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Titrations Volumetric analysis

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Titrations Volumetric analysis Procedures in which we measure the volume of reagent needed to react with an analyte 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 (Thermodynamic) Requires rapid reaction (kinetic) aA + tT → products A: analyte T: titrant slides

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Titrations Buret Evolution 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 slides

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Type of Titrations based on Measuring Techniques Volumetric titrimetry: Measuring the volume of a solution of a known concentration (e.g., mol/L) that is needed to react completely with the analyte. Gravimetric (weight) titrimetry: Measuring the mass of a solution of a known concentration (e.g., mol/kg) that is needed to react completely with the analyte. Coulometric titrimetry: Measuring total charge (current x time) to complete the redox reaction, then estimating analyte concentration by the moles of electron transferred. slides

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Type of Titrations based on Chemical Reactions Acid-Base Titrations, example: H+ + OH– → H2O Precipitation Titrations, example: Ag+(aq) + Cl–(aq) → AgCl(s) Redox Titrations: 5 H2O2 + 2 MnO4– + H+ → 5 O2 + 2 Mn2+ + 8H2O Complexometric Titrations, example: EDTA + Ca2+ → (Ca–EDTA)2+ slides

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Type of Titration Curves Type Example. y-axis x-axis Acid-base HCl/NaOH pH V. NaOH Precipitation Ag+/Cl– pAg+ V. Ag+ Complexation Ca2+/EDTA pCa2+ V. EDTA Redox MnO4–/Fe2+ Potential V. Fe2+ V. = volume Type Example y-axis x-axis Spectro-photometric apotransferrin/ Fe3+ Absorbance V. Thermo- metric H3BO4/ NaOH Temperature slides

6 Expressing concentration
Formality Molarity (V & W) Molality Normality %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 slides

7 relationship between titrant and analyte
# Eqg titrant = # Eqg analyte (V*N)titrant =(V*N)analyte # Eqg titrant = (V*N)titrant #molestitrant=(V*M)titrant #molesanalyte=(V*M)analyte slides

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Standardization: The process by which the concentration of a reagent is determined by reaction with a known quantity of a second reagent Primary standard: The reagent which is ready to be weighted and used prepare a solution with known concentration (standard). Requirements of primary reagent are: - Known stoichiometric composition and reaction - High purity - Nonhygroscopic - Chemically stable both in solid and solution - High MW or FW Secondary standard: A standard solution which is standardized against a primary standard. slides

9 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 - Titration Standardization titrant known concentration titrant unknown concentration analyte unknown concentration analyte known concentration slides

10 Standardization of 0.1 M NaOH
1-selection the PS (e.g. KHP) 2-wheing the PS 10*0.1=mg/ 3-making solution 4-addind suitable indicator 5-titration ml 6-calculation *n=213.8/ n=0.115 N1V1=N2V2 slides

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Blank Titration: Titration procedure is carried out without analyte (e.g., a distilled water sample). It is used to correct titration error. Back titration: A titration in which a (known) excess reagent is added to a solution to react with the analyte. The excess reagent remaining after its reaction with the analyte, is determined by a titration. slides

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Standardization Example: To standardizing a KMnO4 stock solution, the primary standard of g Na2C2O4 is dissolved in mL volumetric flask mL of the Na2C2O4 solution require mL of KMnO4 to reach the titration end point. What is the molarity (M) of MnO4– stock solution? (FW Na2C2O ) Solution: 5C2O42–(aq) + 2MnO4–(aq) + 16H+(aq) → 10CO2(g) + Mn2+(aq) + 8H2O(l) Ans slides

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Unknown Analysis with a Blank Correction Example: A g sample of an iron ore is dissolved in acid, and the iron is converted entirely to Fe2+. To titrate the resulting solution, L of M KMnO4 is required. Also a blank titration require L of KMnO4 solution. What is the % Fe (w/w) in the ore? (AW Fe ) MnO4–(aq) + 5Fe2+ + 8H+(aq) → Mn2+(aq) + 5Fe3+ + 4H2O(l) slides

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Back Titration 1)Add excess of one standard reagent (known concentration) 2)Titrate excess standard reagent to determine how much is left - Add 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 MnO4–(aq) + 5Fe2+ + 8H+(aq) → Mn2+(aq) + 5Fe3+ + 4H2O(l) slides

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Back Titration Example: The arsenic in g sample was pretreated to H3AsO4(aq) by suitable treatment. The mL of M AgNO3 was added to the sample solution forming Ag3AsO4(s): The excess Ag+ was titrated with mL of M KSCN. The reaction was: Calculate the percent (w/w) As2O3(s) (fw g/mol) in the sample. slides

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slides

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Back Titration In a back titration analysis of HCO3-, 25 mL of a bicarbonate solution is reacted with mL of M NaOH. The excess NaOH was titrated with M HCl. This required mL. What is the concentration of bicarbonate in solution? NaOH + HCO3- → Na+ + CO3- + H2O NaOH + HCl → NaCl + H2O slides

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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 slides

19 End point Occurs from the addition of a slight excess of titrant
Marked by a sudden change in the physical property of the solution - Change in color, pH, voltage, current, absorbance of light. - End point ≈ equivalence point Analyte Oxalic acid (colorless) Titrant (purple) After equivalence point occurs, excess MnO4- turns solution purple  Endpoint slides

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Titration Error Difference between endpoint and equivalence point Corrected by a blank titration 1) repeat procedure without analyte 2) Determine amount of titrant needed to observe change 3) subtract blank volume from titration slides

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Calculation of ascorbic acid in Vitamin C tablet: Standardization: Suppose mL of I3- solution is required to react with g of pure ascorbic acid, what is the molarity of the I3- solution? 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 I3-. Find the weight percent of ascorbic acid in the tablet. Starch is used as an indicator: starch + I3-  starch-I3- complex ascorbic acid was oxidized with I3-: 1 mole ascorbic acid  1 mole I3- slides

