analytical chemistry chem 3811 chapters 6 7 n.
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  1. ANALYTICAL CHEMISTRY CHEM 3811CHAPTERS 6 & 7 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university


  3. VOLUMETRIC ANALYSIS - Analysis by volume Titration - Increments of a known reagent solution (the titrant) are added to an unknown solution (the analyte) until the reaction is complete Titrant - Usually in the buret Analyte - Usually in an erlenmeyer flask

  4. VOLUMETRIC ANALYSIS Common Titrations - Acid-Base - Oxidation-Reduction - Complex Formation (Complexometric) - Precipitation Reactions Methods of Determining Analyte Consumption - Color change - Absorbance of light - Sudden change in voltage or current between a pair of electrodes

  5. VOLUMETRIC ANALYSIS Equivalence Point - The quantity of titrant added is the exact amount necessary for stoichiometric reaction with analyte End Point - The quantity of titrant actually measured in an experiment - Ideal end point is the equivalence point

  6. VOLUMETRIC ANALYSIS Titration Error - The difference between end point and equivalence point Blank Titration - Using a solution that contains no analyte (same volume of same solvent but without analyte) - Blank titration is used to estimate titration error

  7. VOLUMETRIC ANALYSIS Primary Standard - Pure and dry reagent that is accurately weighed and used directly to determine the concentration of a solution (e.g. KHP) - Used to standardize solutions Standardization - Titration of a primary standard to determine the concentration of a titrant Standard solution - A solution whose concentration is known

  8. VOLUMETRIC ANALYSIS Direct Titration - Titrant is added to analyte until the end point is observed Titrant (known) + Analyte (unknown) → Product

  9. VOLUMETRIC ANALYSIS Back Titration - Two steps involved 1. A known excess of standard reagent is added to the analyte Reagent 1 (known) + Analyte (unknown) → Product + Excess 1 2. A second standard reagent is used to titrate the excess of the first reagent Reagent 2 (known) + Excess 1 (unknown) → Product Back titration is necessary when - Excess of reagent 1 is required for complete reaction with analyte - End point of back titration is clearer than direct

  10. VOLUMETRIC ANALYSIS Solubility Product - Ksp is the equilibrium constant for the reaction in which a solid salt (an ionic compound) dissolves to give its constituent ions in solution - Solid is in its standard state so is omitted from the Ksp expression Example PbCl2(s) ↔ Pb2+ + 2Cl- Ksp = [Pb2+][Cl-]2 = 1.7 x 10-5

  11. VOLUMETRIC ANALYSIS Solubility Product - The solution is said to be saturated if all the solid is capable of being dissolved PbCl2(s) ↔ Pb2+ + 2Cl- x0 0 0 x0 - x1 x1 2x1 If all solid dissolves x0 - x1 = 0 or x0 = x1

  12. VOLUMETRIC ANALYSIS Common Ion Effect Supposing x2 M NaCl is added to PbCl2 PbCl2(s) ↔ Pb2+ + 2Cl- x0 0 x2 x0 - x1 x1 2x1 + x2 Concentration of Cl- has contributions from both PbCl2 and NaCl

  13. VOLUMETRIC ANALYSIS Common Ion Effect Ksp = [Pb2+][Cl-]2 = (x1)(2x1 + x2)2 - A salt is less soluble if one of its constituent ions is already present in the solution Assuming 2x1 <<< x2 2x1 can be ignored Implies Ksp = (x1)(x2)2

  14. VOLUMETRIC ANALYSIS Mixtures - In precipitation titrations of mixtures the less soluble precipitates first (Smaller Ksp implies less soluble) - If the difference between the Ksp values are sufficiently large, first precipitation nearly completes before second precipitation starts Consider a mixture of PbI2 and PbCl2 PbI2(s) ↔ Pb2+ + 2I- Ksp = 7.9 x 10-9 (precipitates first) PbCl2(s) ↔ Pb2+ + 2Cl- Ksp = 1.7 x 10-5

  15. GRAVIMETRIC ANALYSIS - Analysis by mass - Product should precipitate out Gravimetric Titration - Titrant is measured by mass - Concentrations are expressed as moles/kg solution - Only pipets may be used (burets are not necessary)

  16. GRAVIMETRIC ANALYSIS Precipitation - Falling out of solution Precipitate should - be insoluble - be easily filtered (should have large particles) - have known and constant composition - be very stable to withstand heat - Samples are heated to get rid of traces of solvent Precipitants - Agents that cause precipitation

  17. GRAVIMETRIC ANALYSIS Crystal Growth - Is necessary as particle sizes must be large enough for easy filtration - Two phases exist Nucleation - Dissolved particles (molecules or ions) form small crystalline aggregates capable of growing into larger particles - Solutes are attracted and held on pre-existing surfaces (impurities) Particle Growth - Solute particles add to an existing aggregate to form a crystal - Particle Growth creates larger particles than Nucleation

  18. GRAVIMETRIC ANALYSIS Homogeneous Precipitation - Precipitant is generated slowly by a chemical reaction - Particle Growth dominates over Nucleation due to slow precipitation - Larger particles are formed as a result - Ionic compounds are usually precipitated in the presence of added electrolytes - Most gravimetric precipitations are performed in the presence of electrolytes

  19. COMBUSTION ANALYSIS - Used for the determination of mass percentages and empirical formula of compounds - Samples are burned in excess oxygen and the products are measured - Typically used to measure C, H, N, S and halogens in organic compounds

  20. COMBUSTION ANALYSIS - A combustion train is usually used for analysis of compounds containing only carbon, hydrogen, and oxygen - A first compartment with P4O10 (or Mg(ClO4)2 or CaCl2) traps H2O - A second compartment with NaOH (on asbestos) traps CO2 - A third compartment is a guard tube with both P4O10 and NaOH - Guard tube prevents entry of CO2 and H2O from the atmosphere from the reverse direction - Masses of trapped H2O and CO2 are then determined

  21. COMBUSTION ANALYSIS O2 in O2 out sample P4O10 NaOH P4O10/NaOH H2O trap CO2 trap Guard tube Furnace with catalyst

  22. COMBUSTION ANALYSIS Combustion of a 0.2000-g sample of a compound made up of only carbon, hydrogen, and oxygen yields 0.200 g H2O and 0.4880 g CO2. Calculate the mass and mass percentage of each element present in the 0.2000-g sample. - Convert mass H2O/CO2 to moles using molar mass - Determine moles H/C from number of atoms and moles H2O/CO2 - Convert moles H/C to mass H/C using molar mass - Determine mass O by subtracting total mass H and C from mass sample - Calculate percentages



  25. COMBUSTION ANALYSIS Mass O = 0.2000 g sample – (0.0224 g + 0.1332 g) = 0.0444 g O