ANALYTICAL CHEMISTRY CHEM 3811 CHAPTERS 6 & 7

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