University of Bahrain
Department of Chemistry
Edited by: Dr. Ali Hussain
CONTENTS
Experiment 1
ENTHALPY CHANGE OF REACTION
Experiment 2
MOLECULAR GEOMETRIES OF COVALENT
MOLECULES: LEWIS STRUCTURE AND
VSEPR THEORY
7
Experiment 3
DETERMINATION OF THE DISSOCIATION
CONSTANT AND CONCENTRATION OF
A WEAK ACID
15
Experiment 4
ACID-BASE TITRATION CURVES
Experiment 5
HYDROLYSIS OF SALTS AND THE ACTION OF
A BUFFER SOLUTION
24
Experiment 6
DETERMINATION OF THE SOLUBILITYPRODUCT CONSTANT FOR A SPARINGLY
SOLUBLE SALT
34
RATE OF CHEMICAL REACTION:
THE IODINATION OF ACETONE
38
Experiment 7
3
19
Experiment 8
KINETICS OF THE HYDROLYSIS OF ETHYL ACETATE
BY SODIUM HYDROXIDE BY A CONDUCTIVITY
METHOD
46
Experiment 9
OXIDATION-REDUCTION TITRATION:
DETERMINATION OF OXALATE
2
48
Experiment 1:
ENTHALPY CHANGE OF REACTION
OBJECTIVE
To measure, using a calorimeter, the enthalpy change accompanying the following
displacement reaction:
Zn (s) + Cu2+ (aq) → Cu (s) + Zn2+ (aq)
DISCUSSION
Every chemical change is accompanied by a change in energy, usually in the form
of heat. The energy change of a reaction that occurs at constant pressure is termed
the heat of reaction or the enthalpy change. The symbol ΔH is used to denote the
enthalpy change. If heat is evolved, the reaction is exothermic (ΔH  0); and if
heat is absorbed, the reaction is endothermic (ΔH  0). In this experiment, you will
calculate the enthalpy change of the above displacement reaction by adding an
excess of zinc powder to a measured amount of CuSO4 (aq) and measuring the
temperature change over a period of time.
This quantity of heat is measured experimentally by allowing the reaction to
take place in a thermally insulated vessel called calorimeter. The heat liberated in
the reaction will cause an increase in the temperature of the solution and of the
calorimeter. If the calorimeter were perfect, no heat would be radiated to the
laboratory. The calorimeter you will use in this experiment is shown in Figure 3.1.
Because we are concerned with the heat of the reaction and because some heat
is absorbed by the calorimeter itself, we must know the amount of heat absorbed
by the calorimeter. In this experiment, the temperature of the calorimeter and its
contents is measured before and after the reaction. The change in enthalpy, ΔH, is
equal to the negative product of the heat capacity of the calorimeter and its content
times the temperature change, ΔT,:
ΔH = – (heat capacity of the calorimeter + heat capacity of content)  ΔT
[1]
The heat capacity of the system represents the amount of heat required to raise
the temperature of the system 1oC, and ΔT is the difference between the final and
initial temperatures: ΔT = Tf – Ti.
3
Figure 1.1
A simple calorimeter
However, we will assume here that no heat is gained by the calorimeter and no
heat is lost to the laboratory. Therefore,
ΔH ≈ – (heat capacity of content)  ΔT
[2]
For a pure substance of mass m, the expression for ΔH can be written as:
ΔH = – m  c  ΔT
[3]
The quantity c is known as the specific heat; it is defined as the amount of heat
required to raise the temperature of one gram of a substance 1 oC.
For dilute aqueous solutions, the expression for ΔH can be approximated as:
ΔH = – mwater  cwater  ΔT
[4]
Where mwater is the mass of water and cwater is the specific heat of water which
equals to 4.18 J/goC. The mass of water can be calculated by considering the
density of water equals to 1.00 g/mL.
4
PROCEDURE
1. Pipette 25.0 mL of cupper (II) sulfate solution into the Styrofoam cup.
2. Weigh about 3 grams of zinc powder in the weighing bottle. Since this is an
excess, there is no need to be accurate.
3. Put the thermometer through the hole in the lid, stir and record the
temperature to the nearest 0.1oC every half minute for 2.5 minutes.
4. At precisely 3 minutes, add the zinc powder to the cup.
5. Continue stirring and record the temperature till you get to close to room
temperature. Tabulate your results on the report sheet.
