Page 1
Acid–Base Titrations and Titration Curves
In this section of acid-base reactions, we will study the titrations in a little more detail,
For the addition of 15 – 50 mL NaOH solution, the calculations are again done in an
particularly focusing on how the pH changes during the experiment. The strategy to gain more
identical way. At some point there will only be OH- in the solution so the pH is calculated
insight to these correlations, is building a plot of the pH change vs the volume of the titrant
using the pH + pOH = 14 correlation. Once the data is obtained, we can plot these values to
added.
make a graph:
Strong Acid–Strong Base Titrations
Let’s consider adding a 0.10 M NaOH solution to a 25.0 mL of a 0.10 M HCl solution:
Before the addition, there is only HCl in the solution, therefore, the pH is calculated simply by
taking a negative log of [H+] which is equal to [HCl]:
H+(aq) + Cl–(aq),
HCl(aq)
pH = -log [H+] = -log 0.10 = 1.0
pH after Adding 5.00 mL NaOH
Before determining the pH after the addition of the base, let’s see how many moles of HCl we
have initially in the solution:
Next, we need to calculate the moles of NaOH in the 5.00 mL (0.00500 L) solution:
A few important observations on strong acid-base titrations from the graph:
1) The pH changes slowly in the first half of the titration.
According to the balanced equation, this is how many moles of HCl will react too because of the
2) The equivalence point is at pH 7 because it is reached when the number of moles of
1:1 mole ratio:
base added equals the number of moles of acid initially in the solution. At this point, there
NaOH(aq) + HCl(aq)
NaCl(aq) + H2O(l)
is no excess of H+ ions left in the solution as only contains NaCl which, as a salt of strong
acid and a strong base, has no appreciable effect on pH.
To find the remaining amount of HCl in the solution, we subtract the reacted moles from its
initial moles. Therefore, there will be 0.00250 – 0.000500 mol = 0.00200 mol HCl.
3) Near the equivalence point, the concentration of the H+ and OH– is almost equal and that
is why the addition of even a single drop of titrant cause an abrupt change in the pH.
To find the pH after the addition of 5.00 mL NaOH, we convert the 0.00200 moles to mol/L.
4) After the equivalence point, the predominant species in the solution are the OH– ions
Make sure to include the volume of the NaOH solution:
because of the increasing excess of the added NaOH solution.
V (total) = 25.0 mL + 5.00 mL = 30.0 mL = 0.0300 L
You may wonder; what is the base reacting with after the equivalence point if all the acid
has been neutralized? Remember that in pure water there is always some H+ and OH– ions
and even when the OH– ions start predominating, there is always traces of H+ ions.
Check the corresponding topics at https://general.chemistrysteps.com/general-chemistry-
+
pH = -log [H ] = -log 0.0667 = 1.17
topic-index/for more details on the calculations
gevorg@chemistrysteps.com
Copyright © 2022 Chemistry Steps
Acid–Base Titrations and Titration Curves
Titration of Strong Base with Strong Acid
Page 2
Therefore, the pH is calculated using the Henderson–Hasselbalch Equation like we did it
-
Conceptually, it is the same process except initially there is an excess of OH ions which
for buffer solutions. For example, after the addition of 5.00 mL of NaOH solution, we
switches to the excess of H+ ions. Therefore, the graph is going to be like a mirror image of the
need to determine the concentration of the acid and its conjugate base. The initial moles of
one for titrating a strong acid with a strong base.
the acid and the moles of NaOH in the added 5.00 mL solution:
According to the balanced equation, this is how many moles of CH3CO2H will react too
because of the 1:1 mole ratio.
To find the remaining amount of CH3CO2H in the solution, we subtract the reacted moles
from its initial moles. Therefore, there will be 0.00250 – 0.000500 mol = 0.00200 mol
CH3CO2H.
You can also make a little ICE table to visualize this:
0.000500
0.000500
–
CH3CO2H(aq) + OH (aq)
0.000500
CH3CO2–(aq) + H2O(l)
When solving problems related to strong acid-base titrations, react the acid and the base and
determine the remaining concentration of H+ or OH– ions. Pay attention to the volume of the
solution – you should always add the volume of the titrant.
Titration of a Weak Acid by a Strong Base
Let’s consider the titration of 25.0 mL of 0.100 M acetic acid, CH3CO2H with 0.100 M NaOH:
Before the titration begins, there is only acetic acid in the solution and, as a weak acid, only a
fraction of it is dissociated according to the following equation:
CH3CO2H(aq)
CH3CO2–(aq) + H+(aq)
The pH of a weak acid is calculated as discussed here. In this case, pH = 2.88
pH Before the Equivalence Point
The main difference when comparing the weak acid-base titrations with strong acid-base
Using this strategy, we calculate the pH after the addition of 10 – 25 mL NaOH solution.
titrations is that there is a conjugate base of a weak acid forming here.
Right after passing the equivalence point, let’s say after the addition of 30 mL NaOH, the
solution is no longer a buffer, as it does not contain a weak acid anymore. This is where the
CH3CO2H(aq) + OH–(aq)
CH3CO2–(aq) + H2O(l)
As a result of this, after the addition of NaOH, up until the equivalence point, the solution
turns into a buffer since it contains a weak acid, CH3CO2H, its conjugate base CH3CO2–.
pH is dictated by the acetate ion, and it is calculated like we did for the pH of weak bases.
Once the OH- starts being in excess, the pH then is calculated like we do for strong bases.
gevorg@chemistrysteps.com
Copyright © 2022 Chemistry Steps
Acid–Base Titrations and Titration Curves
Titration of a Weak Acid by a Strong Base
Using the data for the volume of NaOH solution added, and the pH, the following typical
graph is obtained.
