2. Precipitation Gravimetric Analysis: Solid product formed Relatively insoluble Easy to filter High purity Known Chemical composition Precipitation Conditions: Particle Size Small Particles: Clog &pass through filter paper Large Particles: Less surface area for attachment of foreign particles.

3. Crystallization 1. Nucleation 2. Particle Growth Molecules form small Aggregates randomly Addition of more molecules to a nucleus. Supersaturated Solution: More solute than should be present at equilibrium. Supersaturated Solution: Nucleation faster; Suspension (colloid) Formed. Less Supersaturated Solution: Nucleation slower, larger particles formed.

4. How to promote Crystal Growth 1. Raise the temperature Increase solubility Decrease supersaturation 2. Precipitant added slowly with vigorous stirring. 3. Keep low concentrations of precipitant and analyte (large solution volume).

5. Homogeneous Precipitation Precipitant generated slowly by a chemical reaction pH gradually increases Large particle size

6. Precipitation in the Presence of an Electrolyte Consider titration of Ag + with Cl - in the presence of 0.1 M HNO 3. Colloidal particles of ppt : Surface is + vely charged Adsorption of excess Ag + on surface (exposed Cl - ) Colloidal particles need enough kinetic energy to collide and coagulate. Addition of electrolyte (0.1 M HNO 3 ) causes neutralisation of the surface charges. Decrease in ionic atmosphere (less electrostatic repulsion)

7. Net + ve Charge on Colloidal Particle because of Ag + Adsorbed

8. Digestion and Purity Digestion: Period of standing in hot mother liquor. Promotion of recrystallisation Crystal particle size increase and expulsion of impurities. Purity: Adsorbed impurities: Surface-bound Absorbed impurities: Within the crystal Inclusions & Occlusions Inclusion: Impurity ions occupying crystal lattice sites. Occlusion: Pockets of impurities trapped within a growing crystal.

9. Coprecipitation:Adsorption, Inclusion and Occlusion Colloidal precipitates: Large surface area BaSO 4 ; Al(OH) 3 ; and Fe(OH) 3 How to Minimise Coprecipitation: 1. Wash mother liquor, redissolve, and reprecipitate. 2. Addition of a masking agent: Gravimetric analysis of Be 2+, Mg 2+, Ca 2+, or Ba 2+ with N-p-chlorophenylcinnamohydroxamic acid. Impurities are Ag +, Mn 2+, Zn 2+, Cd 2+, Hg 2+, Fe 2+, and Ga 2+. Add complexing KCN.

10. Ca 2+ + 2RH  CaR 2 (s) + 2H + Analyte Precipitate Mn 2+ + 6CN -  Mn(CN) 6 4- Impurity Masking agent Stays in solution Postprecipitation:Collection of impurities on ppt during digestion: a supersaturated impurity e.g., MgC 2 O 4 on CaC 2 O 4. Peptization: Breaking up of charged solid particles when ppt is washed with water. AgCl is washed with volatile electrolyte (0.1 M HNO 3 ). Other electrolytes: HCl; NH 4 NO 3 ; and (NH 4 ) 2 CO 3.

11. Product Composition Hygroscopic substances: Difficult to weigh accurately Some ppts:Variable water quantity as water of Crystallisation. Drying Change final composition by ignition:

12. Thermogravimetric Analysis Heating a substance and measuring its mass as a function of temperature.

14. Example In the determination of magnesium in a sample, 0.352 g of this sample is dissolved and precipitated as Mg(NH 4 )PO 4. 6H 2 O. The precipitate is washed and filtered. The precipitate is then ignited at 1100 o C for 1 hour and weighed as Mg 2 P 2 O 7. The mass of Mg 2 P 2 O 7 is 0.2168 g. Calculate the percentage of magnesium in the sample.

15. Solution: The gravimetric factor is: Relative atomic mass Of Mg FM of Mg 2 P 2 O 7 Note: 2 mol Mg 2+ in 1 mol Mg 2 P 2 O 7.

16. Mass of Mg 2+ = 0.0471 g = 13.45 %

17. Combustion Analysis Determination of the Carbon and Hydrogen content of organic compounds burned in excess oxygen. H 2 O absorption CO 2 Absorption Prevention of entrance of atmospheric O 2 and CO 2. Note: Mass increase in each tube.

