Presentation on theme: "Titrations: Taking Advantage of Stoichiometric Reactions"— Presentation transcript:
1Titrations: Taking Advantage of Stoichiometric Reactions Chapter 11Titrations: Taking Advantage of Stoichiometric Reactions
2Titrimetric methods include a large and powerful group of quantitative procedures based on measuring the amount of a reagent of known concentration that is consumed by the analyte. Titrimetry is a term which 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.
3There are three main types of titrimetry: volumetric titrimetry, gravimetric titrimetry, and coulometrtic titrimetry.Volumetric titrimetry is used to measure the volume of a solution of known concentration that is needed to react completely with the analyte.Gravimetric titrimetry is like volumetric titrimetry, but the mass is measured instead of the volume.Coulometric titrimetry is where the reagent is a constant direct electrical current of known magnitude that consumes the analyte; the time required to complete the electrochemical reaction is measured.The benefits of these methods are that they are rapid, accurate, convenient, and readily available.
4Defining Terms Standard Solution Titration Equivalence Point Back- Titration
5Defining termsStandard Solution: A reagent of a known concentration which is used in the titrimetric analysis.Titration: This is performed by adding a standard solution from aburet or other liquid- dispensing device to a solution of the analyteuntil the point at which the reaction is believed to be complete.
6More Defining TermsEquivalence Point: Occurs in a titration at the point in which theamount of added titrant is chemically equivalent to the amountof analyte in a sample.Back- Titration: This is a process that is sometimes necessary in which anexcess of the standard titrant is added, and the amount of the excess is determined by back titration with a second standard titrant. In this instance the equivalence point corresponds with the amount of initial titrant is chemically equivalent to the amount of analyte plus the amount of back- titrant.
7Equivalence Points and End Points One can only estimate the equivalence point by observing a physical change associated with the condition of equivalence.End point: The point in titration when a physical change occursthat is associated with the condition of chemical equivalence.Indicators are used to give an observable physical change (end point) at or near the equivalence point by adding them to the analyte. The difference between the end point and equivalence point should be very small and this difference is referred to as titration error. To determine the titration error: Et= Vep - VeqEt is the titration errorVep is the actual volume used to get to the end pointVeq is the theoretical value of reagent required to reach the end point
8Primary StandardsA primary standard is a highly purified compound that serves as a reference material in all volumetric and mass titrimetric properties. The accuracy depends on the properties of a compound and the important properties are:1. High purity2. Atmospheric stability3. Absence of hydrate water4. Readily available at a modest cost5. Reasonable solution in the titration medium6. Reasonably large molar massCompounds that meet or even approach these criteria are few, and only a few primary standards are available.
9Standard SolutionsStandard solutions play a key role in titrimetric methods.Desirable Properties of Standard Solutions:Sufficiently stableReact rapidly with analyteReact completely with analyteEndure a selective reaction with analyte
11Example: Calculating the Molarity of Standard Solutions Describe the preparation of a 5.0 L of 0.10 M Na2CO3 ( g/mol)from the primary standard solution.Amount Na2CO3 = n Na2CO3 (mol) = Volume solution x c Na2CO3 (mol/ L)= 5 L x 0.1 mol Na2CO3 = 0.5 mol Na2CO3LMass Na2CO3 = mNa2CO3=0.5 mol Na2CO3 x g Na2CO3 =53 g Na2CO3mol Na2CO3The solution is prepared by dissolving 53 g of Na2CO3 in water and diluting to 5 L.
12Example: Calculating the Molarity using different algebraic relationships How would you prepare 50mL portions of standard solutions that are0.005 M, M, and M in a standard 0.01 M Na+?To solve this the relationship Vconcd x cconcd = Vdil x cdilVconcd = Vdil x cdil = 50mL x mmol Na+ /mL = 25mLcconcd mmol Na+ /mLTo produce 50mL of M Na+, 25mL of the concentration solutionshould be diluted to 50mL.
13How to deal with titration data… The following two examples show the two types of volumetric calculations.The first involves computing the molarity of solutions that have beenstandardized against either a primary standard or another standard solution.The second example involves calculating the amount of analyte in a samplefrom titration data.
14Example: Molarity of solutions that have been standardized A 50mL volume of HCl solution required 29.71mL of M Ba(OH)2 to reach an end point with bromocresol green indicator. Calculate the molarityof the HCl.Stoichiometric ratio= 2 mmol HCl/ 1 mmol Ba(OH)2Amount Ba(OH)2 = mL Ba(OH)2 x mmol Ba(OH)2mL Ba(OH)2Amount HCl = (29.71 x ) mmol Ba(OH)2 x 2 mmol HCl1 mmol Ba(OH)2C HCl = (29.71 x x 2) mmol HCL50mL solution= mmol HCl = MmL solution
15Example: Amount of analyte in sample from titration Titration of g of pure Na2C2O4 (134 g/mol) required mL of KMnO4. What is the molarity of the KMnO4 solution?Stoichiometric ratio = 2 mmol KmnO4/ 5 mmol Na2C2O4Amount Na2C2O4= g Na2C2O4 x 1 mmol Na2C2O40.134 g na2C2O4Amount KMnO4 = mmol Na2C2O4 x 2 mmol KMnO4mmol Na2C2O4C KMnO4 = ( x 2) mmol KMnO4= M43.31 mL KMnO4
16Example: Computing analyte concentrations from titration data A g sample of an iron ore is dissolved in acid. The iron is then reduced to Fe2+ and titrated with 47.22mL of M KMnO4 solution. Calculate the results of this analysis in terms of percent Fe ( g/mol).Stoichiometric ratio = 5 mmol Fe2+/ 1 mmol KMnO4Amount KMnO4 = 47.22mL KMnO4 x mmol KMnO4mL KMnO4Amount Fe2+ = (47.22 x ) mmol KMnO4 x 5 mmol Fe2+1 mmol KMnO4Mass Fe2+ = (47.22 x x 5) mmol Fe2+ x g Fe 2+mmol Fe2+% Fe2+ = (47.22 x x 5 x ) g Fe 2+ x 100% = 36.77%g sample
17Titration CurvesExample of a sigmoidal titration curve once calculations of data have been computed.reference/plambeck/chem1/p01173.htm
18ReferencesSkoog, D., West, D., Holler, F.J., & Crouch, S. (2000). Analytical Chemistry: An Introduction. 7th ed. Thomson Learning, Inc: United States of America.reference/plambeck/chem1/p01173.htm