Classical methods in techniques of analytical chemistry: Titrimetric methods of analysis ERT 207 ANALYTICAL CHEMISTRY SEMESTER 1, ACADEMIC SESSION 2015/16

Overview OVERVIEW OF TITRIMETRY GRADES OF CHEMICALS TITRATION CALCULATIONS PRECIPITATION TITRATIONS - TITRATION CURVE ARGENTOMETRIC TITRATION bblee@unimap

OVERVIEW OF TITRIMETRY Defination: Titrations (or titrimetric method) are based on measuring the amount of a reagent of known concentration that reacts with the unknown. It is the process of determining unknown concentration by adding the small increments of standard solution until the reaction is just complete. A general equation can be expressed: aA + tT → products where, A: analyte; T: titrant bblee@unimap

OVERVIEW OF TITRIMETRY 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. bblee@unimap

OVERVIEW OF TITRIMETRY Type of Titrations based on Chemical Reactions: Acid-Base Titrations: H+ + OH– → H2O K= 1/Kw Precipitation Titrations: Ag+(aq) + Cl–(aq) → AgCl(s) K=1/Ksp Redox Titrations: 5 H2O2 + 2 MnO4– + H+ → 5 O2 + 2 Mn2+ + 8H2O Complexometric Titrations: EDTA + Ca2+ → (Ca–EDTA)2+ bblee@unimap

OVERVIEW OF TITRIMETRY 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. bblee@unimap

OVERVIEW OF TITRIMETRY Type of Titration Curves: bblee@unimap

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 & solution - High MW or FW bblee@unimap

GRADES OF CHEMICALS 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. bblee@unimap

GRADES OF CHEMICALS 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. bblee@unimap

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. bblee@unimap

TITRATION CALCULATIONS (i) Standardization 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) bblee@unimap

TITRATION CALCULATIONS (ii) Unknown Analysis with a Blank Correction 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) bblee@unimap

TITRATION CALCULATIONS Solution: MnO4–(aq) + 5Fe2+ + 8H+(aq) → Mn2+(aq) + 5Fe3+ + 4H2O(l) (iii) Back Titration 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): bblee@unimap

TITRATION CALCULATIONS (iii) Back Titration 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. bblee@unimap

TITRATION CALCULATIONS bblee@unimap

TITRATION CALCULATIONS (iv) Kjeldahl Analysis for Total Nitrogen (TN) (a) KD description: bblee@unimap

TITRATION CALCULATIONS Example 1: 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. bblee@unimap

TITRATION CALCULATIONS Example 2: A 0.7121g sample of wheat flour was analyzed by the Kjeldahl method. The ammonia formed by addition of concentrated base after digestion with H2SO4 was distilled into 25.00 ml of 0.04977 M HCl. The excess HCl was then back-titrated with 3.97 mL of 0.04012 M NaOH. Calculate the percent protein in the flour using the 5.70 factor for cereal. bblee@unimap

TITRATION CALCULATIONS (v) Titration of a Mixture 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. bblee@unimap

TITRATION CALCULATIONS bblee@unimap

PRECIPITATION TITRATIONS A titration in which the reaction between the analyte and titrant involves a precipitation. (I) Titration curve: 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 bblee@unimap

PRECIPITATION TITRATIONS EXAMPLE 3: 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 (a) 0 mL (b) 10.0 mL (c) 25.0 mL (d) 35.0 mL bblee@unimap

PRECIPITATION TITRATIONS (ii) Construct a titration curve: Example: Titration of 50.0 mL of 0.0500 M Cl– with 0.100 M Ag+ pCl pAg bblee@unimap

PRECIPITATION TITRATIONS dy/dx d2y/dx2 (iii) End point determination bblee@unimap

PRECIPITATION TITRATIONS (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+ bblee@unimap

PRECIPITATION TITRATIONS EXAMPLE 6 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–?). bblee@unimap

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 bblee@unimap

ARGENTOMETRIC TITRATION 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–) bblee@unimap

ARGENTOMETRIC TITRATION Fajans Method: An adsorbed/colored indicator. Titrating Cl– and adding dichlorofluoroscein: bblee@unimap

ARGENTOMETRIC TITRATION Mohr Method: A colored precipitate formed by Ag+ with anion, other than analyte, once the Ve reached. Analysing Cl– and adding CrO42–: 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 bblee@unimap

EXAMPLE 1 bblee@unimap

EXAMPLE 2 Amount HCl = = 1.2443 mmol Amount NaOH = = 0.1593 mmol Amount N =Amount HCl – Amount NaOH = (1.2443 – 0.1593) mmol = 1.0850 mmol bblee@unimap

EXAMPLE 2 % N = = 2.1342 % protein = = 12.16 bblee@unimap

EXAMPLE 3 (a) 0 mL Ag+ added (At beginning) [Ag+] = 0, pAg can not be calculated. [Cl–] = 0.0500, pCl = 1.30 bblee@unimap

EXAMPLE 3 (b) 10 mL Ag+ added (Before Ve) bblee@unimap

EXAMPLE 3 (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 bblee@unimap

EXAMPLE 3 (d) 35 mL Ag+ added (At Ve) bblee@unimap

EXAMPLE 4 (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 bblee@unimap

EXAMPLE 4 bblee@unimap