Plot the temperature (y-axis, oC) against time (x-axis, min). Extrapolate the
curve to 3.0 minutes to establish the maximum temperature rise as shown in
Figure 2.3.
Calculate the enthalpy change for the quantity of CuSO4 solution used and for
one mole of Zn and CuSO4, and write the thermochemical equation for the
reaction in the report sheet.
Temp. / oC
Time / sec
Figure 1.2 Temperature as a function of time for the reaction Zn-CuSO4
5
Name …………………………………………… ID ………… Sec …..
Experiment 1: ENTHALPY CHANGE OF REACTION
REPORT SHEET
Time/
min
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
Temp./
o
C
Time/
min
Temp./
o
C
Time/
min
Temp./
o
C
Time/
min
Temp./
o
C
Time/
min
Temp./
o
C
6.0
6.5
7.0
7.5
1. ΔT determined from your curve
_________ oC
2. Heat gained by solution (25.0 g  4.18 J/goC  ΔT oC)
__________________________ J
3. The number of moles of CuSO4 in 25.0 mL
of 1.00 M CuSO4.
_________ mol
4. Heat released per mole of CuSO4 (and Zn) [(2)/(3)]
_________ kJ/mol
5. Complete the thermochemical equation:
Zn (s) + Cu2+ (aq) → Cu (s) + Zn2+ (aq)
ΔH = _______________
6. Compare your result with the accepted value of –217 kJ/mol by calculating the
percentage error in your answer:
% Error = │[(experimental value – accepted value)/accepted value]│  100%.
7. What are the main sources of error in this experiment?
6
Experiment 2:
MOLECULAR GEOMETRIES OF COVALENT
MOLECULES: LEWIS STRUCTURES AND VSEPR
THEORY
OBJECTIVE
To become familiar with Lewis structures, the principles of VSEPR theory, and
the three-dimensional structures of covalent molecules.
DISCUSSION
Types of Bonding Interactions
Whenever atoms or ions are strongly attached to one another, we say that there is a
chemical bond between them. There are three types of chemical bonds: ionic,
covalent, and metallic. The term ionic bond refers to electrostatic forces that exist
between ions of opposite charge. A covalent bond results from the sharing of
electrons between two atoms. The more familiar examples of covalent bonding are
found among nonmetallic elements interacting with one another. This experiment
illustrates the geometric (three-dimensional) shapes of molecules and ions
resulting from covalent bonding among various numbers of elements, and one of
the consequences of geometric structures, polarity. Metallic bonds are found in
metals such as gold, iron and magnesium. In the metals, each atom is bonded to
several neighboring atoms and the bonding electrons are free to move throughout
the metal.
Lewis Symbol
The electrons involved in chemical bonding are the valence electrons, those
residing in the incomplete outer shell of an atom. The Lewis symbol for an element
consists of the chemical abbreviation for the element plus a dot for each valence
electron. The dots are placed on the four sides of the atomic abbreviation. The
number of valence electrons of any representative element is the same as the group
number of the element in the periodic table. The Lewis structure of oxygen, for
example, is shown below
O
7
The Octet Rule
Atoms often gain, lose, or share electrons so as to achieve the same number of
electrons as the noble gas closest to them in the periodic table. According to the
octet rule then, atoms tend to gain, lose, or share electrons until they are
surrounded by eight valence electrons. There are many exceptions to the octet rule,
but it provides a useful framework for many important concepts of bonding.
Covalent Bonding
Lewis reasoned that atoms might acquire a noble-gas electron configuration by
sharing electrons with other atoms to form covalent bonds. The hydrogen
molecule, H2, provides the simplest possible example of a covalent bond. The
attraction between the nuclei and the electrons cause electron density to
concentrate between the nuclei.
Lewis Structures
The formation of a covalent bond can be represented using Lewis symbols as
shown below for H2:
H
+ H
→
H
H
The formation of a bond between two F atoms to give a F2 molecule can be
represented in a similar way:
F
+
F
→
F
F
By sharing the bonding electron pair, each fluorine atom requires eight electrons
(an octet) in its valence shell. It thus achieves the noble-gas electron configuration
of neon. The structures shown here for H2 and F2 are called Lewis structures (or
Lewis electron-dot structures). In writing Lewis structures, we usually show each
electron pair shared between atoms as line, to emphasize it is a bond, and the
unshared electron pairs as dots. Writing them this way, the Lewis structures for H 2
8
and F2, are shown as follows:
H
H
F
F
Multiple Bonds
The sharing of a pair of electrons constitutes a single covalent bond, generally
referred to simply as a single bond. In many molecules, atoms attain complete
octets by sharing more than one pair of electrons between them. When two
electron pairs are shred, two lines (representing a double bond) are drawn. A triple
bond corresponds to the sharing of three pairs of electrons. Such multiple bonding
is found in CO2 and N2.
