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Chapter 07 Let the Titrations Begin

Chapter 07 Let the Titrations Begin. Contents in Chapter07. Overview of Titrimetry Grades of Chemicals Titration Calculations Precipitation Titrations 1) Titration curve 2) Argentometric titration Establish Precipitation Titration Curves with a Spreadsheet. Overview of Titrimetry

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Chapter 07 Let the Titrations Begin

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  1. Chapter 07 Let the Titrations Begin QCA7e Chapter07

  2. Contents in Chapter07 • Overview of Titrimetry • Grades of Chemicals • Titration Calculations • Precipitation Titrations • 1) Titration curve • 2) Argentometric titration • Establish Precipitation Titration Curves with a Spreadsheet QCA7e Chapter07

  3. Overview of Titrimetry • Define Titration: • Titrations (or titrimetric method) are based on measuring the amount of a reagent of known concentration that reacts with the unknown. • A general equation can be expressed: • aA + tT → products • A: analyte • T: titrant QCA7e Chapter07

  4. Revisiting Keywords of Titration • Equivalence point: The point (e.g., volume of titrant) in a titration where (theoretically) stoichiometrically equivalent amounts of analyte and titrant react. • Indicator: A colored compound whose change in color signals the (experimental) end point of a titration. • End point: The point (e.g., volume of titrant) in a (experimental) titration where we stop adding titrant in an experiment. • Titration error: The determinate error in a titration due to the difference between the end point and the equivalence point. QCA7e Chapter07

  5. Type of Titrations based on Chemical Reactions • Acid-Base Titrations, example: • H+ + OH–→ H2O K= 1/Kw • Precipitation Titrations, example: • Ag+(aq) + Cl–(aq)→ AgCl(s) K=1/Ksp • Redox Titrations: • 5 H2O2 + 2 MnO4–+ H+→ 5 O2 + 2 Mn2+ + 8H2O • Complexometric Titrations, example: • EDTA + Ca2+→ (Ca–EDTA)2+ QCA7e Chapter07

  6. 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. QCA7e Chapter07

  7. 5) Type of Titration Curves V. = volume QCA7e Chapter07

  8. 2. Grades of Chemicals • Terms and Definitions • Reagent Grade: The reagents which meets or surpasses the latest American Chemical Society specifications. • 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 - High purity - Nonhygroscopic - Chemically stable both in solid and solution - High MW or FW QCA7e Chapter07

  9. Secondary standard: A standard which is standardized against a primary standard. • Certified reference materials (CRM): A reference material, accompanied by a certificate, which has been analysed by different laboratories to determine consensus levels of the analyte concentration. • NIST Standard Reference Material® (SRM): A CRM issued by NIST that also meets additional NIST-specific certification criteria and is issued with a certificate. • Standardization: The process by which the concentration of a reagent is determined by reaction with a known quantity of a second reagent QCA7e Chapter07

  10. Titration Calculations • Terms and Definitions: • 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. QCA7e Chapter07

  11. 2) Standardization Example: To standardizing a KMnO4 stock solution, the primary standard of 9.1129 g Na2C2O4 is dissolved in 250.0 mL volumetric flask. 10.00 mL of the Na2C2O4 solution require 48.36 mL of KMnO4 to reach the titration end point. What is the molarity (M) of MnO4– stock solution? (FW Na2C2O4 134.0) Solution: 5C2O42–(aq) + 2MnO4–(aq) + 16H+(aq) → 10CO2(g) + Mn2+(aq) + 8H2O(l) Ans QCA7e Chapter07

  12. 3) Unknown Analysis with a Blank Correction Example: A 0.2865 g sample of an iron ore is dissolved in acid, and the iron is converted entirely to Fe2+. To titrate the resulting solution, 0.02653 L of 0.02250 M KMnO4 is required. Also a blank titration require 0.00008 L of KMnO4 solution. What is the % Fe (w/w) in the ore? (AW Fe 55.847) Solution: MnO4–(aq) + 5Fe2+ + 8H+(aq) → Mn2+(aq) + 5Fe3+ + 4H2O(l) Ans QCA7e Chapter07

  13. Example: The arsenic in 1.010 g sample was pretreated to H3AsO4(aq) by suitable treatment. The 40.00 mL of 0.06222 M AgNO3 was added to the sample solution forming Ag3AsO4(s): The excess Ag+ was titrated with 10.76 mL of 0.1000 M KSCN. The reaction was: Calculate the percent (w/w) As2O3(s) (fw 197.84 g/mol) in the sample. 4) Back Titration Solution: mL x M = mmol Total mmol Ag+ added = (mmol Ag+ consumed by SCN–) + (mmol Ag+ consumed by H3AsO4) QCA7e Chapter07

  14. Ans QCA7e Chapter07

  15. Step 1: Kjeldahl digestion (decomposing and dissolving) Step 2: Neutralization by adding base Step 3: Distillation NH3 into excess HCl standard Step 4: Titrating unreacted HCl with NaOH standard • Kjeldahl Analysis for Total Nitrogen (TN) • i) KD description: QCA7e Chapter07

  16. Example: A typical meat protein contains 16.2% (w/w) nitrogen. A 0.500 mL aliquot of protein solution was digested, and the liberated NH3 was distilled into 10.00 mL of 0.02140 M HCl. The unreacted HCl required 3.26 mL of 0.0198 M NaOH for complete titration. Find the concentration of protein (mg protein/mL) in the original sample. Solution: QCA7e Chapter07

  17. Ans QCA7e Chapter07

  18. Example: A solid mixture weighing 1.372 g containing only sodium carbonate (Na2CO3, FW 105.99) and sodium bicarbonate (NaHCO3, FW 84.01) require 29.11 mL of 0.7344 M HCl for complete titration: Find the mass of each component of the mixture. Solution: 6) Titration of a Mixture QCA7e Chapter07

