Volumetric analysis
Chemistry 321, Summer 2014
Volumetric analysis involves titrations
A titration is the use of a known
concentration reagent to determine
the concentration of the analyte.
Typically, a buret is used to
dispense the volume of the
reagent.
Titration is a good technique
Titrations are simple to set up, cheap to run and
yield good results for even low concentration
analytes.
Kf
A (aq) +
analyte
T (aq)
titrant
P
product(s)
The trick is to choose the proper titrant (selective
reagent) – more on this later.
Complications:
Kf, AT
A (aq) +
analyte
These are side
rxns – minimize
by choosing a
good titrant
T (aq)
titrant
+
I (aq)
interferent
Kf, IT
P
product(s)
Make sure Kf,AT >> Kf, IT
Requirements for a good titration
1. A well-defined (stoichiometry known) chemical
reaction
2. A quantitative reaction – proceeds forward easily
(large Kf)
3. A rapid reaction – fast kinetics so measurements
can be made quickly after mixing analyte and
titrant
4. Little to no side reactions – minimize interference
of other reactions by choosing a selective titrant
Requirements for a good titration
5. Easy equivalence point observation
Even as the reaction is occurring between the
analyte and titrant, there must be an easily
measureable property of the solution to
confirm when the proper amount of titrant
has been added to consume the amount of
analyte in the sample aliquot.
NB: The equivalence point is often
called the endpoint of a titration.
Quantitative titration – Equivalence
Consider a simple acid-base titration; they all have
the same net reaction
Kf
H+ (aq)
acid
+ OH– (aq)
base
H2O (l)
neutral
Note that if the product and reactants were switched
around the equilibrium arrow, then the reaction
would be simply the self-ionization of water. That
equilibrium is described by the equilibrium constant
Kw = [H+] [OH–] = 1 × 10–14 at 25°C
Quantitative titration – Equivalence
So, Kf = 1/Kw = 1 × 1014 for a strong acid/base (you can
treat this as an exact number for now)
Recall, large Kf is a good thing for volumetric analysis.
The equivalence point of an acid/base titration is
defined as where the equivalents of base equals the
equivalents of acid. Abbreviated, it is where eqb = eqa.
(The above definition of equivalence point is true of any
titration, with the slight modification, eqanalyte = eqtitrant)
So what is an “equivalent”?
An equivalent (eq) is the effective reactive quantity of
a species in solution, and is calculated by multiplying
the number of moles of the species by the number of
reacting units per mole of the species.
More often, the normality of a solution (N), measured
in eq/L (equivalents per liter), is calculated for
solutions.
Calculating normalities
For 1.0 M H2SO4, assuming a base strong enough to
dissociate both H+ on the acid, the number of
reacting units per mole of sulfuric acid is 2 eq/mol
(exact number).
The normality of that solution is:
1.0 mol/L × 2 eq/mol = 2.0 eq/L
Practice problem: Barium hydroxide (Ba(OH)2) is a
strong diprotic base. Calculate the normality of a
0.50 M solution of barium hydroxide.
Performing a titration
Typically, you know the titrant
concentration, which has been
standardized (concentration
determined to a high degree of
precision and accuracy). With the
buret, you are able to add the
titrant in precisely measurable
volumes. Thus, the quantity you can
control throughout the experiment
is the volume of titrant.
Plotting a
titration
Recall the rules of good graphing – the
x-axis should be the independent
variable (the quantity you have control
over) and the y-axis (the dependent
variable) should be the quantity you
measure as a result of messing with
the independent variable.
Plotting a
titration
Recall that pH = - log [H+]
so it really is
concentration along the
y-axis
Why is this the endpoint?
For a titration, the volume of T added (mL) is a good
independent variable, and the concentration of P or
other species in the flask (derived by pH meter,
spectrophotometry or whatever technique) is a
good dependent variable.
Finding an endpoint on the titration graph
On simple acid/base systems, the endpoint may be estimated from
the inflection point of the graph, but this may prove to be inexact. In
fact, in multiple pKa systems, this may prove to be impossible.
On the next page is a set of graphs; the upper left is the standard
titration graph of a weak acid with a strong base. The upper right
graph shows the first derivative of the first graph (note the y-axis
units are now ΔpH/ΔV). The lower left graph shows the second
derivative of the first graph (Δ2pH/ΔV2).
The inflection point of the titration graph will be the maximum on
the first derivative graph, and a zero of the second derivative graph.
You have received an e-mail with Excel spreadsheets to help
calculate these derivatives, and plot them.
Different types of volumetric methods
1. Acid-base titrations
To monitor the titration, you
can monitor the pH of the
analyte/titrant mixture, or
use a visual indicator
Different types of volumetric methods
2. Complexometric titration
Monitor the titration using
spectrophotometry
(measure absorbance of
the complex as it forms)
Different types of volumetric methods
3. Precipitations
Monitor the titration by
measuring the mass of
precipitate or indirectly by
indicator adsorption
Different types of volumetric methods
4. Oxidation-reduction (redox) titrations
Monitor the titration using a
redox indicator or by
measuring the potential of
the analyte/reagent mixture
Challenge problem
A titration has the following general reaction:
Analyte + Titrant  Product
or
A+TP
Imagine you have a sensor that is monitoring the
progress of a titration, providing a signal that relates to
the chemical reactions occurring in the titration flask.
Additionally, the sensor responds to (i.e., is sensitive
to) the concentration of analyte (A) and titrant (T) with
signal that is linear with the concentration of each but
does not provide any signal for the product (P).
Challenge problem (cont’d)
Sketch a plot of the titration data one would obtain (in
general terms), with the sensor signal as a function of
titrant volume (neglecting the effect of dilution).
Clearly label both axes.
Clearly label the equivalence point.
Assume that the reaction equilibrium constant (Kf) is
large (e.g., ≥ 106)