Transcript Chapter 11 Titrations: Taking Advantage of Stoichiometric Reactions

Chapter 11
Titrations: Taking Advantage
of Stoichiometric Reactions
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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.
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There are three main types of titrimetry: volumetric titrimetry,
gravimetric titrimetry, and coulometrtic titrimetry.
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Volumetric titrimetry is used to measure the volume of a solution of
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Gravimetric titrimetry is like volumetric titrimetry, but the mass is
measured instead of the volume.
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Coulometric titrimetry is where the reagent is a constant direct electrical
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The benefits of these methods are that they are rapid, accurate,
convenient, and readily available.
known concentration that is needed to react completely with the
analyte.
current of known magnitude that consumes the analyte; the time
required to complete the electrochemical reaction is measured.
Defining Terms
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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.
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.
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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
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Vep is the actual volume used to get to the end point
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Veq is the theoretical value of reagent required to reach the end point
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Primary Standards
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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:
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1. High purity
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2. Atmospheric stability
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3. Absence of hydrate water
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4. Readily available at a modest cost
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5. Reasonable solution in the titration medium
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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:
1.
Sufficiently stable
2.
React rapidly with analyte
3.
React completely with analyte
4.
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.
Example: Calculating the Molarity using different algebraic
relationships
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
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Example of a sigmoidal titration curve once calculations of data
have been computed.
www.psigate.ac.uk/newsite/ reference/plambeck/chem1/p01173.htm
References
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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/Aque
ous/Stoich/Aqua02.htm
www.psigate.ac.uk/newsite/
reference/plambeck/chem1/p01173.htm
http://www2.hmc.edu/~karukstis/chem21f20
01/tutorials/tutorialStoichiFrame.html