Molly Baker
November 25, 2005
CHE 221- Analytical Methods & Techniques
Chapter 11 Titrations: Taking Advantage of Stoichiometric Reactions
To handle titrations it is necessary to know and understand some of the basics learned in
general chemistry. To handle data concerning titrations, stoichiometric relationships are
important and deal with chemical equations, balancing equations, and basic algebraic
relationships. First it is important to understand the relationships occurring with the
chemical equations. The most efficient way to describe a chemical reaction is by a
chemical equation, which are the reactants to the products. A proper equation also
indicates the phases of all species. A true equation has the left and right sides equal to one
another, and for a chemist equality is in terms of the conservation of mass - same number
of atoms or moles of each element on both sides of the equation. We also need to
consider conservation of charge when equations are written in terms of species other than
neutral atoms and molecules. A balanced equation is a quantitative relation in terms of
the number of moles (or molecules) of reactants and products. Other useful units of
measure include units of weight or volume.
In general, reactions are often carried out without the use of exact stoichiometric amounts
of reactants for practical reasons. One reasons being that cheaper, more abundant
reagents are used in greater amounts than are stoichiometrically necessary in order to
assure that the more expensive, less abundant reagent is consumed completely. The
limiting reagent, which is the reactant present in limited supply that controls the amount
of product formed in a reaction helps stoichiometric calculations are based on the total
consumption of the limiting reagent. For a reaction described as: a A + b B => products,
then there are three common methods of determining the limiting reagent: 1) actual ratio
< a / b => A is the limiting reactant, 2) actual ratio = a / b => "A and B are present in
the exact stoichiometric amount", and 3) actual ratio > a / b => B is the limiting
reactant.
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.
For one to understand titrimetry there are some key terms that must be known. Standard
solution is a reagent of known concentration that is used to carry out titrimetric analysis.
A titration is performed by adding a standard solution from a buret or other liquiddispensing device to a solution of the analyte until the reaction between the two is
complete. Equivalence point is the point at which the amount of added titrant is
chemically equivalent to the amount of analyte in the sample. Back titration 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.
One can only estimate the equivalence point by observing a physical change associated
with the condition of equivalence. This change is the end point for the titration. 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
Instruments are often used to detect the end point by responding to certain properties of
the solution that change. Examples of these instruments are colorimeters, turbidimeters,
and temperature monitors.
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
Few standards meet all the criteria and sometimes less pure compounds must be used.
The purity of this secondary standard must be established by careful analysis.
Standard solutions are key to titrimetric methods. There are properties that are desirable
such as:
Be sufficiently stable so as to only need to find concentration once
React rapidly with the analyte so that the time required between additions of reagents is
minimized
React completely so end points are realized
Undergo a selective reaction with the analyte that can be a described by a balanced
equation.
However, few reagents meet all these requirements.
There are two basic methods to establish the concentration, direct method and
standardization. With direct method the quantity of the primary standard is carefully
weighed and dissolved in a suitable solvent and diluted to a known volume in a
volumetric flask. Standardization involves the titrant that is to be standardized is used to
titrate. It can be a weighed quantity of a primary standard, a weighed quantity of a
secondary standard, or a measured volume of another standard solution. The latter two
points are subject to larger uncertainty than when using a primary standard.
The concentration of a standard solution is usually expressed in units of molarity, c, or
normality, CN. Molarity gives the number of reagent in 1 liter of solution and normality
gives the number of equivalents of reagent in the same volume.
Some useful algebraic relationships if have chemical species A:
Amount A (mol)= mass A (in grams, g)
molar mass A (in g/ mol)
Amount A (mmol)= mass A (g)
mmolar mass A (g/ mmol)
Amount A (mol) = volume (L) x concentration A (mol/ L)
Amount A (mmol) = volume (mL) x concentration A (mmol/ mL)
Calculating the molarity of a standard solution can help one to determine how to prepare
a solution.
There are two main types of calculations that will be necessary when treating titration
data. 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. Both types are
based on algebraic relationships.
The most widely used end points are changes in color due to the reagent, the analyte, or
the indicator, and a change in potential of an electrode that responds to the concentration
of the reagent or analyte. To help understand the detection of end points, a titration curve
can be constructed. Titration curves are a plot of the reagent volume on the horizontal
axis and some function of the analyte or reagent concentration on the vertical axis.
The first type of titration curve is called a sigmoidal curve and important observations are
confined to a small region. The p- function of the analyte (or reagent) is plotted as a
function of reagent volume. This type of curve is advantageous because of its speed and
convenience. The second type of titration curve is the linear segment curve in which
measurements are made on both sides of, but well away from, the equivalence point. The
vertical axis is an instrument reading that is directly proportional to the concentration of
the analyte or the reagent. This type of curve is advantageous for reactions that are
complete only in the presence of a considerable excess of the reagent or analyte.