18. C, H, N, and S Analyser: Modern Technique Thermal Conductivity, IR,or Coulometry for Measuring products.

19. Sample size usually 2 mg in tin or silver capsule. Capsule melts and sample is oxidised in excess of O 2. ProductsHot WO 3 catalyst: Then, metallic Cu at 850 o C: Dynamic Flash combustion: Short burst of gaseous products

20. Oxygen Analysis: Pyrolysis or thermal decomposition in absence of oxygen. Gaseous products: Nickelised Carbon 1075 o C CO formed Halogen-containing compounds: CO 2, H 2 O, N 2, and HX products HX(aq) titration with Ag + coulometrically. Silicon Compounds (SiC, Si 3 N 4, & Silicates from rocks): Combustion with F 2 in nickel vessel Volatile SiF 4 & other fluorinated products Mass Spectrometry

21. Example 1: Write a balanced equation for the combustion of benzoic acid, C 6 H 5 CO 2 H, to give CO 2 and H 2 O. How many milligrams of CO 2 and H 2 O will be produced by the combustion of 4.635 mg of C 6 H 5 CO 2 H? Solution: C 6 H 5 CO 2 H + 15 / 2 O 2 FW = 122.123 7CO 2 + 3H 2 O 44.010 18.015 4.635 mg of C 6 H 5 CO 2 H = 1 mole C 6 H 5 CO 2 H yields 7 moles CO 2 and 3 moles H 2 O Mass CO 2 = 7 x 0.03795 mmol x 44.010 mg/mmol = 11.69 mg CO 2 Mass H 2 O = 3 x 0.03795 mmol x 18.015 mg/mmol = 11.69 mg H 2 O

22. Example 2: A 7.290 mg mixture of cyclohexane, C 6 H 12 (FW 84.159), and Oxirane, C 2 H 4 O (FW 44.053) was analysed by combustion, and 21.999 mg CO 2 (FW 44.010) were produced. Find the % weight of oxirane in the sample mixture. Solution: C 6 H 12 + C 2 H 4 O + 23 / 2 O 2 8CO 2 + 8H 2 O Let x = mg of C 6 H 12 and y = mg of C 2 H 4 O. X + y = 7.290 mg Also,CO 2 = 6(moles of C 6 H 12 ) + 2(moles of C 2 H 4 O)

23. X + y = 7.290 mg  x = 7.290 - y CO 2  y = mass of C 2 H 4 O = 0.767 mg Therefore, % Weight Oxirane = = 10.52 %

24. The Precipitation Titration Curve Reasons for calculation of titration curves: 1. Understand the chemistry occurring. 2. How to exert experimental control to influence the quality of analytical titration. In precipitation titrations: 1. Analyte concentration 2. Titrant concentration 3. K sp magnitude Influence the sharpness of the end point

25. Titration Curve A graph showing variation of concentration of one reactant with added titrant. Concentration varies over many orders of magnitude P function: pX = -log 10 [X] Consider the titration of 25.00 mL of 0.1000 M I - with 0.05000 M Ag +. I - + Ag +  AgI(s) There is small solubility of AgI: AgI(s)  I - + Ag + K sp = [Ag + ][I - ] = 8.3 x 10 -17

26. I - + Ag +  AgI(s) K =1/K sp = 1.2 x 10 16 V e = Volume of titrant at the equivalent point: Before the Equivalence Point: Addition of 20 mL of Ag + : This reaction: I - + Ag +  AgI(s) goes to completion.  V e = 0.05000 L = 50.00 mL (0.02500 L)(0.1000 mol I - /L)(V e )(0.05000 mol Ag + /L) = mol I - mol Ag +

27. [I - ] due to I - not precipitated by 20.00 mL of Ag +. Fraction of I - reacted: Fraction of I - remaining: Some AgI redissolves: AgI(s)  I - + Ag + Therefore, Fraction Remaining Original Conc. Dilution Factor Original volume of I - Total volume

28.  [I - ] = 3.33 x 10 -2 M   [Ag + ] = 2.49 x 10 -15 M pAg + = -log[Ag + ] = 14.60 The Equivalence Point: All AgI is precipitated AgI(s)  I - + Ag + Then,

29. K sp = [Ag + ][I - ] = 8.3 x 10 -17 And [Ag + ] = [I - ] = x K sp = (x)(x) = 8.3 x 10 -17  X = 9.1 x 10 -9 M  pAg + = -log x = 8.04 At equivalence point: pAg + value is independent of the original volumes or concentrations.

