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What is Analytical Chemistry ? - Analytical chemistry deals with separating, identifying, and quantifying the relative amounts of the components of an.

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Presentation on theme: "What is Analytical Chemistry ? - Analytical chemistry deals with separating, identifying, and quantifying the relative amounts of the components of an."— Presentation transcript:

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2 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.

3 Analytical Chemistry analyze: "what is it? (qualitative analysis) "how much is there? “ (quantitative analysis)

4 The role of analytical chemistry: central science The relationship between analytical chemistry and the other sciences Analytical chemistry 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

5 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.

6 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…

7 Types of Analysis Gravimetric 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

8 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

9 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 10 6 parts per billion (ppb) - X's 10 9 e.g. 22 ppm (w/v) lead 124 ppb (w/w) atrazine in soil

10 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 clip Used for 100 years Mohr (1855) Glass stopcock

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

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

13 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

14 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) (colorless) Equivalence point occurs when 2 moles of MnO 4 - is added to 5 moles of Oxalic acid

15 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

16 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) (colorless) After equivalence point occurs, excess MnO 4 - turns solution purple  Endpoint

17 Titrations Introduction 5.)End point  Titration Error -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 % pure or better  accurately measure concentration Analyte Oxalic acid (colorless) Titrant (purple)

18 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

19 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

20 Titrations Introduction 7.)Back Titration  Add excess of one standard reagent (known concentration) -Completely react all the analyte -Add enough MnO 4 - so all oxalic acid is converted to product  Titrate excess standard reagent to determine how much is left - Titrate Fe 2+ to determine the amount of MnO 4 - 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)

21 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 + I 3 -  starch-I 3 - complex (clear) (deep blue) (ii) Titrate ascorbic acid with I 3 - : 1 mole ascorbic acid  1 mole I 3 -

22 Titrations Titration Calculations 2.)Standardization of Titrant Followed by Analysis of Unknown Standardization: Suppose mL of I 3 - solution is required to react with g of pure ascorbic acid, what is the molarity of the I 3 - solution?

23 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 g was titrated by mL of I 3 -. Find the weight percent of ascorbic acid in the tablet.

24 Titrations Spectrophotometric Titrations 1.)Use Absorbance of Light to Follow Progress of Titration  Example: -Titrate a protein with Fe 3+ 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 Fe 3+ binds protein solution turns red

25 Titrations Spectrophotometric Titrations 1.)Use Absorbance of Light to Follow Progress of Titration  Example: -As more Fe 3+ 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 Fe 3+ continues to bind protein red color and absorbance increases. When all the protein is bound to Fe 3+, no further increase in absorbance.

26 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 K sp 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.

27 Volumetric Procedures and Calculations relate the moles of titrant to the moles of analyte # moles titrant = # moles analyte #moles titrant =(V*M) titrant = #moles analyte =(V*M) analyte

28 EXAMPLE: Describe the preparation of L of M KHP (potassium hydrogen phthalate) solution (f.w ) from primary standard solid.

29 (2.000 L soln )( mol KHP )( g KHP ) #g KHP = (1 L soln ) (1 mol KHP )

30 EXAMPLE: Describe the preparation of L of M KHP (potassium hydrogen phthalate) solution (f.w ) from primary standard solid. (2.000 L soln )( mol KHP )( g KHP ) #g KHP = (1 L soln ) (1 mol KHP ) = g KHP

31 EXAMPLE: Describe the preparation of L of M KHP (potassium hydrogen phthalate) solution (f.w ) from primary standard solid 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 L.

32 EXAMPLE: Describe the preparation of 1 L of 0.1 M NaOH solution (f.w ) from reagent grade solid.

33 (1 L soln )(0.1 mol NaOH )(40.00 g NaOH ) # g NaOH = (1 L soln ) (1 mol NaOH )

34 EXAMPLE: Describe the preparation of 1 L of 0.1 M NaOH solution (f.w ) from reagent grade solid. (1 L soln )(0.1 mol NaOH )(40.00 g NaOH ) # g NaOH = (1 L soln ) (1 mol NaOH ) = 4 g NaOH)

35 EXAMPLE: Describe the preparation of 1 L of 0.1 M NaOH solution (f.w ) from reagent grade solid. (1 L soln )(0.1 mol NaOH )(40.00 g NaOH ) # 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.

36 EXAMPLE: What is the molarity of a NaOH solution when mL of the NaOH solution are required to react with g KHP primary standard solid?