The equivalence point in a titration is characterized by major changes in the relative
concentrations of the reagent and analyte. The changes in relative concentration are due
to the decrease of H3O+, which results in a tenfold increase in OH- concentration.
Therefore the end point detection is based on the large change in the relative
concentration. The large relative concentration changes that occur in the region of
chemical equivalence are shown by plotting the negative logarithm of the analyte or the
reagent concentration against the reagent volume. Titration curves define the properties
required of an indicator and allow us to estimate the error associated with titration
methods.
The titration curve for an acid-base titration is a plot of the solution pH, normally on the
vertical axis, against the volume of titrant added. To truly understand titration curves and
how they are established, an example will follow with detailed explanation..
The example wants to calculate the pH values along the course of a particular titration,
the titration of exactly 100 mL of 0.100 molar acetic acid with 0.100 molar sodium
hydroxide. In the course of the calculations and discussion, other aspects such as
indicator to use, which titrant to use, and other aspects of a titration procedure will be
addressed.
First, consider the titrant solution which is acetic acid and is a weak acid, so a base is
required to obtain any acid-base reaction. To get a pH break which is as large as possible
at the equivalence point we should choose a strong base. The common strong bases are
NaOH and KOH, so one of these will work well. It is normal to choose a concentration of
base that will give us a reasonable volume to add, so that the volume can be accurately
measured and yet not be too large. Here 0.100 molar NaOH is chosen and will therefore
require about 100 mL of it, since the stoichiometric reaction is:
CH3COOH + Na+ + OH- --> CH3COO- + Na+ + H2O.
Once the balanced equation and stoichiometric relationships are established, the next step
is to deal with the pH of the starting solution.
At the start of the titration the solution contains only the weak acid CH3COOH, for which
Ka = 1.75 x 10-5 = [H3O+][CH3COO-]/[CH3COOH]; Ka = [H3O+]2/[CH3COOH]
We know that [H3O+] is approximately equal to [CH3COO-], from the stoichiometric
dissociation of the weak acetic acid and the fact that it is the major source of hydrated
protons in the solution. Solving:
[H3O+]2 = 1.75 x 10-5 x 0.1, [H3O+] = 1.32 x 10-3, pH = 2.88
Next, one must consider before the equivalence point. The pH slowly rises as the NaOH
added reacts with the acetic acid. The acetic acid has been one-quarter titrated when 25
mL of the NaOH solution have been added. At this point 1/4 of the original moles of
CH3COOH have been titrated to CH3COO-, while 3/4 of the original moles of CH3COOH
remain, so 3[CH3COO-] = [CH3COOH] and
Ka = 1.75 x 10-5 = [H3O+][CH3COO-]/[CH3COOH] = [H3O+]/3
[H3O+] = 3 x 1.75 x 10-5 = 5.25 x 10-5, pH = 4.28
Now consider the situation when half of the acetic acid has been titrated. On the volume
axis, that would be at 1/2 x 100 mL = 50 mL of added NaOH. In the solution at that
point, half of the original acetic acid has been titrated and half has not, so mol CH3COOH
= mol CH3COO- and, since both are present in the same solution, [CH3COOH] =
[CH3COO-]. As a consequence, Ka = 1.75 x 10-5 = [H3O+][CH3COO-]/[CH3COOH] and
[H3O+] = 1.75 x 10-5, pH = 4.76.
The acetic acid is three-quarters titrated when 75 mL of the NaOH solution have been
added. At this point 3/4 of the original moles of CH3COOH have been titrated to
CH3COO-, while 1/4 of the original moles of CH3COOH remain, so [CH3COO-] =
3[CH3COOH] and
Ka = 1.75 x 10-5 = [H3O+][CH3COO-]/[CH3COOH] = 3[H3O+]/1
[H3O+] = (1/3)(1.75 x 10-5) = 5.83 x 10-6, pH = 5.23
Therefore, at the equivalence point, this is reached after 100 mL of the NaOH solution
have been added, just enough to react with all of the acetic acid present. Since the
reaction is that of a weak acid plus a strong base to yield a weak base and water, the
solution at the equivalence point is simply a solution of the weak base CH3COONa,
sodium acetate. As stated previously the equivalence point no excess of either weak acid
or strong base can be present. The situation is therefore just as it would be for a solution
of trhe weak base sodium acetate in water. Since there were originally 100 mL of 0.1
molar CH3COOH, or 10 mmol CH3COOH, there are now 10 mmol of CH3COONa.
These are contained in 200 mL of solution because we started with 100 mL and added
another 100 mL, so the formal concentration of acetate is 10 mmol/200 mL = 0.05 molar.
