ANALYTICAL CHEMISTRY CHEM 3811 CHAPTERS 6 & 7 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of...
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Transcript of ANALYTICAL CHEMISTRY CHEM 3811 CHAPTERS 6 & 7 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of...
ANALYTICAL CHEMISTRY
CHEM 3811
CHAPTERS 6 & 7
DR. AUGUSTINE OFORI AGYEMANAssistant professor of chemistryDepartment of natural sciences
Clayton state university
CHAPTERS 6 & 7
VOLUMENTRIC, GRAVIMETRICAND
COMBUSTION
ANALYSIS
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
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
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
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
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
VOLUMETRIC ANALYSIS
Direct Titration
- Titrant is added to analyte until the end point is observed
Titrant (known) + Analyte (unknown) → Product
VOLUMETRIC ANALYSIS
Back Titration- Two steps involved
1. A known excess of standard reagent is added to the analyteReagent 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
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
ExamplePbCl2(s) ↔ Pb2+ + 2Cl-
Ksp = [Pb2+][Cl-]2 = 1.7 x 10-5
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
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
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 ignoredImplies Ksp = (x1)(x2)2
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
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)
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
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
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
COMBUSTION ANALYSIS
- Used for the determination of mass percentages and empiricalformula 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
- A combustion train is usually used for analysis of compoundscontaining 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 atmospherefrom the reverse direction
- Masses of trapped H2O and CO2 are then determined
COMBUSTION ANALYSIS
Furnace with catalyst
P4O10 NaOHO2 in
sample
H2O trap CO2 trap
COMBUSTION ANALYSIS
O2 outP4O10/NaOH
Guard tube
Combustion of a 0.2000-g sample of a compound made up ofonly carbon, hydrogen, and oxygen yields 0.200 g H2O and0.4880 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
COMBUSTION ANALYSIS
OHmol0.0111g18.02
mol1xg0.200OHmol 22
Hmol0.0222OHmol1
Hmol2xOHmol0.0111Hmol
22
H0.0224Hmol1
H01.1xHmol0.0222Hmass g
g
COMBUSTION ANALYSIS
22 COmol0.01109g44.01
mol1xg0.4880COmol
Cmol0.01109COmol1
Cmol1xCOmol0.01109Cmol
22
Cg0.1332Cmol1
Cg12.01xCmol0.01109Cmass
COMBUSTION ANALYSIS
Mass O = 0.2000 g sample – (0.0224 g + 0.1332 g)
= 0.0444 g O
% .6066100%x0.2000
g 0.1332C%
% .211100%x0.2000
g 0.0224H%
% .222100%x0.2000
g 0.0444O%
COMBUSTION ANALYSIS