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What is Analytical Chemistry ?
- Analytical chemistry deals with separating, identifying, and quantifying the relative amounts of the components of an analyte.
- Analyte = the thing to analyzed; the component(s) of a sample that are to be determined.
Analytical Chemistry
analyze :
" what is it? (qualitative analysis)
" how much is there?“ (quantitative analysis)
The role of analytical chemistry: central scienceThe relationship between analytical chemistry and the other sciences
Analyticalchemistry
Chemistry : Biological, Inorganic, Organic, Physical
Physics : Astrophysics, Astronomy, Biophysics
Biology : Botany, Genetics, Microbiology, Molecular biology, Zoology
Geology : Geophysics, Geochemistry, Paleontology, Paleobiology
Environmental science : Ecology, Meteorology, Oceanography
Medicine : Clinical, Medicinal, Pharmacy, Toxicology
Material science : Metallurgy, Polymers, Solid state
Engineering : Civil, Chemical, Electronical, Mechanical
Agriculture : Agronomy, Animal, Crop, Food, Horticulture, Soil
Social Science : Archeology, Anthropology, Forensics
Several different areas of analytical chemistry:
1. Clinical analysis - blood, urine, feces, cellular fluids, etc., for use in diagnosis.
2. Pharmaceutical analysis - establish the physical properties, toxicity, metabolites, quality control, etc.
3. Environmental analysis - pollutants, soil and water analysis, pesticides.
4. Forensic analysis - analysis related to criminology; DNA finger printing, finger print detection; blood analysis.
5. Industrial quality control - required by most companies to control
product quality.
6. Bioanalytical chemistry and analysis - detection and/or analysis of biological components (i.e., proteins, DNA, RNA, carbohydrates, metabolites, etc.).
This often overlaps many areas. Develop new tools for basic and clinical research.
History of Analytical Methods
Classical methods: early years (separation of analytes) via precipitation, extraction or distillation
Qualitative: recognized by color, boiling point, solubility, taste
Quantitative: gravimetric or titrimetric measurements
Instrumental Methods: newer, faster, more efficient
Physical properties of analytes: conductivity, electrode potential, light emission absorption, mass to charge ratio and fluorescence, many more…
Types of Analysis
•Gravimetric Methodsmeasure the mass of an analyte (or something chemically equivalent to the analyte)
•Titrimetric (Volumetric) Methodsmeasure the quantity of a reagent needed to completely react the analyte
•Electroanalytical Methodsmeasure the change in the electrical potential, current, resistance or charge produced by an analyte
•Spectroscopic Methodsmeasure the interaction between electromagnetic radiation (light, UV, IR, etc.) and the analyte
•Chemical Separationsseparate and measure the analyte of interest by chemical means (chromatography)
•Other Methods
Process of Analysis
1.Define the information you need2.Select an analysis method
3.Obtain a sample & 'clean' it up 4.Prepare the sample, solutions and
standards5.Do the analysis!
6.Account for interferences7.Calculate results and estimate
reliability8.Convert results to information
Expressing Analysis Results
percent composition (% composition) - X's 100%W/W %W/V %V/V
part per thousand (ppt) - X's 1000parts per million (ppm) - X's 106
parts per billion (ppb) - X's 109
e.g.22 ppm (w/v) lead
124 ppb (w/w) atrazine in soil
mass of analyte
mass (or volume) of sample
Titrations Introduction
1.) Buret Evolution Primary tool for titration
Descroizilles (1806)Pour out liquid
Gay-Lussac (1824)Blow out liquid
Henry (1846) Copper stopcock
Mohr (1855) Compression clipUsed for 100 years
Mohr (1855) Glass stopcock
Principles ofVolumetric Analysis
titration
titrant
analyte
indicator
equivalence point vs. end point
titration error
blank titration
Principles ofVolumetric Analysis
standardization
standard solution
Methods for establishing concentration
direct method
standardization
secondary standard solution
Titrations Introduction
2.) Volumetric analysis Procedures in which we measure the volume of
reagent needed to react with an analyte
3.) Titration Increments of reagent solution (titrant) are added
to analyte until reaction is complete.- Usually using a buret
Calculate quantity of analyte from the amount of titrant added.
