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 science The relationship between analytical chemistry and the other sciences 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 Analytical chemistry

Several different areas of analytical chemistry: Clinical analysis - blood, urine, feces, cellular fluids, etc., for use in diagnosis. 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 Methods Titrimetric (Volumetric) Methods measure the mass of an analyte (or something chemically equivalent to the analyte) Titrimetric (Volumetric) Methods measure the quantity of a reagent needed to completely react the analyte Electroanalytical Methods measure the change in the electrical potential, current, resistance or charge produced by an analyte Spectroscopic Methods measure the interaction between electromagnetic radiation (light, UV, IR, etc.) and the analyte Chemical Separations separate and measure the analyte of interest by chemical means (chromatography) Other Methods

1.Define the information you need 2.Select an analysis method Process of Analysis 1.Define the information you need 2.Select an analysis method 3.Obtain a sample & 'clean' it up 4.Prepare the sample, solutions and standards 5.Do the analysis! 6.Account for interferences 7.Calculate results and estimate reliability 8.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 1000 parts 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

Titrations Introduction 1.) Buret Evolution Primary tool for titration Gay-Lussac (1824) Blow out liquid Mohr (1855) Compression clip Used for 100 years Descroizilles (1806) Pour out liquid Henry (1846) Copper stopcock Mohr (1855) Glass stopcock

Principles of Volumetric Analysis titration titrant analyte indicator equivalence point vs. end point titration error blank titration

Principles of Volumetric Analysis standardization standard solution Methods for establishing concentration direct method secondary standard solution

Titrations Introduction Controlled Chemical Reaction 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 Analyte Oxalic acid (colorless) Titrant (purple) Equivalence point occurs when 2 moles of MnO4- is added to 5 moles of Oxalic acid

CuCl Titration with NaOH Before any addition of NaOH 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 After the addition of 8 drops of NaOH End Point Before any addition of NaOH

Titrations Introduction 5.) End point Occurs from the addition of a slight excess of titrant - Endpoint does not equal equivalence point Analyte Oxalic acid (colorless) Titrant (purple) After equivalence point occurs, excess MnO4- turns solution purple  Endpoint

Titrations Introduction 5.) End point Titration Error Primary Standard - Difference between endpoint and equivalence point - Corrected by a blank titration i. repeat procedure without analyte ii. Determine amount of titrant needed to observe change iii. 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 Analyte Oxalic acid (colorless) Titrant (purple)

Titrations Introduction 6.) Standardization Titration 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!!! Titration Standardization titrant known concentration titrant unknown concentration analyte unknown concentration analyte known concentration

Principles of Volumetric 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 Analyte Oxalic 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: 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) titrant (colorless) (red) 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) When all the protein is bound to Fe3+, no further increase in absorbance. As Fe3+ continues to bind protein red color and absorbance increases.

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.

Volumetric Procedures and Calculations relate 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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-function pX = - 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 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 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 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*MAgNO3 MNaBr unreacted = ------------------------------------- VNaBr + VAgNO3

EXAMPLE: Derive a curve for the titration of 50. 00 mL of 0 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 (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 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 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 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*MNaBr MAgNO3 unreacted = ----------------------------------- VAgNO3 + VNaBr

EXAMPLE: Derive a curve for the titration of 50. 00 mL of 0 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*MNaBr MAgNO3 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 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 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 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