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Titration of a Mixture Example: A solid mixture weighing g containing only sodium carbonate (Na2CO3, FW ) and sodium bicarbonate (NaHCO3, FW 84.01) require mL of M HCl for complete titration: Find the mass of each component of the mixture. mass moles Total mixture 1.372 Na2CO3 x x/ g/mol NaHCO3 x 1.372 – x/84.01 g/mol slides

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Ans Ans slides

24 Direct and back (indirect) titration of Aspirin
slides

25 Precipitation Titrations
A titration in which the reaction between the analyte and titrant involves a precipitation. Ag+(aq) + Cl–(aq)  AgCl(s) AgCl(s)  Ag+(aq) + Cl–(aq) Ksp = 1.8×10–10 s = [Ag+]=[Cl–] [Ag+]=[Cl–]=1.35x10–5 pAg = 4.89 pCl = 4.89 slides

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Titration curve of 50.0 mL of M Cl– with M Ag+ pCl pAg slides

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Example: For the titration of 50.0 mL of M Cl– with M Ag+. The reaction is: Ag+(aq) + Cl–(aq)  AgCl(s) K = 1/Ksp = 1/(1.8×10–10) = 5.6 x 109 Find pAg and pCl of Ag+ solution added 0 mL (b) 10.0 mL (c) 25.0 mL (d) 35.0 mL Solution: (a) 0 mL Ag+ added (At beginning) [Ag+] = 0, pAg can not be calculated. [Cl–] = , pCl = 1.30 (b) 10 mL Ag+ added (Before Ve) N1V1=N2V2 50*0.05=0.1*V2 In eq point, 25 ml √Ksp=[Ag+]=[Cl-]=1.34*10-5 (d) 35 mL Ag+ added (After Ve) slides

28 Diluting effect of the titration curves
25.00 mL M I– titrated with M Ag+ 25.00 mL M I– titrated with M Ag+ 25.00 mL M I– titrated with M Ag+ slides

29 Ksp effect of the titration curves
25.00 mL M halide (X–) titrated with M Ag+ slides

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slides

31 Titration of a mixture (uncertainty concerned)
40.00 mL of M KI M KCl, titrated with M Ag+ 20.00 mL of M KI titrated with M Ag+ slides

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Example: A mL solution containing Br– and Cl– was titrated with M AgNO3. Ksp(AgBr)=5x10–13, Ksp(AgCl)=1.8x10–10. Which analyte is precipitated first? The first end point was observed at mL. Find the concentration of the first that precipitated (Br– or Cl–?). The second end point was observed at mL. Find the concentration of the second that precipitated (Br– or Cl–?). Solution: (a) Ag+(aq) + Br–(aq)  AgBr(s) K = 1/Ksp(AgBr) = 2x1012 Ag+(aq) + Cl–(aq)  AgCl(s) K = 1/Ksp(AgCl) = 5.6x109 Ans: AgBr precipitated first slides

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Argentometric Titration Define Argentometric Titration: A precipitation titration in which Ag+ is the titrant. Argentometric Titration classified by types of End-point detection: – Volhard method: A colored complex (back titration) – Fajans method: An adsorbed/colored indicator – Mohr method: A colored precipitate slides

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Mohr method     The Mohr method was first published in 1855 as a method for chloride analysis. ·                   In the precipitation of chloride by silver ion, chromate ion (CrO42) is used as an indicator in the formation of Ag2CrO4, a reddish-brown precipitate formed when excess Ag+ is present.  Ag Cl  AgCl(s) white precipitate  2Ag CrO42  Ag2CrO4(s) red precipitate CrO42 Ksp= 1.8 x (S = 1.34 x 10-5 M) Ksp= 1.2 x (s = 6.7 x 10-5M) slides

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 The titrations are performed only in neutral or slightly basic medium to prevent silver hydroxide formation (at pH > 10). 2Ag+ + 2OH  2AgOH(s)  Ag2O(s) + H2O precipitate  to prevent chromic acid formation (at pH < 7).  CrO42 + H3O HCrO4 + H2O 2 CrO42 + 2 H3O H2CrO4 + H2O slides

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Volhard METHOD     Back titration for determination of Cl-..  First published in 1874.    Reactions: Ag Cl  AgCl(s) Ksp = 1.82 x 10-10 (excess) white precipitate SCN + Ag+  AgSCN(s) Ksp = 1.1 x 10-12 titrant white precipitate SCN + Fe3+  FeSCN Kf = 1.4 x 10+2 Indicator red complex H+, Fe3+ slides

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 The titration is usually done in acidic pH medium to prevent precipitation of iron hydroxides, Fe(OH)3. Fe3+ +3(OH)- ⇄ Fe(OH)3 Ksp=1*10-39 If [Fe3+]=0.001 M pH=????? slides

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 Since SAgSCN <SAgCl , equilibrium will shift to the right causing a negative error for the chloride analysis. Ag Cl  AgCl(s) Ksp = 1.82 x 10-10 SCN + Ag+  AgSCN(s) Ksp = 1.1 x 10-12 SCN- + AgCl  AgSCN + Cl- To eliminate this error, AgCl must be filtered or add nitrobenzene before titrating with thiocyanate; nitrobenzene will form an oily layer on the surface of the AgCl precipitate, thus preventing its reaction with thiocyanate. slides

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Fajans Method: An adsorbed/colored indicator. Titrating Cl– and adding dichlorofluoroscein as indicator: Before Ve (Cl– excess) Greenish yellow solution After Ve (Ag+ excess) slides

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5) Applications of argentometric titrations: slides


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