Requires large equilibrium constant Requires rapid reaction
- Titrant is rapidly consumed by analyte
Controlled Chemical Reaction
Titrations
Introduction
4.) Equivalence point Quantity of added titrant is the exact amount necessary for stoichiometric
reaction with the analyte- Ideal theoretical result
AnalyteOxalic acid(colorless)
Titrant(purple)
(colorless) (colorless)
Equivalence point occurs when 2 moles of MnO4- is added to 5 moles of Oxalic acid
Titrations
Introduction
5.) End point What we actually measure
- Marked by a sudden change in the physical property of the solution- Change in color, pH, voltage, current, absorbance of light,
presence/absence ppt.
CuCl Titration with NaOH
Before any addition of NaOH After the addition of 8 drops of NaOH
End Point
Titrations
Introduction
5.) End point Occurs from the addition of a slight excess of titrant
- Endpoint does not equal equivalence point
AnalyteOxalic acid(colorless)
Titrant(purple)
(colorless) (colorless)
After equivalence point occurs, excess MnO4- turns solution purple Endpoint
Titrations
Introduction
5.) End point Titration Error
- Difference between endpoint and equivalence point- Corrected by a blank titration
i. repeat procedure without analyteii. Determine amount of titrant needed to observe changeiii. subtract blank volume from titration
Primary Standard- Accuracy of titration requires knowing precisely the
quantity of titrant added.- 99.9% pure or better accurately measure concentration
AnalyteOxalic acid(colorless)
Titrant(purple)
Titrations
Introduction
6.) Standardization Required when a non-primary titrant is used
- Prepare titrant with approximately the desired concentration- Use it to titrate a primary standard- Determine the concentration of the titrant- Reverse of the normal titration process!!!
titrant known concentration
analyte unknown concentration
titrant unknown concentration
analyte known concentration
Titration Standardization
Principles ofVolumetric Analysis
primary standard
1 .High purity
2 .Stability toward air
3 .Absence of hydrate water
4 .Available at moderate cost
5 .Soluble
6 .Large F.W.
secondary standard solution
Titrations
Introduction
7.) Back Titration Add excess of one standard reagent (known concentration)
- Completely react all the analyte- Add enough MnO4
- so all oxalic acid is converted to product
Titrate excess standard reagent to determine how much is left- Titrate Fe2+ to determine the amount of MnO4
- that did not react with oxalic acid
- Differences is related to amount of analyte - Useful if better/easier to detect endpoint
AnalyteOxalic acid(colorless)
Titrant(purple)
(colorless) (colorless)
Titrations
Titration Calculations
1.) Key – relate moles of titrant to moles of analyte
2.) Standardization of Titrant Followed by Analysis of Unknown
Calculation of ascorbic acid in Vitamin C tablet:
(i) Starch is used as an indicator: starch + I3- starch-I3
- complex(clear) (deep blue)
(ii) Titrate ascorbic acid with I3-:
1 mole ascorbic acid 1 mole I3-
Titrations
Titration Calculations
2.) Standardization of Titrant Followed by Analysis of Unknown
Standardization: Suppose 29.41 mL of I3- solution is required to react with 0.1970 g of
pure ascorbic acid, what is the molarity of the I3- solution?
Titrations
Titration Calculations
2.) Standardization of Titrant Followed by Analysis of Unknown
Analysis of Unknown: A vitamin C tablet containing ascorbic acid plus an inert binder was ground to a powder, and 0.4242g was titrated by 31.63 mL of I3
-. Find the weight percent of ascorbic acid in the tablet.
Titrations
Spectrophotometric Titrations
1.) Use Absorbance of Light to Follow Progress of Titration Example:
- Titrate a protein with Fe3+ where product (complex) has red color- Product has an absorbance maximum at 465 nm- Absorbance is proportional to the concentration of iron bound to protein
Analyte(colorless)
(red)titrant(colorless)
As Fe3+ binds protein
solution turns red
Titrations
Spectrophotometric Titrations
1.) Use Absorbance of Light to Follow Progress of Titration Example:
- As more Fe3+ is added, red color and absorbance increases, - When the protein is saturated with iron, no further color can form- End point – intersection of two lines (titrant has some absorbance at 465nm)
As Fe3+ continues to bind proteinred color and absorbance increases.
When all the protein is bound to Fe3+,no further increase in absorbance.
Titrations
Precipitation Titration Curve
1.) Graph showing how the concentration of one of the reactants varies as titrant is added.
Understand the chemistry that occurs during titration Learn how experimental control can be exerted to influence the quality of
an analytical titration- No end point at wrong pH- Concentration of analyte and titrant and size of Ksp influence end point- Help choose indicator for acid/base and oxidation/reduction titrations
Sharpness determined by titration condition
Monitor pH, voltage, current, color, absorbance, ppt.
