1. Introduction
This experiment deals with properties of acids and base, including differences between strong
and weak acids and how these differences affect the course of titrations. It also demonstrates how
titrations are used to characterize the amount of an acid or base in an unknown solution and
explores the pH of salt solutions.
Acid Dissociation
The strength of an acid depends on the fraction of molecules that dissociate when the acid
dissolves in water. The general equation for acid dissociation is:
Equation 1:
HA (aq) + H2O (l) ⇋ H3O+
(aq) + A−
(aq)
where HA(aq) is the acidic form of the molecule, and A-
(aq) is the conjugate base. Similar to all
equilibrium equations we can define an equation for the equilibrium constant of this reaction.
Since this equation represent the dissociation of an acid, the resulting equilibrium constant is
called the acid-dissociation constant, Ka, and has the general equation:
Equation 2:
Ka = {[A−]×[H3O+]}÷[ H A ]
The dissociation of strong acids, such as hydrochloric acid, is essentially complete, with all of
the acid form (HCl) dissociating to form hydronium and chloride ions.
Equation 3:
HCl (aq) + H2O (l) ⇋ H3O+
(aq) + Cl−
(aq)
Because the dissociation reaction of a strong acid is a reaction that favors product formation, the
resulting acid dissociation constant (Ka) is large.
Equation 4:
Ka = {[Cl−]×[H3O+]}÷[HCl] >> 1
On the other hand, weak acids, such as acetic acid (CH3COOH) dissociate only partially.
Equation 5:
CH3COOH (aq) + H2O (l) ⇋ H3O+
(aq) + CH3COO−
(aq)
2. Because weak acids do not fully dissociate, the reaction favours the reactants and the resulting
acid dissociation constant is small.
Equation 6:
Ka = {[CH3COO−
]×[H3O+
]}÷[CH3COOH] < 1
Keep these differences in mind when comparing different acids. A larger value of Ka
corresponds to greater dissociation and therefore a stronger acid and a smaller value of Ka
corresponds to lower dissociation and therefore a weaker acid.
Dissociation of Water and pH
Water can act as both an acid and a base. In pure water, protons are transferred between water
molecules as shown in the following equation:
Equation 7:
2H2O (l) ⇋ H3O+
(aq) + OH−
(aq)
As with any equilibrium equation we can assign an equilibrium constant. The equilibrium
constant for the dissociation of water is called Kw and has a value of 1.0 x 10-14
at 298 K.
Equation 8:
Kw = [H3O+
] [OH−
] = 1.0 × 10−14
Concentrations of hydronium and hydroxide ions are always related through Kw. When the
solution is neutral both concentrations are 1.0 x 10-7
M. When the solution is acidic, the
concentration of H3O+
is greater than 1.0 x 10-7
M, and the concentration of OH-
is lower than
1.0 x 10-7
M. When the solution is basic, the concentration of H3O+
is less than 1.0 x 10-7
M, and
the concentration of OH-
is greater than 1.0 x 10-7
M.
When describing solutions as being acidic, basic or neutral, scientists often use pH. In science
“p” is equal to the negative logarithm (base 10) of the value in question, and so pH is the
negative logarithm of the concentration of hydronium ion.
Equation 9:
pH = −log[H3O+
]
Acid-Base Pairs and the pH of Salt Solutions
Acids and bases exist as pairs; every acid has a conjugate base and every base has a conjugate
acid. Let’s take acetic acid as an example. Upon interaction with water, acetic acid will
3. dissociate to produce its conjugate base, acetate ion (CH3COO-
), and hydronium ion. Because
this reaction produces hydronium ions, the solution is acidic.
Equation 10:
CH3COOH (aq) + H2O (l) ⇋ H3O+
(aq) + CH3COO−
(aq)
If the conjugate base, acetate ion (CH3COO−
), is added to water, water would act as an acid and
transfer H+
to the acetate ion to reform some acetic acid and form hydroxide ion. Because this
reaction produces hydroxide ion, the solution will be basic.
