Titration Curve of a Weak Acid
Amina Khalifa El-Ashmawy, Ph.D.
Collin College
Department of Chemistry
Introduction:
Titration is an analytical process whereby two reactant solutions are carefully reacted in
controlled environment. The reactant that is under study is called the analyte while the reactant
added to the analyte is called the titrant. Titrations are usually carried out for acid-base
neutralization. These can be conducted in two ways: 1) with an indicator, and 2) with a pH
meter or probe. The equivalence point is when the reaction solution contains amounts of acid and
base consistent with the stoichiometric ratio. The indicator, if carefully chosen, changes color
about the same time as the equivalence point is reached. The color change of the indicator is the
end point. You have already experimented with indicator titrations.
Titration Curves:
When a pH meter is used to monitor the reaction, the experimenter records the pH of the solution
as a function of total volume of titrant added. The volume of titrant is the independent variable,
the one that is controlled. The pH is the dependent variable as it is the one measured as the
outcome of manipulating the independent variable. Once the volume of titrant – pH data is
plotted, the resulting graph is called a titration curve.
The plot generally makes an S-shaped or inverted S-shaped curve depending on whether the
titrant is an acid or base. If the analyte is an acid and titrant is base, the titration curve will be an
S-shaped curve (figure 1). If the analyte is a base and titrant is acid, the titration curve will be an
inverted S-shaped curve (figure 2).
Figure 1. Titration curve of a acid with base
Copyright© 2011 Amina Khalifa El-Ashmawy
Figure 2. Titration curve of a base with acid
14
12
Ka = 1.0X10-10
10
Ka = 1.0X10-8
Series1
8
Series2
pH
Ka = 1.0X10-6
Series3
6
Series4
Ka = 1.0X10-4
Series5
Series6
4
Ka = 1.0X10-2
2
Strong Acid
0
0
10
20
30
40
50
60
70
Volumeofof1.00
1.00
NaOH
Volume
MM
of NaOH
Figure 3. Titration curves of different monoprotic weak acids with strong base.
The slope of the flat regions on the left and right tells you something about the acid or base. The
flatter the region the stronger the substance (as in the figure above); the more incline there is in a
region the weaker the substance. Once a little bit of titrant, whether it is a strong acid or strong
base, has been added to the weak substance, the conjugate species is generated, which results in a
buffer solution. The point of inflection in the curve is the equivalence point. The pH at the
equivalence point depends on the chemical species that remain in solution. For example, in the
titration of HCl (strong acid) with NaOH (strong base), the species in solution at the equivalence
point are Na+ and Cl- in H2O. Let’s explore what might happen if you think of H2O as H+OH-.
Copyright© 2011 Amina Khalifa El-Ashmawy
The reaction of ions with water is called hydrolysis. The Na+ might combine with OH- from the
water while Cl- might combine with H+ to give:
Na+(aq) + H2O(l)  NaOH(aq) + H+(aq)
and
Cl-(aq) + H2O(l)  HCl(aq) + OH-(aq).
Looking closely at what we are suggesting for the products, assuming the reactions occur, we see
that we are forming NaOH (a strong base) and HCl (a strong acid). By definition, strong species
completely ionize or dissociate in solution and will not come together as written in the above
equations. Hence, there is no increase in either the H+ or OH- in solution. The solution pH at the
equivalence point will be 7.000.
Na+(aq) + H2O(l)  No Reaction
and
Cl-(aq) + H2O(l)  No Reaction.
Now, let’s take a look at what happens at the equivalence point of a weak acid with strong base.
In the case of titrating formic acid (HCOOH) with NaOH, the pH at the equivalence point will
depend on the ions present, which are Na+ and HCOO-. We can write equations for the
hydrolysis by these ions as we did before. The equations would be
Na+(aq) + H2O(l)  No Reaction
and
HCOO-(aq) + H2O(l)  HCOOH(aq) + OH-(aq).
In the second equation we are forming a weak acid, which will exist in its molecular form in
solution. This will result in equilibrium, and we see that there is increased OH- concentration in
solution. Hence, the pH at the equivalence point of a weak acid and a strong base will be higher
than 7.000 due to hydrolysis of the conjugate base.
- HCOO
HCOO
(aq)
(aq) ++HH
2O(l)
2O(l)
-HCOOH
HCOOH(aq)
OH(aq)
(aq).
(aq) + +OH
The pH of such a solution is dictated by the Kb of the conjugate base HCOO-. Its value depends
on the Ka of the weak acid HCOOH and can be calculated by
Kb = Kw/Ka.
Polyprotic Acids:
Polyprotic acids ionize one step at a time. For a weak diprotic acid, the equilibria reactions are:
HCOO
HCOOH
H2A(aq)
HA-(aq)(aq)
+ H+3O+OH
. (aq)
(1)
(aq) ++HH
2O(l)
2O(l)
Copyright© 2011 Amina Khalifa El-Ashmawy
HCOO
HCOOH
HA-(aq)
O(l)
A2-(aq) +(aq)H3+O+OH
. (aq)
(2)
(aq) ++H2H
2O(l)
The neutralization of such acids also occurs in steps where the first proton is neutralized, then the
second proton is neutralized.