VolumetricProcedures and
Calculationsrelate the moles of titrant to the moles of
analyte
#moles titrant = # moles analyte
#molestitrant=(V*M)titrant
=
#molesanalyte=(V*M)analyte
EXAMPLE: Describe the preparation of 2.000 L of 0.1000 M KHP (potassium hydrogen phthalate) solution (f.w. 204.23) from primary standard solid.
EXAMPLE: Describe the preparation of 2.000 L of 0.1000 M KHP (potassium hydrogen phthalate) solution (f.w. 204.23) from primary standard solid.
)2.000 Lsoln)(0.1000 molKHP)(204.32 gKHP(#g KHP-------------------------------------------- =
)1 L soln) (1 mol KHP(
EXAMPLE: Describe the preparation of 2.000 L of 0.1000 M KHP (potassium hydrogen phthalate) solution (f.w. 204.23) from primary standard solid.
)2.000 Lsoln)(0.1000 molKHP)(204.32 gKHP(#g KHP-------------------------------------------- =
)1 L soln) (1 mol KHP(
=40.85 g KHP
EXAMPLE: Describe the preparation of 2.000 L of 0.1000 M KHP (potassium hydrogen phthalate) solution (f.w. 204.23) from primary standard solid.
40.85 g of primary standard grade KHP is weighed on a balance and transferred to a 2-L volumetric flask. Carbonate free water is added to the flask
until it reaches the bottom of the neck of the flask. The solution is then mixed. More
carbonate free water is now added o bring the volume up to the line on the neck of the flask.
That line represents a volume of 2.000 L.
EXAMPLE: Describe the preparation of 1 L of 0.1 M NaOH solution (f.w. 40.00) from reagent grade solid.
EXAMPLE: Describe the preparation of 1 L of 0.1 M NaOH solution (f.w. 40.00) from reagent grade solid.
)1 Lsoln)(0.1 molNaOH)(40.00 gNaOH( #g NaOH = --------------------------------------
(1 L soln) (1 mol NaOH)
EXAMPLE: Describe the preparation of 1 L of 0.1 M NaOH solution (f.w. 40.00) from reagent grade solid.
)1 Lsoln)(0.1 molNaOH)(40.00 gNaOH( #g NaOH = --------------------------------------
(1 L soln) (1 mol NaOH)
=4 g NaOH(
EXAMPLE: Describe the preparation of 1 L of 0.1 M NaOH solution (f.w. 40.00) from reagent grade solid.
)1 Lsoln)(0.1 molNaOH)(40.00 gNaOH( #g NaOH = --------------------------------------
(1 L soln) (1 mol NaOH)
=4 g NaOH(
4 g of NaOH are weighed and added to 1 L of carbonate free water.
EXAMPLE: What is the molarity of a NaOH solution when 37.85 mL of the NaOH solution are required to react with 0.7565 g KHP primary
standard solid?
EXAMPLE: What is the molarity of a NaOH solution when 37.85 mL of the NaOH solution are required to react with 0.7565 g KHP primary
standard solid? )0.7565 gKHP)(1 molKHP)(1 molNaOH)(1000 mLsoln(M NaOH------------------------------------------------------------- = ) 37.85 mLsoln)(204.32 gKHP)(1 molKHP)(1 Lsoln(
EXAMPLE: What is the molarity of a NaOH solution when 37.85 mL of the NaOH solution are required to react with 0.7565 g KHP primary
standard solid? )0.7565 gKHP)(1 molKHP)(1 molNaOH)(1000 mLsoln(M NaOH------------------------------------------------------------- = ) 37.85 mLsoln)(204.32 gKHP)(1 molKHP)(1 Lsoln(
= 0.09782 molar
EXAMPLE: What is the % KHP in an unknown if 42.06 mL of the above NaOH soln are required to
titrate 1.545 g of unknown?
EXAMPLE: What is the % KHP in an unknown if 42.06 mL of the above NaOH soln are required to
titrate 1.545 g of unknown? )42.06mLsoln)(0.09782molNaOH)(1 mol KHP(
% KHP---------------------------------------------------------- = ) 1.545 g sample) (1 L soln) (1 mol NaOH(
)204.32 gKHP( )1 L soln(--------------------------------X 100
) 1 mol KHP)(1000 mL soln(
EXAMPLE: What is the % KHP in an unknown if 42.06 mL of the above NaOH soln are required to
titrate 1.545 g of unknown? )42.06mLsoln)(0.09782molNaOH)(1 mol KHP(
% KHP---------------------------------------------------------- = ) 1.545 g sample) (1 L soln) (1 mol NaOH(
)204.32 gKHP( )1 L soln(--------------------------------X 100
) 1 mol KHP)(1000 mL soln(
=54.41 % KHP
EXAMPLE: What is the molarity of an HCl solution if it took 39.72 mL of the above NaOH solution to titrate 25.00
mL HCl solution?