O
C
N
O
N
Drawing Lewis Structures
Lewis structures are useful in understanding the bonding in many compounds and
are frequently used when discussing the properties of molecules. To draw Lewis
structures, we follow a regular procedure:
1- Sum the valence electrons from all atoms. Use the periodic table to help
determine the number of valence electrons on each atom. For an anion, add
an electron to the total for each negative charge. For a cation, subtract an
electron for each positive charge.
2- Write the symbols for the atoms to show which atoms are attached to
which, and connect them with a single bond (a dash, representing two
electrons). Chemical formulas are often written in the order in which the
atoms are connected to the molecule or ion, as in HCN. When a central
atom has a group of other atoms bonded to it, the central atom is usually
written first, as in CO 32 or BF3. In other cases you may need more
information before you can draw the Lewis structure.
3- Complete the octets of the atoms bonded to the central atom. (Remember,
however, that hydrogen can have only two electrons).
4- Place any leftover electrons on the central atom, even if doing so results in
more than an octet.
5- If there are not enough electrons to give the central atom an octet, try
multiple bonds. Use one or more of the unshared pairs of electrons on the
atoms bonded to the central atom to form double or triple bonds.
9
Bond Polarity and Dipole Moments
A covalent bond between two different kinds of atoms is usually a polar bond.
This is because the two different atoms have different electronegativities and the
electrons in the bond are then not shared equally by the two bound atoms. This
creates a bond dipole moment, μ, which is a charge separation, Q, over a distance,
r:
μ = Qr
The bonding electrons have an increased attraction to the more electronegative
atom, thus creating an excess of electron density (δ‾) near it and a deficiency of
electron density (δ+) near the less electronegative atom. We symbolize the bond
dipole (a vector) with an arrow and a cross, , with the point of the arrow
representing the negative and the cross the positive end of the dipole. Thus, in the
polar covalent molecule H—Cl, because the chlorine is more electronegative than
hydrogen, the bond dipole is as illustrated below:
δ+
δ‾
H  Cl
The bond dipole is a vector (it has a magnitude and direction), and the dipole
moments of polyatomic molecules are the vector sums of the individual bond
dipoles. Therefore, H2O has a dipole moment (a polar molecule) and CCl4 does not
(a nonpolar molecule).
VSEPR Theory
In covalent molecules, atoms are bonded together by sharing pairs of valence-shell
electrons. Electron pairs repel one another and try to stay out of each other’s way.
The best arrangement of a given number of electron pairs is the one that
minimizes the repulsions among them. This simple idea is the basis of valenceshell electron pair repulsion theory, or the VSEPR model. Thus, as illustrated in
Table 2.1, two electron pairs are arranged linearly, three pairs are arranged in a
trigonal planar fashion, four are arranged tetrahedrally, five are arranged in a
trigonal bipyramidal geometry, and six are arranged octahedrally. The shape of a
molecule or ion can be related to these five basic arrangements of electron pairs.
Predicting Molecular Geometries
When we draw Lewis structures, we encounter two types of valence-shell electron
pairs: bonding pairs, which are shared by atoms in bonds, and nonbonding pairs
(or lone pairs) such as in the Lewis structure for NH3. Because there are four
electron pairs around the N atom, the electron-pair repulsions will be minimized
10
when the electron pairs point toward the vertices of a tetrahedron (see Table 2.1).
The arrangement of electron pairs about the central atom of an ABn molecule is
Table 2.1 Electron-Pair Geometries as a Function of the Number of
Electron Pairs.
11
Nonbonding pair
H
N
H
H
Bonding pairs
called its electron-pair geometry. However, when we use experiments to
determine the structure of a molecule, we locate atoms, not electron pairs. The
molecular geometry of a molecule (or ion) is the arrangement of atoms in space.
We can predict the molecular geometry of a molecule from its electron-pair
geometry. In NH3, the three bonding pairs point toward three vertices of a
tetrahedron and the lone pair is located at a fourth vertex. The arrangement of the
atoms in NH3 is thus a trigonal pyramid, and the sequence of steps to arrive at this
prediction is illustrated in Figure 2.1. We see that the trigonal-pyramidal
molecular geometry of NH3 is a consequence of its tetrahedral electron-pair
geometry. When describing the shapes of molecules, we always give the
molecular geometry rather than the electron-pair geometry.