H2A(aq) + NaOH(aq)  NaHA(aq) + H2O. (1)
NaHA(aq) + NaOH(aq)  Na2HA(aq) + H2O.
(2)
Each reaction will have its own, distinct equivalence point. The pH at each equivalence point
depends on the respective Ka value. The titration curve looks like two S-shaped curves connected
to each other.
Figure 4. Titration curve of a diprotic acid with strong base
Features of the Titration Curve:
The equivalence point is an important feature of a titration curve because it allows us to
determine the concentration of an unknown acid or base. At the equivalence point, the reactants
are in stoichiometric amounts such that the solution contains [H+] = [OH-] prior to any hydrolysis
that may occur. If you titrate a monoprotic acid of unknown concentration, you can determine its
concentration stoichiometrically from the amount of base needed to reach the equivalence point.
For example, you want to titrate a 20.00 mL aliquot (liquid sample to be analyzed) of unknown
monoprotic acid. If it takes 18.78 mL of 0.10 M KOH to reach the equivalence point in the
titration of a 20.00 mL aliquot of the unknown monoprotic acid, you can determine that
Mol KOH = (18.78 mL)(1 L/1000 mL)(0.10 mol/L) = 0.001878 mol KOH.
Note: we will not round any numbers in the calculations to avoid rounding errors.
Since the reaction is:
HA(aq) + KOH(aq)  KA(aq) + H2O(l),
there is a 1 to 1 molar ratio of KOH to HA. We then know that the 20.00 mL aliquot of acid
contained 0.001878 mol HA. Therefore, the concentration of the acid is
Copyright© 2011 Amina Khalifa El-Ashmawy
(0.001878 mol HA)/[(20.00 mL)(1 L/1000 mL)] = 0.094 M HA.
Another important feature is the point of half-neutralization. With the concentration of the
unknown weak acid determined, let’s calculate what is present in solution at half-neutralization.
You have:
(0.02000 L HA)(0.094 mol/L) = 0.001878 mol HA.
At half-neutralization, we know that moles KOH = ½ moles HA = 0.000939 moles. From the
moles of KOH and the known molarity of KOH, we can calculate the volume of KOH required
for half-neutralization.
(0.000939 mol KOH)/(0.10mol/L) = 0.00939 L
HA(aq)
+ KOH(aq)  KA(aq) + H2O(l)
Initial Moles: 0.001878
0.000939
Change:
-0.000939
-0.000939
Final Moles: 0.000939
0.00
0.00
0.00
+0.000939
+0.000939
0.000939
0.000939
From the reaction chart above, we see that KOH is the limiting reactant and will completely
react. There remains (0.001878 – 0.000939) mol HA, or 0.000939 mol HA. The other 0.000939
mol HA already reacted to produce KA, which contains the common ion A -. In solution at halfneutralization, then, you have
[HA] = (0.000939 mol)/(0.02939 L) = 0.0319 M and
[A-] = (0.000939 mol)/(0.02939 L) = 0.0319 M.
The [H3O+], hence the pH of the solution, can be determined by setting up the RICE (Reaction,
Initial concentration, Change, Equilibrium concentration) table and the Ka expression for HA.
HA +
0.0319
-X
0.0319 –X
H2 O
A- +
0.0319
+X
0.0319+X
H 3 O+
0
+X
X
For weak acids we can approximate [HA] ~ 0.0319 M and [A-] ~ 0.0319 M. Plugging into the Ka
expression we get
(X) (0.0319)
[H3O+] [A-]
Ka =
=
= X = [H3O+]
(0.0319)
[HA]
Copyright© 2011 Amina Khalifa El-Ashmawy
So, we see that at half-neutralization [H3O+] = Ka and pH = pKa.
The Problem:
You will be given an unknown weak polyprotic acid to titrate with 0.10 M NaOH using a pH
probe. From the titration curve you will determine the Ka values for each proton and the
unknown acid concentration. Possible unknowns are malonic, succinic, phosphoric, glutaric or
adipic acids.
Questions to answer prior to performing the lab:
Following are some questions to consider before starting the lab.
What is a diprotic acid? Triprotic acid?
Write the ionization equations for the protons of a triprotic acid.
Do you need to use an indicator when you are doing a pH titration?
What is the pH of 0.10 M HCl? 0.10 M CH3CO2H? 0.10 M H2SO4?
How many trials of each titration should you carry out in order to obtain dependable
results?
Critical Data/Discussion to Include in Your Lab Report:
Data table
Titration curve for each titration trial
Calculation of acid concentration
Calculation of both Ka values
Discussion of why the second Ka is smaller than the first Ka value (removing a proton
from an anionic species as opposed to removing a proton from an electrically neutral
species.)
Discussion of the possible identity of the unknown acid based on the Ka values calculated
Work cited
Copyright© 2011 Amina Khalifa El-Ashmawy