Acid-Base Indicators
Precipitation Titration Curve
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000
M AgNO3.
titration curve => pAg vs. vol. AgNO3 added
Precipitation Titration Curve
p-functionpX = - log10[X]
precipitation titration curve
four types of calculations
initial point
before equivalence point
equivalence point
after equivalence point
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.
titration curve => pAg vs. vol. AgNO3 added
initial point
after 0.0 mL of AgNO3 added
at the initial point of a titration of any type, only analyte is present, no titrant is present, therefore pAg can not be calculated.
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.
before equivalence point
pAg can be accurately calculated only after some AgBr has started to form. This may take a few mL of titrant
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.
before equivalence point
after 5.0 mL of AgNO3 added
VNaBr*MNaBr - VAgNO3*MAgNO3MNaBr unreacted------------------------------------- =
VNaBr + VAgNO3
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.
before equivalence pointafter 5.0 mL of AgNO3 added
)50.00mL*0.00500M)- (5.00mL*0.01000M(MNaBr --------------------------------------------------------------- =
)50.00 + 5.00(mL
MNaBr unreacted = 3.64 X 10-3M
Ksp = [Ag+][Br-] = 5.2 X 10-13M2
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.
equivalence point
at 25.00 mL of AgNO3 added
becomes when [Ag+] = [Br-]
]Ag+[2 = 5.2 X 10-13M2
]Ag+ = [7.21 X 10-7M
pAg = 6.14
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.
after equivalence point
After the equivalence point there is very little change in the amount of precipitate present (except very close to the
equivalence point)
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.
after equivalence point
thus, at 25.10 mL of AgNO3 added
VAgNO3*MAgNO3 - VNaBr*MNaBrMAgNO3 unreacted----------------------------------- =
VAgNO3 + VNaBr
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.
after equivalence point
thus, at 25.10 mL of AgNO3 added
VAgNO3*MAgNO3 - VNaBr*MNaBrMAgNO3 unreacted---------------------------------- =
VAgNO3 + VNaBr
) 25.10 mL * 0.01000) - (50.00 mL * 0.00500(MAgNO3 -------------------------------------------------------------- =
)25.10 + 50.00(mL
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.
after equivalence point
thus, at 25.10 mL of AgNO3 added
) 25.10 mL * 0.01000 M) - (50.00 mL * 0.00500 M(
MAgNO3 -------------------------------------------------------------- = unreacted (25.10 + 50.00)mL
MAgNO3 unreacted = 1.33 X 10-5M
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.
after equivalence point
thus, at 25.10 mL of AgNO3 added
MAgNO3 unreacted = 1.33 X 10-5M
]Ag+[total = [Ag+]AgNO3 unreacted + [Ag+]dissolved
AgBr
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.
after equivalence point
thus, at 25.10 mL of AgNO3 added
]Ag+[total = [Ag+]AgNO3 unreacted + [Ag+]dissolved
AgBr
]Ag+[total = 1.33 X 10-5M + [Ag+]dissolved AgBr
]Ag+[total ~ 1.33 X 10-5M
pAg = 4.88
0
Vol of titrantpAg5.00 9.84
25.00 6.1425.10 4.88
Precipitation Titration
0
2
4
6
8
10
12
0.00 5.00 10.00 15.00 20.00 25.00 30.00
Vol of AgNO3 added
pA
g
Vol of titrantpAg5.00 9.84
10.00 9.68 2.60 0.0025
15.00 9.47 2.81 0.001538
20.00 9.13 3.15 0.000714
21.00 9.03 3.25 0.000563
22.00 8.90 3.38 0.000417
23.00 8.72 3.56 0.000274
24.00 8.41 3.87 0.000135
24.50 8.11 4.17 6.71E-05
25.00 6.1425.10 4.8826.00 3.88 0.0001316
27.00 3.59 0.0002597
28.00 3.41 0.0003846
29.00 3.30 0.0005063
30.00 3.20 0.000625
35.00 2.93 0.0011765
40.00 2.78 0.0016667
45.00 2.68 0.0021053
Precipitation Titration
0
2
4
6
8
10
12
0.00 10.00 20.00 30.00 40.00 50.00
Vol of AgNO3 added
pA
g