The equilibrium constants are Ka = 1.75 x 10-5 = [H3O+][CH3COO-]/[CH3COOH] and Kb
= 5.77 x 10-10 = [OH-][CH3COOH]/[CH3COO-]. It can no longer be assumed that [H3O+]
is approximately equal to [CH3COO-], because the solution is now basic, containing the
weak base acetate ion. The major source of the [H3O+] is not the dissociation of
CH3COOH since there is virtually no CH3COOH left to dissociate. However, it can now
be assumed that [OH-] is approximately equal to [CH3COOH], because the major source
of hydroxide ion is the weak base CH3COO- which hydrolyzes, giving CH3COOH and
OH- in a stoichiometric 1:1 ratio, following the reaction equilibrium CH3COO- + H2O <-> CH3COOH + OH- . To a good approximation, Kb = [OH-]2/[CH3COO-]. Since
[CH3COO-] is about 0.05 molar, [OH-]2 = Kb[CH3COO-], [OH-] = 5.37 x 10-6
pOH = -log(5.37 x 10-6) = 5.27, pH = 14.00 - 5.27 = 8.73
The equivalence point pH having been determined as 8.73, an indicator can now be
chosen for this titration. Those indicators given in a table of acid-base indicators for
which pKa is approximately equal to the pH at the equivalence point, 8.73, are thymol
blue, whose pKa is 8.9, and cresol purple, whose pKa is 8.3. Either would be satisfactory
for this titration. For thymol blue, the color change would be from the yellow color of the
acid form to the blue color of the base form.
It is important to take notice that the pH slowly rises throughout that part of the titration
curve prior to the equivalence point. If an indicator such as bromocresol green (pKa = 4.7)
had been chosen it would have changed from its acid color, yellow, to green well before a
chemist was 3/4 of the way to the equivalence point. The titration results would be
meaningless, because the endpoint at which color change occurs would not be easily
relatable to the equivalence point.
Beyond the Equivalence Point
As NaOH addition continues beyond the equivalence point, the pH continues to rise.
When twice as much NaOH has been added as is required, in other words when 200 mL
of the NaOH solution have been added, the resulting solution is then a mixture of the
strong base NaOH and the weak base CH3COONa. In any mixture of a strong base and a
weak base of comparable concentration, the strong base controls the pH, so that [OH-] is
approximately equal to the excess concentration of added NaOH. There were added 200
mL x 0.1 molar = 20 mmol NaOH, but of this 10 mmol was used to react with the 10
mmol of acetic acid originally present. As a consequence, there are now 20 - 10 = 10
mmol NaOH present in 100 + 200 = 300 mL solution. Then [OH-] = 10 mmol/300 mL =
0.033 molar,
[OH-] = 3.3 x 10-2, pOH = 1.48, pH = 14.00 - 1.48 = 12.52
An example of what a titration curve would like, is shown in the following figure.
www.psigate.ac.uk/newsite/ reference/plambeck/chem1/p01173.htm
The titration curve of an acid with a base will vary not only as the amount of the acid
present varies, but also with the strength of the acid, as the Figure below shows. The
plateau region of the titration curve shifts to higher pH values, and the magnitude of the
pH break at the equivalence point therefore decreases, as the acid becomes weaker.
www.psigate.ac.uk/newsite/ reference/plambeck/chem1/p01173.htm
The following example is the calculations for the titration of a weak base with a strong
acid. A total of 50 mL of aqueous ammonia, NH3, solution which is approximately 0.01
molar is to be titrated. Let us consider what it should be titrated with, what indicator
should be used, and what color change would then be seen.
To titrate this weak base, a strong acid could be used, such as HCl, with a convenient
concentration of about 0.01 molar. For ammonium ion, Ka = 5.60 x 10-10 =
[H3O+][NH3]/[NH4+].
At the equivalence point, the solution is essentially a solution of NH4Cl. On hydrolysis
one ammonium ion yields one hydrogen ion and one ammonia; NH4+ --> NH3 + H3O+.
To a good approximation, [H3O+] = [NH3], so
Ka = [H3O+]2/[NH4+]; [H3O+] = Ka[NH4+]. There were 0.01 mmol/L x 50 mL = 0.5
mmol NH4+; this is in 50 + 50 = 100 mL solution so [NH4+] = 0.5 mmol/100 mL = 0.005
mmol/L NH4+.
[H3O+]2 = (5.60 x 10-10)(5 x 10-3) = 2.80 x 10-12
[H3O+] = 1.67 x 10-6, pH = 5.78
Since the pH should be approximately equal to pKa of the indicator, a reasonable choice
would be either chlorphenol red or methyl red. The change is from base color to acid
color, which is from red to yellow for chlorphenol red and yellow to red for methyl red.
This chapter deals with some of the basics in titrations that have been shown in many of
the labs performed this semester in class. The equations involving the titrant and the
analyte, algebraic relationships, and how to treat the data are important when dealing
with titrations. By understanding the information in this chapter are some of the
fundamentals for further chapters and more complex titrations that are encountered.
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