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CHAPTER 11

CHAPTER 11. Titrations: Taking Advantage of Stoichiometric Reactions. Terms Defined.

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CHAPTER 11

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  1. CHAPTER 11 Titrations: Taking Advantage of Stoichiometric Reactions

  2. Terms Defined • Titrimetry – includes a group of analytical methods based on determining the quantity of a reagent of known concentration that is required to react completely with the analyte. The reagent may be a standard solution of a chemical or electric current of known magnitude. • Volumetric Titrimetry – a type of titrimetry in which the standard reagent is measured volumetrically.

  3. Gravimetric Titrimetry – titrimetry in which the standard reagent is measured by mass instead of volume. • Standard Solution – a reagent of known concentration that is used in titrimetric analysis. A primary standard is an ultrapure compound that serves as a reference material in titrations.

  4. Equivalence point – point in a titration when the amount of added titrant is chemically equivalent to the amount of anlayte. • End point – point in a titration when a physical change occurs that is associated with the condition of chemical equivalence.

  5. Back Titration – a process in which the excess of a standard solution used to consume an analyte is determined by titration with a second standard solution. Back titrations are usually required when the rate of reaction between the analyte and reagent is slow or when the standard solution lacks stability.

  6. Equivalence and End Points • Experimentally, the equivalence point is difficult to observe. We near the equivalence point when we observe physical changes in the reaction with the aid of an indicator or an instrument such as a pH meter. (Titration methods are used throughout the lab sessions. Of particular interest are Lab 27C-7 Standardization of Sodium Hydroxide against Potassium Hydrogen Phalate and 27 J-1Potentiometric Titration)

  7. Titration Curves • Titration curves can be used to determine the end point of a titration • One such curve is the Sigmoidal curve. It is a useful tool to illustrate changes in concentration and pH of the analyte. (Fig 11-2 a)

  8. Primary Standard Requirements • High purity • Thermally stable • Anhydrous • Large molar mass • Fast stoichiometric reaction with analyte • 99.999% reaction completion • Available at moderate cost

  9. Concentration The concentration of a standard solution used in titration is vital to the accuracy of the titrimetric method. In order to obtain the proper concentration, either the direct method or standardization is used. • Direct method – carefully weighed quantity of primary standard is dissolved in solvent and diluted to known volume

  10. Standardization – the titrant to be standardized is used to titrate either a) a weighed quantity of primary standard b) a weighed quantity of a secondary standard – a compound whose purity has been established by chemical analysis and that serves as the reference material c) a measured volume of another standard solution.

  11. When discussing concentration, it is important to note that concentration is most often expressed in terms of molarity, M, which is moles/Liter or mmol/mL. Moles, molar mass and volumes can be determined by understanding certain relationships between molarity and the standards and analytes used as follows:

  12. Moles A = Volume A (L) X Concentration A (mol/L) Mass A = Moles A X Molar mass (g/mol)

  13. Examples p. 261 11-6 • 2.00 L X 2.76 x10-3 mol/L X 1000 mmol/mol = 5.52 mmol • 750 mL X 0.0416 mol/mL = 31.3 mmol • 4.20 g CuSO4/ 106 g soln X 1.00 g soln/mL soln X 1mmol/0.1596 g CuSO4 X 250 mL soln = 6.58 X 10-3 mmol • 3.50 L X 0.276 mol/L X 1000 mmol/mol = 966 mmol p. 261 11-8 • 26.0 mL X 0.150 mmol/mL X 0.342 g/mol X 1000 mg/g = 1.33 X 103 mg • 2.92 L X 5.23 x10-3 mol/L X 34.02 g/mol X 1000 mg/g = 520 mg

  14. Examples Page 262 11-22 Back titration VHClO4 = 27.43 mL HClO4 = 1.0972 mL HClO4 VNaOH 25.00 mL NaOH mL NaOH Volume of HClO4 needed to titrate 0.4793 g of Na2CO3 is: 40.00 mL HClO4 – 8.70 mLNaOH X 1.0972 mL HClO4 mL NaOH = 30.45 mL

  15. Thus: 0.4793 g Na2CO3/ 30.45 mL HClO4 X 1 mmol Na2CO3/0.10599 gNa2CO3X 2mmol HClO4/ mmol Na2CO3 = 0.2970 M HClO4 Page 263 11-27 MAg2O3 = 197.84 g/mol Amount Ag+ required for Ag3AsO4= total mmol Ag+ - # mmol KSCN 40.00mL X 0.07891 M – 11.27 mL X 0.1000M = 2.0294 mmol Ag+ 2.0294 mmol Ag+ X 1mmolAs/3mmol Ag X 1mmol Ag2O3/2mmol Ag X 0.19784 g Ag2O3/mmolAs2O3

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