37 ( g KHP )(1 mol KHP )(1 mol NaOH )(1000 mL soln ) M NaOH = (37.85 mL soln )( g KHP )(1 mol KHP )(1 L soln )

38 EXAMPLE: What is the molarity of a NaOH solution when mL of the NaOH solution are required to react with g KHP primary standard solid? ( g KHP )(1 mol KHP )(1 mol NaOH )(1000 mL soln ) M NaOH = (37.85 mL soln )( g KHP )(1 mol KHP )(1 L soln ) = molar

39 EXAMPLE: What is the % KHP in an unknown if mL of the above NaOH soln are required to titrate g of unknown?

40 (42.06mL soln )( mol NaOH )(1 mol KHP ) % KHP = (1.545 g sample ) (1 L soln ) (1 mol NaOH ) ( g KHP ) (1 L soln ) X 100 (1 mol KHP )(1000 mL soln )

41 EXAMPLE: What is the % KHP in an unknown if mL of the above NaOH soln are required to titrate g of unknown? (42.06mL soln )( mol NaOH )(1 mol KHP ) % KHP = (1.545 g sample ) (1 L soln ) (1 mol NaOH ) ( g KHP ) (1 L soln ) X 100 (1 mol KHP )(1000 mL soln ) = % KHP

42 EXAMPLE: What is the molarity of an HCl solution if it took mL of the above NaOH solution to titrate mL HCl solution?

43 Acid-Base Indicators

44 Precipitation Titration Curve EXAMPLE: Derive a curve for the titration of mL of M NaBr with M AgNO 3. titration curve => pAg vs. vol. AgNO 3 added

45 Precipitation Titration Curve p-functionpX = - log 10 [X] precipitation titration curve four types of calculations initial point before equivalence point equivalence point after equivalence point

46 EXAMPLE: Derive a curve for the titration of mL of M NaBr with M AgNO 3. titration curve => pAg vs. vol. AgNO 3 added initial point after 0.0 mL of AgNO 3 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.

47 EXAMPLE: Derive a curve for the titration of mL of M NaBr with M AgNO 3. before equivalence point pAg can be accurately calculated only after some AgBr has started to form. This may take a few mL of titrant

48 EXAMPLE: Derive a curve for the titration of mL of M NaBr with M AgNO 3. before equivalence point after 5.0 mL of AgNO 3 added V NaBr *M NaBr - V AgNO3 *M AgNO3 M NaBr unreacted = V NaBr + V AgNO3

49 EXAMPLE: Derive a curve for the titration of mL of M NaBr with M AgNO 3. before equivalence point after 5.0 mL of AgNO 3 added (50.00mL* M) -(5.00mL* M) M NaBr = ( )mL M NaBr unreacted = 3.64 X M K sp = [Ag + ][Br - ] = 5.2 X M 2

50 EXAMPLE: Derive a curve for the titration of mL of M NaBr with M AgNO 3. equivalence point at mL of AgNO 3 added becomes when [Ag + ] = [Br - ] [Ag + ] 2 = 5.2 X M 2 [Ag + ] = 7.21 X M pAg = 6.14

51 EXAMPLE: Derive a curve for the titration of mL of M NaBr with M AgNO 3. 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)

52 EXAMPLE: Derive a curve for the titration of mL of M NaBr with M AgNO 3. after equivalence point thus, at mL of AgNO 3 added V AgNO3 *M AgNO3 - V NaBr *M NaBr M AgNO3 unreacted = V AgNO3 + V NaBr

53 EXAMPLE: Derive a curve for the titration of mL of M NaBr with M AgNO 3. after equivalence point thus, at mL of AgNO 3 added V AgNO3 *M AgNO3 - V NaBr *M NaBr M AgNO3 unreacted = V AgNO3 + V NaBr (25.10 mL * ) - (50.00 mL * ) M AgNO3 = ( )mL

54 EXAMPLE: Derive a curve for the titration of mL of M NaBr with M AgNO 3. after equivalence point thus, at mL of AgNO 3 added (25.10 mL * M) - (50.00 mL * M) M AgNO3 = unreacted ( )mL M AgNO3 unreacted = 1.33 X M

55 EXAMPLE: Derive a curve for the titration of mL of M NaBr with M AgNO 3. after equivalence point thus, at mL of AgNO 3 added M AgNO3 unreacted = 1.33 X M [Ag + ] total = [Ag + ] AgNO3 unreacted + [Ag + ] dissolved AgBr

56 EXAMPLE: Derive a curve for the titration of mL of M NaBr with M AgNO 3. after equivalence point thus, at mL of AgNO 3 added [Ag + ] total = [Ag + ] AgNO3 unreacted + [Ag + ] dissolved AgBr [Ag + ] total = 1.33 X M + [Ag + ] dissolved AgBr [Ag + ] total ~ 1.33 X M pAg =

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