Following are the steps used to predict molecular geometries with the VSEPR
model:
1- Sketch the Lewis structure of the molecule or ion.
2- Count the total number of electron pairs around the central atom, and
arrange them in a way that minimizes electron-pair repulsions (Table 2.1).
3- Describe the molecular geometry in terms of the angular arrangement of the
bonding pairs, which corresponds with the arrangement of bound atoms.
4- A double or triple bond is counted as one bonding pair when predicting
geometry.
Figure 2.1
12
Name …………………………………………… ID ……….…… Sec …..
Experiment 2: MOLECULAR GEOMETRIES OF COVALENT MOLECULES:
LEWIS STRUCTURES AND VSEPR THEORY
REPORT SHEET
A. Using an appropriate set of models, make molecular models of the compounds
listed below and complete the table.
Molecular
Formula
No. of
bond pairs
No. of lone
pairs (around
central atom)
Molecular
geometry
BeCl2
BF3
SnCl2
CH4
NH3
H2O
PCl5
SF4
BrF3
XeF2
SF6
IF5
XeF4
13
Bond
angle(s)
Dipole
moment
(yes or no)
Name …………………………………………… ID ………….… Sec …..
Experiment 2: MOLECULAR GEOMETRIES OF COVALENT MOLECULES:
LEWIS STRUCTURES AND VSEPR THEORY
REPORT SHEET
B. Using an appropriate set of models, make molecular models of the compounds
listed in the sheet given to you by instructor and complete the table.
Molecular
Formula
No. of
bond pairs
No. of lone
pairs (around
central atom)
Molecular
geometry
14
Bond
angle(s)
Dipole
moment
(yes or no)
Experiment 3:
DETERMINATION OF THE DISSOCIATION
CONSTANT AND CONCENTRATION
OF A WEAK ACID
OBJECTIVE
To become familiar with the operation of a pH meter and quantitative equilibrium
constants.
DISCUSSION
According to the BrØnsted-Lowry acid-base theory, the strength of an acid is
related to its ability to donate protons. All acid-base reactions are then
competitions between bases of various strengths for these protons. For example,
the strong acid HCl reacts with water according to Equation [1]:
HCl (aq) + H2O (l) → H3O+ (aq) + Cl‾ (aq)
[1]
or, in terms of the net ionic equation,
HCl (aq) → H+ (aq) + Cl‾ (aq)
[2]
This acid is a strong acid and is completely dissociated–in another words, 100%
dissociated–in dilute aqueous solution. Consequently, the H+ concentration of 0.1
M HCl is 0.1 M.
By contrast, acetic acid, HC2H3O2, is a weak acid and is only slightly
dissociated, as shown in Equation [3]:
HC2H3O2 (aq) ⇌ H+ (aq) + C2H3O2‾ (aq)
[3]
Its acid dissociation constant, as shown by Equation [4], is therefore small:
[H  ][C2 H3O 2  ]
Ka =
[HC 2 H3O 2 ]
15
[4]
Acetic acid only partially dissociates in aqueous solution, and an appreciable quantity of
undissociated acetic acid remains in solution.
For the general weak acid HB, the dissociation reaction and dissociation
constant expressions are
HB (aq) ⇌ H+ (aq) + B‾ (aq)
[5]
[H  ][ B ]
[HB]
[6]
– log [H+] = pH
[7]
Ka =
Recall that pH is defined as
Solving Equation [6] for [H+] and substituting this quantity into Equation [7]
yields:
pH = pKa – log
[ HB ]
[ B ]
[8]
where pKa = – log Ka
If we titrate the weak acid HB with a base, there will be a point in the titration at
which the number of moles of base added is half the number of moles of acid
initially presents. This is the point at which 50% of the acid has been titrated to
produce B‾ and 50% remains as HB. At this point [HB] = [B‾], the ratio [HB]/[B‾]
= 1, and log [HB]/[B‾] = 0. Hence, at this point in a titration, that is, at half the
equivalence point, Equation [8] becomes
pH = pKa
[9]
By titrating a weak acid with a strong base and recording the pH versus the
volume of the base added, we can determine the ionization constant of the weak
acid. From the resultant titration curve we obtain the ionization constant, as
explained in the following paragraph.