Equation 11:
CH3COO−
(aq) + H2O (l) ⇋ OH−
(aq) + CH3COOH (aq)
Similar to the Ka discussed above that defines the equilibrium constant for the reaction of an acid
with water, the reaction between a base and water is described by the Kb. The Kb for the reaction
between acetate ions and water (equation [11]) is defined as:
Equation 12:
Kb = [CH3COOH]×[OH−
]÷[CH3COO−
]
Acetic acid and acetate are an acid-base pair. The Ka of the acid form of a molecule is related to
the Kb of its basic form through the dissociation constant of water, Kw.
Equation 13:
Kw = Ka × Kb = 1.0 × 10−14
Through this relationship it is clear the stronger the acid (larger Ka), the weaker its conjugate
base (smaller Kb), and conversely the weaker the acid (smaller Ka) the stronger its conjugate base
(larger Kb).
Acid-Base Titrations
In this experiment you will titrate both a strong acid and a weak acid using a strong base, sodium
hydroxide. The general equation for the reaction between an acid and sodium hydroxide is:
Equation 14:
HA (aq) + NaOH (aq) ⇋ H2O (l) + NaA (aq)
This is a neutralization reaction. A plot of pH as a function of the volume of NaOH added to a
weak acid solution is shown in Figure 1. This type of plot is called a titration curve. In the
above equation HA is being consumed by NaOH to form water and the sodium salt of the
4. conjugate base (NaA). When the number of moles of NaOH added is equal to the number of
moles of HA present, the equivalence point of the titration has been reached. Because all of the
moles of HA have been converted into its conjugate base A−
, the pH of the solution at the
equivalence point is determined by the acid-base properties of its conjugate base A−
. The
equivalence point of a titration can be used to determine the concentration of the original acid
solution using the volume and concentration of OH−
added and the initial volume of the acid
solution.
Figure 1: A weak acid-strong base titration curve
In this experiment, all of the titrations are of an acid with a strong base; however a base can also
be titrated with an acid. For more information, see Ch. 15 of the class textbook.
Buffers and pKa
A buffer is a solution that is resistant to changes in pH when acid or base is added. Buffers are
composed of a weak acid and its conjugate base or a weak base and its conjugate acid. To better
understand how a buffer works, consider the titration of a weak acid (HA) with a strong base
(NaOH). As shown in equation [14], NaOH reacts with HA, transforming it into a salt (NaA)
without producing H3O+
or OH-
. The strong base is neutralized, resulting in a small change in
pH. The buffering region of a weak acid-strong base titration curve is shown in Figure 1. Within
this region, the weak acid (HA) and its conjugate base (A-
) are both present in the solution in
significant quantities – any added strong base is neutralized by reaction with the acid component,
HA (equation [15]) and any added strong acid is neutralized by reaction with the conjugate base,
A-
(equation [16]). In this way the solution is resistant to changes in pH.
5. Equation 15:
HA (aq) + OH−
(aq) ⇌ A−
(aq) + H2O (l)
Equation 16:
A−
(aq) + H3O+
(aq) ⇌ HA (aq) + H2O (l)
A buffer always consists of a weak acid (pKa < 1) and its conjugate weak base, or a weak base
(pKb < 1) and its conjugate weak acid. The conjugate base of a strong acid is such a weak base
that is has essentially no ability to accept protons from water. Consider the example of the strong
acid HCl. The equilibrium constant for the reaction of its conjugate, Cl-
, with water (equation
[17]) is such a small value (Kb << 10−14
) that this reaction does not proceed to any appreciable
extent. The chloride ion is such a weak base that it is essentially not a base at all.
Equation 17:
Cl−
(aq) + H2O (l) ⇌ HCl (aq) + OH−
(aq)
Therefore, the reaction between a strong acid (or a strong base) with water is essentially a one-
way reaction, leading to formation of its conjugate base (or acid) which has no capacity to act as
a base so a buffer is not formed during its titration. As a result, titration curves for strong acids or
bases have different shapes than the corresponding titration curves for weak acids or bases.
In addition to the equivalence point and the buffer region of the titration curve, another important
point in a titration curve is the pH of the solution halfway to equivalence. At this point, half of
the initial moles of acid (HA in equation [14]) have been converted to their conjugate base (A–
in
equation [14]) and so the concentration of the two are equal:
Equation 18:
[HA] = [A−
]
When equation [18] is substituted into the Ka expression in equation [2] the concentrations of the
acid and conjugate base cancel and you are left with the expression:
Equation 19:
Ka = [H3O+
]
Taking the negative logarithm (base 10) of both sides, the pH of the solution is equal to the pKa
of the acid being titrated at the half-equivalence point.
