PbNO 3 (aq) and NaCl(aq) are mixed together, some of Pb 2+ and Cl - ions would form PbCl 2 (s). Conc. of ions would decrease until K sp is reached.

5. p.05 Solubility Product (K sp ) and Solubility Solubility of salt = conc. of salt dissolved (mol dm -3 ), or = mass of salt dissolved per volume (g dm -3 ) e.g.Given that K sp of AgBr = 7.7  10 -13 mol 2 dm -6, calculate the solubility of AgBr in g dm -3 in water. 7.7  10 -13 = [Ag + ][Br - ]  [Ag + ] = [Br - ] = 8.77  10 -7 mol dm -3  [AgBr] dissolved = 8.77  10 -7 mol dm -3  Solubility of AgBr = (8.77  10 -7 )(107.9+80.0) g dm -3 = 1.65  10 -4 g dm -3

6. p.06 What is the solubility of AgBr in 0.001M NaBr(aq)? AgBr(s) Ag + (aq) + Br - (aq) X at start 0.001 at eqm X – a a 0.001 + a 7.7  10 -13 = a (0.001 + a) a = 7.7  10 -10  [AgBr] dissolved = 7.7  10 -10 mol dm -3  Solubility of AgBr = (7.7  10 -10 )(107.9+80.0) g dm -3 = 1.45  10 -7 g dm -3 Solubility of AgBr is reduced by “Common Ion Effect”.

7. p.07 Calculate K sp from Solubility? e.g.Given that solubilty of copper (I) bromide (CuBr) = 2.0  10 -4 mol dm -3, calculate the K sp of CuBr. K sp = (2.0  10 -4 ) 2 = 4.0  10 -8 mol 2 dm -6 CuBr(s) Cu + (aq) + Br - (aq) at eqm X - 2.0  10 -4 2.0  10 -4

8. p.08 Predict “precipitation” with K sp ? e.g.Given: K sp of Ca(OH) 2 = 8.0  10 -6 mol 3 dm -9. If 2 dm 3 0.20 M NaOH(aq) + 1 dm 3 0.1M CaCl 2 (aq), Will precipitation occur? Ca(OH) 2 (s) Ca 2+ (aq) + 2OH - (aq) at start 0.1330 0.033 ionic product = [Ca 2+ ][OH - ] 2 = (0.033)(0.133) 2 = 5.84  10 -4 >> K sp of Ca(OH) 2  Eqm will shift BW and precipitation will occur.

9. Assignment p.09 Next …. Redox Eqm [p.186 – 192] Study all examples in p.173 - 176 p.179 Q.10, p.228 Q.18-20 [due date: 30/4(Thur)] Pre-Lab: Expt. 14 Determination of K sp of Ca(OH) 2

10. p.10 Q: Calculate the solubility of Fe(OH) 2, in g dm -3, in 0.10M Fe 2 SO 4 (aq). Given that K sp of Fe(OH) 2 = 7.9  10 -16 mol 3 dm -9. Fe(OH) 2 (s) Fe 2+ (aq) + 2OH - (aq) X at start 0. 10 at eqm X – a 0.10 + a 2a 7.9  10 -16 = (0. 10 + a)(2a) 2 a = 4.44  10 -8  [Fe(OH) 2 ] dissolved = 4.44  10 -8 mol dm -3  Solubility of Fe(OH) 2 = (4.44  10 -8 )(55.8+2  17) g dm -3 = 4.00  10 -6 g dm -3

11. p.11 Given:K a1 of H 2 CO 3 = 4.3  10 -7 M K a2 of H 2 CO 3 = 4.8  10 -11 M HCO 3 - CO 3 2- + H 3 O + CO 3 2- + H 2 O H 2 CO 3 + H 2 O H 2 CO 3 + OH - + H 2 O + OH - + H 3 O + K1K1 K3K3 K2K2 K4K4 K 1 = K a2 of H 2 CO 3 = 4.8  10 -11 M K2 =K2 =K2 =K2 = K b1 of CO 3 2- 1 = KwKwKwKw K a of HCO 3 - = KwKwKwKw K a2 of H 2 CO 3 = 4800 M -1 K3 =K3 =K3 =K3 = K a1 of H 2 CO 3 1 = 2.33  10 6 M -1 K4 =K4 =K4 =K4 = K b of HCO 3 - = KwKwKwKw K a1 of H 2 CO 3 = 2.33  10 -8 M

12. p.12 Titration Curve ? Given K b of HCO 3 - = 2.22  10 -8 M K b of CO 3 2- = 2.13  10 -4 M CO 3 2- + H 2 O HCO 3 - + OH - 0.0448 M 0.0704 M pOH = – log(2.13  10 -4 ) + log 0.0704 0.0448 = 3.87 x = pH = 10.13 14 7 vol. of HCl added / cm 3 pH11.2 40.0 y x all HCO 3 - reacted, H 2 CO 3 exists H 2 CO 3 + H 2 O HCO 3 - + H 3 O + 4.50  10 -7 = a2a2 (0.0443 – a) a = 1.41  10 -4 0.0443 – a a a [H 2 CO 3 ] = 2.88  10 -3 (25+40)/1000 = 0.0443 M y = pH = 3.85 25cm 3 0.0448M CO 3 2- and 0.0704M HCO 3 -

13. p.13 Expt. 14To determine K sp of Ca(OH) 2 temp. dependent! Stock Solution 1 1g Ca(OH) 2 (s) in 50cm 3 D.I. H 2 O well diluted sample (ten fold) 8 drops supernatant liquid 1 diluted by adding 72 drops D.I. H 2 O 2 25 drops + 1 drop phenolphthalein 3 titrate against std HCl with plastic pipette 4 OH -

14. p.14 Example: 10 drops of HCl reached the end pt … Given that [HCl] = 0.01084 M no. of mol of OH - = [OH - ](25 drops) no. of mol of H + = (0.01084)(10 drops)  [OH - ](25) = (0.01084)(10) [OH - ] = 4.34  10 -3 M Ca(OH) 2 (s) Ca 2+ (aq) + 2OH - (aq) at start 0.5(0.0434) 0.0434 K sp = [0.5(0.0434)](0.0434) 2 = 4.09  10 -5 mol dm -3 [OH - ] before dilution = 10(4.34  10 -3 ) = 0.0434 M