From the titration curve shown in Figure 4.1, we see at the point denoted as half
equivalence point, where [HB] = [B‾], the pH is 4.3. Thus from Equation [9], at
this point pH = pKa = 4.3, or Ka = 510‾5.
16
Figure 3.1
Exemplary titration curve
for the titration of a weak
acid HB with a strong base
PROCEDURE
A. Determination of Ka of Unknown Weak Acid
With the aid of a pipette bulb, pipette a 50-mL aliquot of your unknown
monoprotic weak acid solution into a 100-mL beaker and carefully immerse the
previously rinsed electrodes of the pH meter in this solution. Remember that the
glass electrode is very fragile; it breaks easily! Do not touch the bottom of the
beaker with the electrodes!! Also, the pH meter has been standardized therefore
do not readjust the knobs! Add a small stirring paddle and stand the beaker on a
magnetic stirrer. Measure the pH of the solution and record it in the report sheet.
Begin your titration by adding 2 mL of your standardized base (0.200 M) from a
burette and record the volume of the titrant and pH. Repeat with successive
additions shown in the table of your report sheet. From these data, plot a titration
curve of pH versus mL titrant added. From the curve, calculate the ionization
constant of the unknown acid.
B. Determination of the Concentration of Unknown Weak Acid
Using the volume of the base at the equivalence point, its molarity, and the fact
that you used 50.0 mL of acid, calculates the concentration of the unknown
monoprotic weak acid. Recall that the net ionic equation for the reaction of weak
acid HB with strong base is given by
HB (aq) + OH‾ (aq) → B‾ (aq) + H2O (l)
17
[10]
Name …………………………………………… ID ………… Sec …..
Experiment 3: DETERMINATION OF THE DISSOCIATION CONSTANT AND
THE CONCENTRATION OF A WEAK ACID
REPORT SHEET
mL NaOH
0.0
2.0
4.0
6.0
8.0
10.0
11.0
12.0
13.0
14.0
15.0
16.0
17.0
18.0
19.0
pH
mL NaOH
pH
mL NaOH
pH
27.5
28.0
28.5
29.0
29.5
30.0
31.0
32.0
33.0
34.0
35.0
36.0
37.0
20.0
20.5
21.0
21.5
22.0
22.5
23.0
23.5
24.0
24.5
25.0
25.5
26.0
26.5
27.0
Volume of NaOH at equivalence point = __________ mL
Volume of NaOH at one half equivalence point = __________ mL
pKa = _________ , Ka = _________
MHB = (VNaOH  MNaOH) / VHB = ________________________________ M
18
Experiment 4:
ACID-BASE TITRATION CURVES
OBJECTIVE
To track the change in pH with an acid-base titration curves and to gain familiarity
with acid-base indicators.
DISCUSSION
The acid-base titration can be examined by tracking the change in pH with an
acid-base titration curve, a plot of pH versus volume of titrant added. We will
consider three types of reactions: (1) titrations involving a strong acid and a strong
base, (2) titrations involving a weak acid and a strong base, and (3) titrations
involving a strong acid and a weak base.
Strong AcidStrong Base Titrations
The reaction between HCl, a strong acid, and NaOH, a strong base, can be
represented, in terms of the net ionic equation, by
H+ (aq) + OH‾ (aq) → H2O (l)
[1]
Consider the addition of 0.10 M NaOH solution (from a burette) to a conical flask
containing 25 mL of 0.10 M HCl. Before the addition of NaOH, the pH of the acid
is given by log (0.10), or 1.00. When NaOH is added, the pH of the solution
increases slowly at first. Near the equivalence point the pH begins to rise steeply,
and at the equivalence point (that is, the point at which equimolar amounts of acid
and base have reacted) the curve rises almost vertically. In a strong acid-strong
base titration, both the hydrogen ion and hydroxide ion concentrations are very
small at the equivalence point (approximately 110‾14 M); consequently, the
addition of a single drop of the base can cause a large increase in [OH‾] and in the
pH of the solution. Beyond the equivalence point, the pH again increases with the
addition of NaOH.
19
Weak AcidStrong Base Titrations
Consider the neutralization reaction between acetic acid (a weak acid) and NaOH
(a strong base):
HC2H3O2 (aq) + OH‾ (aq) → C2H3O2‾ (aq) + H2O (l)
[2]
The acetate ion undergoes hydrolysis as follows:
C2H3O2‾ (aq) + H2O (l) ⇌ HC2H3O2 (aq) + OH‾ (aq)
[3]
Therefore, at the equivalence point, when we only have sodium acetate present,
the pH will be greater than 7 as a result of the excess OH– ions formed.
