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C. Y. Yeung (CHW, 2009)

Acid-Base Eqm (7): Double Indicator Titration & Solubility Product (K sp ). p.01. 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.

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C. Y. Yeung (CHW, 2009)

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  1. Acid-Base Eqm (7): Double Indicator Titration & Solubility Product (Ksp) p.01 Double Indicator Titration [Phenolphthalein & Methyl Orange] decreasing Kb When titrate against standard HCl(aq): NaOH + HCl  NaCl + H2O Na2CO3 + HCl  NaHCO3 + NaCl NaHCO3 + HCl  NaCl + CO2 + H2O For mixtures containing TWO BASES. (e.g. NaOH / Na2CO3 / NaHCO3) C. Y. Yeung (CHW, 2009)

  2. All CO32- converted to HCO3-. 11.2 cm3 0.1M HCl: Phenolphthalein changes colour. 28.8 cm3 0.1M HCl: Methyl Orange changes colour. p.02 All HCO3- converted to CO2 and H2O. E.g. 25cm3 mixture containing NaHCO3 & Na2CO3 11.2 cm3 0.1M HCl to be neutralized first no. of mol of Na2CO3 = 1.1210-3 mol NaHCO3 28.8 cm3 0.1M HCl NaCl + CO2 + H2O total no. of mol of NaHCO3 = 2.8810-3 mol  Original no. of mol of NaHCO3 = 2.8810-3 – 1.1210-3 = 1.76 10-3 mol • [Na2CO3] = 0.0448 M [NaHCO3] = 0.0704 M

  3. p.03 18.5 cm3 0.05M HCl: Phenolphthalein changes colour. 10.0 cm3 0.05M HCl: Methyl Orange changes colour. to be neutralized first no. of mol of NaOH = 9.2510-4 – 510-4 = 4.2510-4mol 18.5 cm3 0.05M HCl 25cm3 mixture containing NaOH & Na2CO3 NaCl + NaHCO3 10.0 cm3 0.05M HCl NaCl + CO2 + H2O total no. of mol of NaHCO3 = 510-4 mol p. 171 Check Point 18-4 • [NaOH] = 0.017 M [Na2CO3] = 0.020 M  Original no. of mol of Na2CO3 = 510-4 mol

  4. p.04 e.g. sparingly soluble salt (very low solubility in water!) “K” is very small! PbCl2(s) Pb2+(aq) + 2Cl-(aq) Ksp = [Pb2+(aq)] [Cl-(aq)]2 Equilibrium between (s) and (aq) Key points : 1. When PbCl2 is added to water, a very small amount of ions will be formed. (as Ksp is very small) 2. If extra ions (e.g. Cl-) are added to the solution, eqm will shift BW, and more PbCl2(s) will be formed. 3. If PbNO3(aq) and NaCl(aq) are mixed together, some of Pb2+ and Cl- ions would form PbCl2(s). Conc. of ions would decrease until Ksp is reached.

  5. p.05  Solubility of AgBr = (8.77 10-7)(107.9+80.0) g dm-3 = 1.65 10-4 g dm-3 Solubility Product (Ksp) 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 Ksp of AgBr = 7.710-13 mol2 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

  6. p.06 AgBr(s) Ag+(aq) + Br-(aq) 0.001 at start X a 0.001 + a at eqm X – a  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”. What is the solubility of AgBr in 0.001M NaBr(aq)? 7.710-13 = a (0.001 + a) a = 7.710-10  [AgBr] dissolved = 7.7 10-10 mol dm-3

  7. p.07 CuBr(s) Cu+(aq) + Br-(aq) at eqm X - 2.010-4 2.010-4 2.010-4 Calculate Ksp from Solubility? e.g. Given that solubilty of copper (I) bromide (CuBr) = 2.010-4 moldm-3, calculate the Ksp of CuBr. Ksp = (2.010-4)2 = 4.010-8 mol2 dm-6

  8. p.08 Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) 0.133 at start 0.033 0 Predict “precipitation” with Ksp? e.g. Given: Ksp of Ca(OH)2 = 8.010-6 mol3 dm-9. If 2 dm3 0.20 M NaOH(aq) + 1 dm3 0.1M CaCl2(aq), Will precipitation occur? ionic product = [Ca2+][OH-]2 = (0.033)(0.133)2 = 5.84 10-4 >> Ksp of Ca(OH)2  Eqm will shift BW and precipitation will occur.

  9. p.09 Next …. Redox Eqm [p.186 – 192] Assignment 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 Ksp of Ca(OH)2

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