  19. Ans Ans QCA7e Chapter07

  20. Precipitation Titrations • A titration in which the reaction between the analyte and titrant involves a precipitation. 1) Titration curve i) Guidance in precipitation titration calculation • Find Ve (volume of titrant at equivalence point) • Find y-axis values: - At beginning - Before Ve - At Ve - After Ve QCA7e Chapter07

  21. Example: For the titration of 50.0 mL of 0.0500 M Cl– with 0.100 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: QCA7e Chapter07

  22. (a) 0 mL Ag+ added (At beginning) [Ag+] = 0, pAg can not be calculated. [Cl–] = 0.0500, pCl = 1.30 (b) 10 mL Ag+ added (Before Ve) QCA7e Chapter07

  23. (c) 25 mL Ag+ added (At Ve) AgCl(s) Ag+(aq) + Cl–(aq) Ksp = 1.8×10–10 s = [Ag+]=[Cl–] Ksp = 1.8×10–10 = s2 [Ag+]=[Cl–]=1.35x10–5 pAg = 4.89 pCl = 4.89 QCA7e Chapter07

  24. (d) 35 mL Ag+ added (After Ve) QCA7e Chapter07

  25. pCl pAg ii) Construct a titration curve Example: Titration of 50.0 mL of 0.0500 M Cl– with 0.100 M Ag+ QCA7e Chapter07

  26. dy/dx d2y/dx2 iii) End point determination QCA7e Chapter07

  27. iv) Diluting effect of the titration curves 25.00 mL 0.1000 M I– titrated with 0.05000 M Ag+ 25.00 mL 0.01000 M I– titrated with 0.005000 M Ag+ 25.00 mL 0.001000 M I– titrated with 0.0005000 M Ag+ QCA7e Chapter07

  28. v) Ksp effect of the titration curves 25.00 mL 0.1000 M halide (X–) titrated with 0.05000 M Ag+ QCA7e Chapter07

  29. vi) Titration of a mixture (uncertainty concerned) • 40.00 mL of 0.0502 M KI + 0.0500 M KCl, titrated with 0.0845 M Ag+ • 20.00 mL of 0.1004 M KI titrated with 0.0845 M Ag+ QCA7e Chapter07

  30. Example: A 25.00 mL solution containing Br– and Cl– was titrated with 0.03333 M AgNO3. Ksp(AgBr)=5x10–13, Ksp(AgCl)=1.8x10–10. • Which analyte is precipitated first? • The first end point was observed at 15.55 mL. Find the concentration of the first that precipitated (Br– or Cl–?). • The second end point was observed at 42.23 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 QCA7e Chapter07

  31. (b) Ans (c) Ans QCA7e Chapter07

  32. Argentometric Titration • General information: • 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 QCA7e Chapter07

  33. 2) Volhard method: A colored complex (back titration). Analysing Cl– for example: Step 1: Adding excess Ag+ into sample Ag+ + Cl– → AgCl(s) + left Ag+ Step 2: Removing AgCl(s) by filtration/washing Step 3: Adding Fe3+ into filtrate (i.e., the left Ag+) Step 4: Titrating the left Ag+ by SCN–: Ag+ + SCN– → AgSCN(s) Step 5: End point determination by red colored Fe(SCN)2+ complex. (when all Ag+ has been consumed, SCN– reacts with Fe3+) SCN– + Fe3+ → Fe(SCN)2+(aq) Total mol Ag+ = (mol Ag+ consumed by Cl–) + (mol Ag+ consumed by SCN–) QCA7e Chapter07

  34. Fajans Method: An adsorbed/colored indicator. Titrating Cl– and adding • dichlorofluoroscein for example: Cl– Before Ve (Cl– excess) Greenish yellow solution AgCl(s) 1st layer Ag+ In– pink AgCl(s) After Ve (Ag+ excess) 1st layer QCA7e Chapter07

  35. 4) Mohr Method: A colored precipitate formed by Ag+ with anion, other than analyte, once the Ve reached. Analysing Cl– and adding CrO42–for example: Precipitating Cl–: Ag+ + Cl– → AgCl(s) Ksp = 1.8 x 10–10 End point determination by red colored precipitate, Ag2CrO4(s): 2Ag+ + CrO42– → Ag2CrO4(s) Ksp = 1.2 x 10–12 QCA7e Chapter07

  36. 5) Applications of argentometric titrations: QCA7e Chapter07

  37. Establish Precipitation Titration Curves with a Spreadsheet • Example: • M+(aq) + X–(aq) MX(s) K = 1/Ksp Titrant Analyte 筆算法 工作表 已知值 已知值 已知值 已知值 已知值 已知值 應變數 自變數 自變數 應變數 應變數 應變數 QCA7e Chapter07

  38. 1) Derive the equation: moles of M in precipitate moles of X in precipitate Mass balance: The moles of an element in all species in a mixture equal to the total moles of that element delivered to the solution. QCA7e Chapter07

  39. 2) Apply the equation: QCA7e Chapter07

  40. Example: Construct a titration curve for the titration of 50.0 mL of 0.0500 M Cl– with 0.100 M Ag+. (Ksp of AgCl = 1.8×10–10) QCA7e Chapter07

  41. EXCEL 981_Ch07_Titration_X_with_Ag QCA7e Chapter07

  42. Homework: Problem 7-30/p.139, Due 2009/12/9 請用學號作為檔名上載 Examples: All Exercise： A-E, G Problems: 1-7, 11-14, 31, 36 End of Chapter07 QCA7e Chapter07

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