chapter03-Fundamentals-of-Titrimetric-Analysis分析化學(xué)課件

1、chapter03-Fundamentals-of-Titrimetric-Analysis分析化學(xué)課件chapter03-Fundamentals-of-Titr3.1 General Principles3.1.1 Some General Aspect In a titration, the test substance (analyte) reacts with a reagent added as a solution of known concentration-refered to as a standard solution, and it is generally added

2、 from a buret. The added solution is called titrant. The volume of the titrant required just to completely react with the analyte is measured. From the required volume and the concentration of the standard solution, the amount of analyte can be calculated. The analytical techniques of this kind are

3、called titrimetric analysis or volumetric analysis.moving2022/9/213.1 General Principles3.1.1 1. Standard solution A reagent added as a solution of known concentration-refered to as a standard solution.2. Stoichiometric point (equivalence point) The point at which the titrant added just completely r

4、eacts with the analyte.3. End point The point at which the color of the indicator changes and the titration is stopped.2022/9/211. Standard solution A reag4. Indicator The material used to indicate the end point. 5. Titration error The difference between the end point and the equivalence point.movin

5、g2022/9/214. Indicator The material u3.1.2 Classification of Titration Methods1. Acid-base titrations These methods are based on the proton transferring reactions. Both acids and bases can be titrated. The involved reactions may essentially be expressed as: H+A- = HA where A- is a Brosted base. The

6、titrant used are generally standard solutions of strong electrolytes, such as hydrochloric acid and sodium hydroxide.moving2022/9/213.1.2 Classification of Titr2. Precipitation titrations Some of the precipitation reactions can be used in titrimetric procedures, such as Ag+ + X - = AgX where X- can

7、be chloride, bromide, iodide or thiocyanide( - SNC).2022/9/212. Precipitation titrations 3. Complexometric titrations The commonly used complexing agent in complexometric titrations is sodium salt of ethylenediaminetetraacetic acid (EDTA), which reacts with a number of metal ions, forming stable com

8、plexes: M2+ + Y4- = MY2- where M2+ is the bivalent metal ions, and Y4- represents the EDTA anion.2022/9/213. Complexometric titrations 4. Oxidation-reduction(redox) titrations Chemical reactions involving oxidation-reduction are widely used in titrimetric analysis. For example, iron, in the +2 oxida

9、tion state, can be titrated with standard KMnO4 solution under acidic condition.2 MnO4- + 5 C2O42- + 16 H+ = 2 Mn2+ + 10 CO2 + 8 H2O2022/9/214. Oxidation-reduction(redox) 3.1.3 Requirements for Reactions Used in Titrimetric Analysis1. The reaction must proceed according to a definite chemical equati

10、on without side reaction, which means that the reaction follow the definite stoichiometric relationship.2. The reaction must proceed to virtual completion at the equivalence point. The equilibrium constant of the reaction should be very large（KMY, high; KSp low) . The above two will provide a basis

11、for stoichiometric calculation.3. Some means must be available for determining when the equivalence point is reached, such as an indicator or some instrumental methods to tell the end point.4. The reaction should be rapid.2022/9/213.1.3 Requirements for Reacti3.2 Standard SolutionsA standard solutio

12、n is a solution with known concentration; it can sometimes be prepared by dissolving an accurately weighed quantity of a highly pure material called primary standard and diluting to an accurately known volume in a volumetric flask.2022/9/213.2 Standard SolutionsA standA primary standard should fulfi

13、ll: (1) It should be sufficiently pure to meet the demand for accuracy in a quantitative procedure. In general, the total amount of impurities should not exceed 0.01 to 0.02%. (2) It should have a definite composition, just as shown in its formula. (3) It should be stable. The reagent should not be

14、so hygroscopic that it takes up water during weighing and should not react with O2 or CO2. (4) High molar mass in order to minimize the consequence of errors in weighing.2022/9/21A primary standard should fulfOnly a few chemical reagents can meet these requirements like K2Cr2O7; Na2C2O4; H2C2O4 2H2O

15、 Na2B4O7 10H2O; CaCO3; NaCl; Na2CO3 .More commonly, a solution with approximately desired concentration is prepared, then it is standardized by titrating a weighed portion of a primary standard. For example, in acid-base titrations, solutions of hydrochloric acid and sodium hydroxide are so prepared

16、, and it is customary to standardize one of them against a secondary standard to obtain the concentration of the other solution.2022/9/21Only a few chemical reagents c3.3 Ways of Expressing Concentration of Standard Solution3.3.1Concentration of Amount Substance the amount of substance in unit volum

17、e Molarity-a kind of concentration expression. If nB moles of substance B are contained in V liters of solution, the concentration of substance B is given by cB = nB/V, The commonly used unit is mol/L. The number of elemental unit of substance B contained in a one mole system equals to the atom numb