6. Equation 20:
pKa = pH
Determining the Concentration of a Weak Acid
Titrations can be used industrially to determine concentrations of samples and assess purity of
compounds. In this virtual lab, titration with NaOH will be used as an analytical chemistry tool
to determine the concentration of an unknown weak acid, as well as its pKa and molar mass.
Acid-base indicators
pH can be measured in many different ways. One way is to use an acid-base indicator, which is a
molecule where the colour of the acidic form of the molecule is different from that of its
conjugate base. In reaction [18], HIn represents the acidic form of the indicator and In-
represents
its conjugate base.
Equation 21:
HIn (aq) + H2O (l) ⇌ H3O+
(aq) + In−
(aq)
acidic colour basic colour
Like any acidic molecule, this reaction has an acid dissociation constant.
Equation 22:
Ka = {[In−]×[H3O+]}÷[HIn]
The colour change of an indicator occurs when the major species in solution changes from the
acidic form to the basic form and vice versa. When the concentrations of the acidic and basic
forms of the indicator are equal, it is the midpoint of the range of the colour change and Ka =
[H3O+
]. Therefore, the colour change of the indicator is centered over the pKa of the indicator,
and generally occurs over one pH unit on either side.
Indicators are useful tools for determining the equivalence point of a titration, but a suitable
indicator must be chosen such that the colour change (end point of indicator) occurs as close as
possible to the equivalence point. You will explore end point detection by using various
indicators in the titrations.
pH meters
A pH meter is a convenient means of measuring pH accurately. This device has two electrodes;
both are immersed in the solution. One of them is sensitive to H3O+
so the electrical potential
between the electrodes varies according to the pH. The meter is calibrated by immersing the
electrodes in a solution having a known H3O+
concentration. Knowing the measured voltage at
7. this known pH value allows the meter to report the pH for any other voltage it measures in the
solution. In this virtual experiment, the pH values of the solution during the titrations are given.
The pH data will provide information about the course of the acid-base reaction.
Guide to Report Sheet
Part A: Strong Acid-Strong Base Titration
1. Find the equivalence point of the strong acid-strong base titration curve
Plot in Excel the titration curve (pH versus volume of NaOH added) as shown in Figure 3 for the
titration of a strong acid with a strong base.
As shown in Figure 3, draw straight lines through the two horizontal portions of the curve, then
draw a straight vertical line through the vertical portion of the curve. Find the midpoint of the
vertical line by measuring halfway between the two horizontal lines. What is the pH and volume
of NaOH at the equivalence point? (Note: The volume and pH at the equivalence point
determined using the titration curve is likely to be in between experimental data points.) Note:
your TA will discuss how to do this in your meeting. Come prepared with your data
plotted.
2. Determining the concentration of the hydrochloric acid solution from titration curve
You will determine the concentration of the HCl solution as practice using the equivalence point
from the titration curve and end point of the titration. You will use this same procedure to
determine the unknown acid concentration in Part D.
8. In the titration of a monoprotic acid such as HCl with NaOH, one mole of acid reacts with one
mole of base, so the equivalence point is reached when the moles of base added equal the moles
of acid present in the original solution.
nacid = nbase
Determine the moles of the NaOH solution using the volume of base required to reach the
equivalence point and molarity. Find the molarity of the acid by dividing by the volume of acid
used in the titration.
Cacid = nacidVacid
Repeat the calculation using the volume of NaOH added at the endpoint of the titration detected
by the colour change of the indicator.
One objective of this lab is to compare the end point determined using the pH indicator with the
equivalence point determined using the titration curve derived from the pH meter readings.
Part B: Weak Acid-Strong Base Titration
1. Find the equivalence point of the weak acid-strong base titration curve
Plot the titration curve (pH versus volume of NaOH added) as shown above for the titration of
your strong acid with the strong base.
2. Determining the concentration of the weak acid solution
Use your titration curve to determine the moles of weak acid found to be present in your sample
and then use the initial volume of the solution (before the titration was performed) to find the
initial concentration of weak acid.