Page 3
Titration of a Weak Base by a Strong Acid
Suppose we have 25.0 mL of 0.100 M NH3 titrated with a 0.100 M solution of HCl.
HCl(aq) + NH3(aq)
NH4Cl(aq)
Before the titration begins, there is only ammonia in the solution, and as a weak base,
it reacts with water producing OH– ions making the solution basic. The pH is calculated
based on the Kb of ammonia like we do for weak bases. In this case, pH = 11.13.
NH3(aq) + H2O(l)
NH4+(aq) + OH–(aq) basic
Once some of the ammonia is converted into its conjugate acid NH4+, the solution turns
into a buffer since it contains a weak base and its conjugate acid.
HCl(aq) + NH3(aq)
NH4Cl(aq)
Like in the titration of a weak acid by a strong base, the pH at half the equivalence point
So, what key information and patterns do we see in the graph?
is going to be equal to the pKa of the acid because the logarithm term in the Henderson–
Hasselbalch Equation becomes zero.
Notice first, that the pH of the initial solution is not as low as it was in the case of a strong
acid. The equivalence point is still reached when the moles of the base is equal to the moles
of the acid in the solution. That is when all the acetic acid has been neutralized.
What is interesting, and different from the strong acid-base titrations, is that the pH at the
equivalence point is not 7. It is 8.75, and this is because of the accumulated acetate ion which
In this case, it is the ammonium ion with pKa = 9.25.
is formed as the acetic acid reacts with NaOH. It is a salt of a weak acid, and a strong base,
The equivalence point is reached when the moles of the added HCl are equal to the
therefore, it increases the pH. Go over the acid-base properties of salts, if this is still not clear.
initial moles of the ammonia in the solution, and that corresponds to 25.0 mL of HCl.
CH3CO2H(aq) + OH–(aq)
CH3CO2–(aq) + H2O(l) basic
The only predominant species in the solution becomes the NH4Cl and the pH is
dictated by the hydrolysis of the ammonium ion which produces some H3O+ thus
making the solution acidic, so the pH < 7 at the equivalence point.
Remember that in the reaction of HCl and NaOH, there was NaCl forming which is a salt of
strong and a strong base, therefore has no effect on the pH.
NaOH(aq) + HCl(aq)
NaCl(aq) + H2O(l) pH neutral salt
NH4+(aq) + H2O(l)
NH3(aq) + H3O+(aq) acidic
After the equivalence point, when all the ammonia has been reacted with the acid, the
only predominant species in the solution becomes the excess HCl which makes the
At half the equivalence point, the pH is equal to the pKa of the acid. This is when the
concentration of the acetate ion becomes equal to the remaining 50% of the acid.
According to the balance chemical equation, there would be 0.00125 mol acetate ion
formed, and therefore, the Henderson–Hasselbalch is:
solution very acidic, and the pH reaches up to 1.30.
To make keeping track easier, the graph and the key points are on the next page.
The calculations for pH for each period of the titration are given in the articles below.
Strong Acid–Strong Base Titrations
Titration of a Weak Base by a Strong Acid
Titration of a Weak Acid by a Strong Base
Titration of Polyprotic Acids
gevorg@chemistrysteps.com
Copyright © 2022 Chemistry Steps
Acid–Base Titrations and Titration Curves
Page 4
Titration of a Polyprotic Acids
Titration of a Weak Base by a Strong Acid
The data is used to obtain a representative graph for the titration of a weak base by a strong acid.
The first part of the titration represents the reaction between H2SO3 and NaOH, and the
second part is for the reaction between the conjugate base HSO3– and NaOH.
H2SO3(aq) + OH–(aq)
HSO3–(aq) + H2O(l)
HSO3–(aq) + OH–(aq)
SO32–(aq) + H2O(l)
The shape of the titration curve is like merging two separate ones for monoprotic acids:
The key information and patters:
The pH starts high and goes down as the acid is added.
The equivalence point is reached when the moles of the acid are equal to the moles of the base in
the solution. That is when all the ammonia has been neutralized.
What is different from the strong acid-base titrations, is that the pH at the equivalence point is
not 7. It is 5.28, and this because of the accumulated ammonium. It is a salt of a strong acid, and
The pH at the first half equivalence point is equal to the pKa of sulfurous acid.
a weak base, therefore, it makes the solution acidic thus decreasing the pH. Go over the acid-base
Remember, this is because the pH of buffer solutions, when [HA] = [A–] becomes equal
properties of salts, if this is still not clear.
to the pKa since the logarithm term becomes zero.
Titration of a Polyprotic Acids
At the first equivalence point, pH is below 7 because of the acidic nature of HSO3–.
For example, H2SO3 is a diprotic acid, and its dissociation can be divides in two parts each
associated with a different dissociation constant:
The pH at the second half equivalence point corresponds to the pKa of the conjugate
base HSO3–.
H2SO3(aq) + H2O(l)
–
HSO3 (aq) + H2O(l)
HSO3–(aq) + H3O+(aq) Ka1 = 1.6 x 10-2
2-
+
-8
SO3 (aq) + H3O (aq) Ka2 = 6.4 x 10
When titrated with a base, one of the protons reacts with the hydroxide first to form the
conjugate base HSO3–, which then reacts with the base forming the SO32- conjugate base.
Let’s say we are titrating 25.0 mL of 0.100 M H2SO3 with 0.100 M NaOH.
Notice that for both equivalence points, the same volume of the NaOH (25.0 mL) is
needed because the HSO3– is produced from H2SO3 in a 1:1 mole-ratio.
The information from titration curves, and especially the correlation between the half
equivalence point and the pH can be used to estimate the pKa values for the polyprotic
acid and identify an unknown acid.
gevorg@chemistrysteps.com
Copyright © 2022 Chemistry Steps