What is Analytical Chemistry ? - Analytical chemistry deals with separating, identifying, and...

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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