3. Determination of the acid dissociation constant Ka of weak acid
In the exercises below, you will experimentally determine a value for Ka of your assigned weak
acid by averaging the values of Ka calculated using the data from three different points along
your titration curve. In these exercises, you will use the initial concentration of the weak acid
which you determined above.
0% to the equivalence point
Calculate the acid-dissociation constant (Ka) of the weak acid using the initial pH of the acid
(before adding any NaOH) and its concentration. The dissociation of a weak acid in water is
shown below, where A-
is the conjugate base of the weak acid:
9. HA (aq) + H2O (l) ⇌ H3O+
(aq) + A-
(aq)
and an acid-dissociation constant (Ka), given by the equation below.
Ka = {[A−]×[H3O+]}÷[ H A ] < 1
Determine Ka using an ICE (Initial / Change / Equilibrium) table as shown below
HA + H2O ⇌ A-
+ H3O+
Initial [HA]i - 0 0
Change - x - + x + x
Final [HA] i - x - x x
Calculate Ka, using the initial pH of the solution and the initial concentration of HA.
50% to the equivalence point
When you are halfway to the equivalence point in a titration (half of the number of moles of
NaOH have been added), [HA] is equal to the concentration of [A–
]. Therefore these values
cancel in the Ka expression leaving Ka = [H3O+
]. Taking the negative logarithm of both sides
gives pKa = pH, and so the pKa of the acid can be determined by extrapolating the pH value at
50% to the equivalence point from the titration curve.
75% to the equivalence point
Calculating Ka at 75% of the volume of base to reach the equivalence point requires three steps:
1. Determine the number of moles of HA and A-
in solution after the addition of base. It is
easier to work in moles at this point because the volume of the solution is changing with
the addition of base.
2. Convert the moles of acid and conjugate base found into concentrations using the total
volume of solution.
3. Find equilibrium concentrations of HA and A-
by using an ICE table.
10. Step 1: During the titration, the reaction of NaOH with HA produces A-
and water. Use the table
shown below to determine the number of moles of these species present in solution. At this point
in the calculation it is easier to work in moles.
HA + OH- ⇌ A-
+ H2O
Start of
titration
(mol HA)init (mol NaOH)added 0 -
Change - (mol NaOH)added - (mol NaOH)added + (mol NaOH)added -
After
titration
(mol HA)init - (mol
NaOH)added
=(mol HA)after
0
(mol NaOH)added =
(mol A-
)after
-
Step 2: Using the moles of HA and A-
found above, calculate the concentrations of each using
the total volume at that point in the titration.
Step 3: Now that you have concentration data for both HA and A-
you can re-establish
equilibrium using an ICE table:
HA + H2O ⇌ A-
+ H3O+
Initial [HA]new - [A-
]new 0
Change - x - + x + x
Final [HA]new - x - [A-
]new + x x
Use the pH of the solution at 75% to the equivalence point (from the titration curve) to determine
x and calculate Ka.
Determine the mean of your three experimentally determined Ka values . Use the literature value
of your assigned weak acid from your textbook to determine the percent accuracy of your
experimentally determined Ka value:
%Accuracy = [truevalue − experimentalvalue]÷[truevalue] × 100 %
Part C: Acid-base properties of Salt Solutions
The pH of two salt solutions, ammonium chloride and sodium acetate, was measured in the
Virtual Lab. Give the observed pH of the solution and the acid-base reaction that is responsible
for the observed pH.
11. Part D: Unknown Solid Acid
1. Find the equivalence point of the unknown weak acid-strong base titration curve
Plot the titration curve (pH versus volume of NaOH added) as shown above for the titration of
your strong acid with the strong base.
2. Determining the concentration of the weak acid solution
Use your titration curve to determine the moles of weak acid found to be present in your sample
and then use the initial volume of the solution (before the titration was performed) to find the
initial concentration of weak acid.
3. Determining the molar mass of the solid
Using the volume of your weak acid solution titrated and the mass of the unknown solid,
calculate the molar mass of the unknown. (Note: remember you used only 25.00 mL of the
100.00 mL solution you prepared.)
4. Determination of the acid dissociation constant Ka of the unknown weak acid
As described in Part B, use the half-equivalence point to determine the pKa of your unknown
acid.