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Precipitation

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

candace-roberson
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Precipitation

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  1. 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.

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

  3. 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).

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

  5. Precipitation in the Presence of an Electrolyte Consider titration of Ag+ with Cl- in the presence of 0.1 M HNO3. 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 HNO3) causes neutralisation of the surface charges. Decrease in ionic atmosphere (less electrostatic repulsion)

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

  7. 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 Inclusions & Occlusions Within the crystal Absorbed impurities: Inclusion: Impurity ions occupying crystal lattice sites. Occlusion: Pockets of impurities trapped within a growing crystal.

  8. Coprecipitation: Adsorption, Inclusion and Occlusion Colloidal precipitates: Large surface area BaSO4; 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 Be2+, Mg2+, Ca2+, or Ba2+ with N-p-chlorophenylcinnamohydroxamic acid. Impurities are Ag+, Mn2+, Zn2+, Cd2+, Hg2+, Fe2+, and Ga2+. Add complexing KCN.

  9. Ca 2+ + 2RH  CaR2(s) + 2H+ Analyte Precipitate Mn2+ + 6CN- Mn(CN)64- Impurity Masking agent Stays in solution Postprecipitation: Collection of impurities on ppt during digestion: a supersaturated impurity e.g., MgC2O4 on CaC2O4. Peptization: Breaking up of charged solid particles when ppt is washed with water. AgCl is washed with volatile electrolyte (0.1 M HNO3). Other electrolytes: HCl; NH4NO3; and (NH4)2CO3.

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

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

  12. Example In the determination of magnesium in a sample, 0.352 g of this sample is dissolved and precipitated as Mg(NH4)PO4.6H2O. The precipitate is washed and filtered. The precipitate is then ignited at 1100 oC for 1 hourand weighed as Mg2P2O7. The mass of Mg2P2O7 is 0.2168 g. Calculate the percentage of magnesium in the sample.

  13. Solution: Relative atomic mass Of Mg The gravimetric factor is: FM of Mg2P2O7 Note: 2 mol Mg2+ in 1 mol Mg2P2O7.

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

  15. Combustion Analysis Determination of the Carbon and Hydrogen content of organic compounds burned in excess oxygen. H2O absorption Prevention of entrance of atmospheric O2 and CO2. CO2 Absorption Note: Mass increase in each tube.

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

  17. Sample size usually 2 mg in tin or silver capsule. Capsule melts and sample is oxidised in excess of O2. Dynamic Flash combustion: Short burst of gaseous products Products Hot WO3 catalyst: Then, metallic Cu at 850 oC:

  18. Oxygen Analysis: Pyrolysis or thermal decomposition in absence of oxygen. Nickelised Carbon Gaseous products: CO formed 1075 oC Halogen-containing compounds: CO2, H2O, N2, and HX products HX(aq) titration with Ag+ coulometrically. Silicon Compounds (SiC, Si3N4, & Silicates from rocks): Combustion with F2 in nickel vessel Volatile SiF4 & other fluorinated products Mass Spectrometry

  19. Example 1: Write a balanced equation for the • combustion of benzoic acid, C6H5CO2H, to give CO2 • and H2O. How many milligrams of CO2 and H2O will be • produced by the combustion of 4.635 mg of C6H5CO2H? Solution: C6H5CO2H + 15/2O2 FW = 122.123 7CO2 + 3H2O 44.010 18.015 4.635 mg ofC6H5CO2H = 1 moleC6H5CO2H yields 7 moles CO2 and 3 moles H2O Mass CO2 = 7 x 0.03795 mmol x 44.010 mg/mmol = 11.69 mg CO2 Mass H2O = 3 x 0.03795 mmol x 18.015 mg/mmol = 11.69 mg H2O

  20. Example 2: A 7.290 mg mixture of cyclohexane, C6H12 • (FW 84.159), and Oxirane, C2H4O (FW 44.053) was • analysed by combustion, and 21.999 mg CO2 • (FW 44.010) were produced. Find the % weight of • oxirane in the sample mixture. Solution: C6H12 + C2H4O + 23/2O2 8CO2 + 8H2O Let x = mg ofC6H12and y = mg ofC2H4O. X + y = 7.290 mg Also,CO2 = 6(moles of C6H12) + 2(moles of C2H4O)

  21. CO2 X + y = 7.290 mg  x = 7.290 - y  y = mass of C2H4O = 0.767 mg Therefore, % Weight Oxirane = = 10.52 %

  22. 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. Ksp magnitude Influence the sharpness of the end point

  23. Titration Curve A graph showing variation of concentration of one reactant with added titrant. Concentration varies over many orders of magnitude P function: pX = -log10[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+ Ksp = [Ag+][I-] = 8.3 x 10-17

  24. I- + Ag+ AgI(s)K =1/Ksp = 1.2 x 1016 Ve = Volume of titrant at the equivalent point: (0.02500 L)(0.1000 mol I-/L) (Ve)(0.05000 mol Ag+/L) = mol I- mol Ag+  Ve = 0.05000 L = 50.00 mL Before the Equivalence Point: Addition of 20 mL of Ag+: This reaction: I- + Ag+ AgI(s) goes to completion.

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

  26.  [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,

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

  28. After the Equivalence Point: Note: Ve = 50.00 mL [Ag+] is in excess after the equivalence point. Suppose that52.00 mLis added: Volume of excess Ag+ Therefore, 2.00 mL excess Ag+ Total volume of solution Original Ag+ Concentration Dilution Factor

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

  30. Titration Curves: Effect of Diluting the reactants • 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+

  31. Titrations involving 1:1 stoichiometry of reactants Equiv. Point: Steepest point in titration curve Other stoichiometric ratios: 2Ag+ + CrO42- Ag2CrO4(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

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

  33. Titration of a Mixture Less soluble precipitate forms first. Titration of KI & KCl solutions with AgNO3 Ksp (AgI)<< Ksp (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.

  34. Titration Curve for 40.00 mL of 0.05000 M KI and 0.05000 M KCl with 0.084 M AgNO3.

  35. 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. Midpoint of the second steep section. AgCl End Point:

  36. 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 Example:Some Cl- attached to AgI ppt and carries down an equivalent amount of Ag+. Coprecipitation error lowers the calculated concentration of the second precipitated halide. High nitrate concentration to minimise coprecipitation. NO3- competes with Cl- for binding sites.

  37. Separation of Cations by Precipitation Consider a solution of Pb2+ and Hg22+: Each is 0.01 M PbI2(s) ⇌ Pb2+ + 2I- Ksp = 7.9 x 10 -9 Hg2I2(s) ⇌ Hg22+ + 2I- Ksp = 1.1 x 10 -28 Smaller Ksp Considerably Less soluble Is separation of Hg22+ from Pb2+“complete”? Is selective precipitation of Hg22+ with I- feasible?

  38. Can we lower [Hg22+ ] to 0.010 % of its original value without precipitating Pb2+? From 0.010 M to 1.0 x 10–6 M? Add enough I- to precipitate 99.990 % Hg22+. Hg2I2(s) ⇌ Hg22+ + 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

  39. Will this [I-] = 1.0 x 10 –11 M precipitate 0.010 M Pb2+? Q = 1.0 x 10-24 << 7.9 x 10–9 = Ksp for PbI2 Therefore, Pb2+ will not precipitate. Prediction: All Hg22+ will virtually precipitate before any Pb2+ precipitates on adding I-.

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