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Chapter 11. Titrations: Taking Advantage of Stoichiometric Reactions.

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### Chapter 11

Titrations: Taking Advantage of Stoichiometric Reactions

Titrimetric 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.
There 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.

Defining Terms
• Standard Solution
• Titration
• Equivalence Point
• Back- Titration
Defining terms
• Standard Solution:A reagent of a known concentration which is used in the titrimetric analysis.
• Titration: This is performed by adding a standard solution from a
• buret or other liquid- dispensing device to a solution of the analyte
• until the point at which the reaction is believed to be complete.
More Defining Terms
• Equivalence Point: Occurs in a titration at the point in which the
• amount of added titrant is chemically equivalent to the amount
• of analyte in a sample.
• Back- Titration: This is a process that is sometimes necessary in which an
• excess 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.
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 - Veq

Et is the titration error

Vep is the actual volume used to get to the end point

Veq is the theoretical value of reagent required to reach the end point

Equivalence 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 occurs
• that is associated with the condition of chemical equivalence.
Primary Standards
• A 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 purity
• 2. Atmospheric stability
• 3. Absence of hydrate water
• 4. Readily available at a modest cost
• 5. Reasonable solution in the titration medium
• 6. Reasonably large molar mass

Compounds that meet or even approach these criteria are few, and only a few primary standards are available.

Standard Solutions

Standard solutions play a key role in titrimetric methods.

Desirable Properties of Standard Solutions:

• Sufficiently stable
• React rapidly with analyte
• React completely with analyte
• Endure a selective reaction with analyte
Example of titration and set up

http://wine1.sb.fsu.edu/chm1045/notes/Aqueous/Stoich/Aqua02.htm

Example: Calculating the Molarity of Standard Solutions

Describe the preparation of a 5.0 L of 0.10 M Na2CO3 (105.99 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 Na2CO3

L

Mass Na2CO3 = mNa2CO3=0.5 mol Na2CO3 x 105.99 g Na2CO3 =53 g Na2CO3

mol Na2CO3

The solution is prepared by dissolving 53 g of Na2CO3 in water and diluting to 5 L.

How would you prepare 50mL portions of standard solutions that are

0.005 M, 0.002 M, and 0.001 M in a standard 0.01 M Na+?

To solve this the relationship Vconcd x cconcd = Vdil x cdil

Vconcd = Vdil x cdil = 50mL x 0.005 mmol Na+ /mL = 25mL

cconcd 0.01mmol Na+ /mL

To produce 50mL of 0.005 M Na+, 25mL of the concentration solution

should be diluted to 50mL.

How 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 been

standardized against either a primary standard or another standard solution.

The second example involves calculating the amount of analyte in a sample

from titration data.

Example:Molarity of solutions that have been standardized

A 50mL volume of HCl solution required 29.71mL of 0.01963 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)2

Amount Ba(OH)2 = 29.71 mL Ba(OH)2 x 0.01963 mmol Ba(OH)2

mL Ba(OH)2

Amount HCl = (29.71 x 0.01963) mmol Ba(OH)2 x 2 mmol HCl

1 mmol Ba(OH)2

C HCl = (29.71 x 0.01963 x 2) mmol HCL

50mL solution

= 0.023328 mmol HCl = 0.0233M

mL solution

Example: Amount of analyte in sample from titration

Titration of 0.2121 g of pure Na2C2O4 (134 g/mol) required 43.31 mL of KMnO4. What is the molarity of the KMnO4 solution?

Stoichiometric ratio = 2 mmol KmnO4/ 5 mmol Na2C2O4

Amount Na2C2O4= 0.2121 g Na2C2O4 x 1 mmol Na2C2O4

0.134 g na2C2O4

Amount KMnO4 = 0.2121 mmol Na2C2O4 x 2 mmol KMnO4

0.134 5 mmol Na2C2O4

C KMnO4 = ( 0.2121 x 2) mmol KMnO4

0.134 5 = 0.01462M

43.31 mL KMnO4

Example:Computing analyte concentrations from titration data

A 0.8040g sample of an iron ore is dissolved in acid. The iron is then reduced to Fe2+ and titrated with 47.22mL of 0.02242 M KMnO4 solution. Calculate the results of this analysis in terms of percent Fe (55.847 g/mol).

Stoichiometric ratio = 5 mmol Fe2+/ 1 mmol KMnO4

Amount KMnO4 = 47.22mL KMnO4 x 0.02242 mmol KMnO4

mL KMnO4

Amount Fe2+ = (47.22 x 0.02242) mmol KMnO4 x 5 mmol Fe2+

1 mmol KMnO4

Mass Fe2+ = (47.22 x 0.02242 x 5) mmol Fe2+ x 0.055847 g Fe 2+

mmol Fe2+

% Fe2+ = (47.22 x 0.02242 x 5 x 0.055947) g Fe 2+ x 100% = 36.77%

0.8040 g sample

Titration Curves
• Example of a sigmoidal titration curve once calculations of data have been computed.

www.psigate.ac.uk/newsite/ reference/plambeck/chem1/p01173.htm

References
• Skoog, D., West, D., Holler, F.J., & Crouch, S. (2000). Analytical Chemistry: An Introduction. 7th ed. Thomson Learning, Inc: United States of America.
• http://wine1.sb.fsu.edu/chm1045/notes/Aqueous/Stoich/Aqua02.htm
• www.psigate.ac.uk/newsite/ reference/plambeck/chem1/p01173.htm
• http://www2.hmc.edu/~karukstis/chem21f2001/tutorials/tutorialStoichiFrame.html