18、er contained in 0.012 kg of 12C. nB is proportional to the number of elemental unit of substance B contained in the system (NB). That is, nB NB.2022/9/213.3 Ways of Expressing Concen When mole is used to express the amount of substance, the elemental unit must be pointed out. The elemental unit can

19、be atom, molecule, ion, other particles or specific combinations. For instance: The amount of 98.08 g of H2SO4 = 1 mol. The relationship between the amount of substance (n) and the mass (m) may be expressed as n = m / M , where M is the molar mass. Then the amount of substance for solute can be calc

20、ulated from its mass, and the concentration of a solution can further be obtained.2022/9/21 When mole is used to expreExample A H2SO4 liquid, density is 1.84 g/mL, content of H2SO4 is 95.0%. Calculate the molarity of the H2SO4.Solution: n of H2SO4 per liter: n = m / M = 1.841000 0.950 / 98.08 =17.9

21、mol c = n / V = 17.9 mol/L2022/9/21Example Solution: 2022/9/213.3.2 Titer1. Usually expressed as Ta/b such as TFe / KMnO4, the unit is g/mL.2. TFe/KMnO4 = 0.005682 g/mL implies that 1 mL of the standard KMnO4 solution is equivalent to 0.005682 g of iron. That is, 1 mL of this solution can oxidize 0.

22、005682 g of Fe2+ to Fe3+.3. It is convenient to obtain the mass of the analyte from titer according to the following relation: m = TV, where m is the mass of the analyte, and V is the volume of the standard solution required in titration.Titerthe mass of analyte that is chemically equivalent to 1 mL

23、 of the titrant.2022/9/213.3.2 Titer1. Usually express4. c is the molar concentration of the titrant, mol/L; a is the coefficient of analyte, b is the coefficient of the titrant, M is the molar mass of the analyte, g/mol.2022/9/214. c is the molar concentratio3.4 Calculations in Titrimetric Analysis

24、3.4.1 Stoichiometric Relationships in Titrimetry1. Direct titration(one reaction) reaction formula between titrant and analyte： a A + b B = c C + d D if it is at the equivalence point, a moles of A just completely react with b moles of B, where A is the analyte and B is the titrant：2022/9/213.4 Calc

25、ulations in TitrimetrRelationships in titrimetry nA / nB = a / b ， nA = (a / b) nB (c V )A = (a / b ) (c V )Bor： (c V )A = (a / b)(W / M)B In acid-base titrations，the concentration of standard solutions are around 0.1 mol/L , the volumes of titrant used are usually 2030mL, thus the mass of the analy

26、te needed can be estimated。2022/9/21Relationships in titrimetry Example A 0.3000g portion of pure oxalic acid(H2C2O4 2H2O) can be neutralized by 22.59 mL of KOH solution.(a) Calculate the concentration of KOH solution; (b) Suppose that 20.0 mL of H2SO4 solution of titrated with the KOH solution requ

27、iring 25.34 mL, Calculate the concentration of H2SO4 solution. 2022/9/21Example A 0.3000g portion ofSolution:The reaction in titration is : H2C2O4 + 2OH - = C2O42- + 2H2OThus: n(KOH) = 2 n(H2C2O4 2H2O) M(H2C2O4 2H2O) = 126.1 g/mol n(H2C2O4 2H2O) = m(H2C2O4 2H2O) / M(H2C2O4 2H2O) = 0.3000/126.1 = 2.3

28、79 10-3 molThen: n (KOH) = 2n(H2C2O4 2H2O) = 4.758 10-3 mol c (KOH) = n / V = 0.2106 mol/L2022/9/21Solution:The reaction in titra2. Diluting a solution(1) In diluting a solution, the amount of the solute keeps constant, the volume and concentration changes. (2) So: c1V1 = c2V23. Indirect titration（t

29、wo or more reactions） The stoichiometric relationship between the reactions should be found out from the overall reactions.2022/9/212. Diluting a solution(1) In d ExampleIn standardization of Na2S2O3 solution by using KBrO3 as a primary standard under acidic condition(two steps) .（can not be direct

30、titrated for complex products)(1) KBrO3 reacts with an excess KI to release I2 in a acidic solution: BrO3- + 6 I - + 6H+ = 3 I2 + Br- + 3H2O n(BrO3- ) = 1 / 3 n(I2)(2) The released I2 is titrated with Na2S2O3 solution 2 S2O32- + I2 = 2 I - + S4O62- n(I2 ) = 1/2 n(S2O32-)(3) the relationship between n of BrO3- and S2O32- n(BrO3 - ) = 1/3 n(I2 ) = 1/6 n(S2O32-) (cV)S2O32- = 6(W/M)BrO3-2022/9/21 ExampleIn standardization of ExampleDetermination of Ca2+ by KMnO4 (several steps)