30. After the Equivalence Point After the Equivalence Point: [Ag + ] is in excess after the equivalence point. Note: V e = 50.00 mL Suppose that 52.00 mL is added: Therefore, 2.00 mL excess Ag+ Original Ag + Concentration Dilution Factor Volume of excess Ag + Total volume of solution

31. [Ag + ] = 1.30 x 10 -3 M pAg + = -log[Ag + ] = 2.89 Shape of the Titration Curve: Steepest slope:has maximum value Equivalence point: point of maximum slope Inflection point:

32. Titration Curves: Effect of Diluting the reactants 1.0.1000 M I - vs 0.05000 M Ag+ 2. 0.01000 M I - vs 0.005000 M Ag+ 3. 0.001000 M I - vs 0.0005000 M Ag+

33. Titrations involving 1:1 stoichiometry of reactants Equiv. Point: Steepest point in titration curve Other stoichiometric ratios: 2Ag + + CrO 4 2-  Ag 2 CrO 4 (s) 1. Curve not symmetric near equiv. point 2. Equiv. Point: Not at the centre of the steepest section of titration curve 3. Equiv. Point: not an inflection point In practice: Conditions chosen such that curves are steep enough for the steepest point to be a good estimate of the equiv. point

34. Effect of K sp on the Titration Curve AgI is least soluble Sharpest change at equiv. point Least sharp, but steep enough for Equiv. point location K = 1/K sp largest 

35. Titration of a Mixture Less soluble precipitate forms first. Titration of KI & KCl solutions with AgNO 3 K sp (AgI) << K sp (AgCl) First precipitation of AgI nearly complete before the second (AgCl) commences. When AgI pption is almost complete, [Ag+] abruptly increases and AgCl begins to precipitate. Finally, when Cl - is almost completely consumed, another abrupt change in [Ag+] occurs.

36. Titration Curve for 40.00 mL of 0.05000 M KI and 0.05000 M KCl with 0.084 M AgNO 3.

37. I - end point: Intersection of the steep and nearly horizontal curves. Note: Precipitation of AgI not quite complete when AgCl begins to precipitate. End of steep portion better approximation of the equivalence point. AgCl End Point: Midpoint of the second steep section.

38. The AgI end point is always slightly high for I - /Cl - mixture than for pure I -. 1. Random experimental error: both +tive and –tive. 2. Coprecipitation: +ve error High nitrate concentration to minimise coprecipitation. Example: Some Cl - attached to AgI ppt and carries down an equivalent amount of Ag +. NO 3 - competes with Cl - for binding sites. Coprecipitation error lowers the calculated concentration of the second precipitated halide.

39. Separation of Cations by Precipitation Consider a solution of Pb 2+ and Hg 2 2+ : Each is 0.01 M PbI 2 (s) ⇌ Pb 2+ + 2I - Hg 2 I 2 (s) ⇌ Hg 2 2+ + 2I - K sp = 7.9 x 10 -9 K sp = 1.1 x 10 -28 Smaller K sp Considerably Less soluble Is separation of Hg 2 2+ from Pb 2+ “complete”? Is selective precipitation of Hg 2 2+ with I - feasible?

40. Can we lower [Hg 2 2+ ] to 0.010 % of its original value without precipitating Pb 2+ ? From 0.010 M to 1.0 x 10 –6 M? Add enough I - to precipitate 99.990 % Hg 2 2+. Hg 2 I 2 (s) ⇌ Hg 2 2+ + 2I - Initial Concentration: 0 0.010 0 Final Concentration: solid 1.0 x 10 -6 x  (1.0 x 10 -6 )(x) 2 = 1.1 x 10 -28 X = [I - ] = 1.0 x 10 –11 M  X = [I - ] = 1.0 x 10 –11 M

41. [I - ] = 1.0 x 10 –11 M Will this [I - ] = 1.0 x 10 –11 M precipitate 0.010 M Pb 2+ ? Q = 1.0 x 10 -24 << 7.9 x 10 –9 = K sp for PbI 2 Therefore, Pb 2+ will not precipitate. Prediction: All Hg 2 2+ will virtually precipitate before any Pb 2+ precipitates on adding I -.