Strong AcidWeak Base Titrations
Consider the titration of HCl, a strong acid, with NH3, a weak base:
H+ (aq) + NH3 (aq) → NH4+ (aq)
[4]
The pH at the equivalence point is less than 7 due to the hydrolysis of the NH4+:
NH4+ (aq) ⇌ NH3 (aq) + H+ (aq)
[5]
Because of the volatility of an aqueous ammonia solution, it is more convenient to
add hydrochloric acid from a burette to the ammonia solution.
Acid-Base Indicators
The equivalence point, as we have seen, is the point at which the number of moles
of OH‾ ions added to a solution is equal to the number of moles of H + ions
originally present. To determine the equivalence point in a titration, then, we must
know exactly how much volume of a base to add from a burette to an acid in a
flask. One way to achieve this goal is to add a few drops of an acid-base indicator
to the acid solution at the start of the titration. An indicator is usually a weak
organic acid or a base that has distinctly different colors in its nonionized and
ionized forms. These two forms are related to the pH of the solution in which the
indicator is dissolved. The end point of a titration occurs when the indicator
changes color. However, not all indicators change color at the same pH. So the
choice of indicator for a particular titration depends on the nature of the acid and
base used in the titration (that is, whether they are strong or weak). By choosing
the proper indicator for a titration, we can use the end point to determine the
equivalence point.
20
Let us consider a weak monoprotic acid that we will call HIn. To obtain an
effective indicator, HIn and its conjugate base, In‾, must have distinctly different
colors. In solution, the acid ionizes to a small extent:
HIn (aq) ⇌ H+ (aq) + In‾ (aq)
[6]
If an indicator is in a sufficiently acidic medium, the equilibrium, according to Le
Châtelier’s principle, shifts to the left and the predominant color of the indicator is
that of the nonionized form (HIn). On the other hand, in a basic medium the
equilibrium shifts to the right and the color of the solution will be due mainly to
that of the conjugate base (In‾).
The end point of an indicator does not occur at a specific pH; rather, there is a
range of pH within which the end point will occur. In practice, we choose an
indicator whose end point lies on the steep part of the titration curve. Because the
equivalence point also lies on the steep part of the curve, this choice ensures that
the pH at the equivalence point will fall within the range over which the indicator
changes color. Table 5.1 lists a number of indicators commonly used in acid-base
titrations. The choice of a particular indicator depends on the strength of the acid
and base to be titrated.
Table 4.1 Some Common Acid-Base Indicators
Color
Indicator
in acid
in base
pH range
thymol blue
bromophenol blue
methyl orange
methyl red
chlorophenol blue
bromothymol blue
cresol red
red
yellow
orange
red
yellow
yellow
yellow
yellow
bluish purple
yellow
yellow
red
blue
red
phenolphthalein
colorless
reddish pink
1.2−2.8
3.0−4.6
3.1−4.4
4.2−6.3
4.8−6.4
6.0−7.6
7.2−8.8
8.3−10.0
21
PROCEDURE
1- Pipette 50.0 mL of 0.100 M HCl into a 100 mL beaker, add a small stirring
paddle and stand the beaker on a magnetic stirrer.
2- Carefully clamp the electrode of a pH meter (which must have been
calibrated) so that the bulb is completely immersed in the acid and is clear
of the stirring paddle.
3- Fill a burette with 0.200 M NaOH solution and clamp it over the beaker.
4- Measure the pH of the acid.
5- Add the alkali from the burette in steps as shown in the report sheet and
record the pH at each addition after thorough mixing.
6- Repeat steps 1-5 with 0.100 M NH3 (in a beaker) and 0.200 M HCl (in a
burette).
7- Plot the two curves with the pH on the vertical scale and the volume of the
solution added from the burette on the horizontal scale. Recall that the
titration of weak acid (acetic acid) and NaOH was performed in Experiment
3.
8- From the three plots, determine, at the equivalence point, the volume of the
titrant, the pH, and the acidity (or basicity) of the solution.
9- Using Table 4.1, determine what indicator(s) among methyl red,
bromothymol blue and phenolphthalein can be used to locate the
equivalence point in the three titrations performed.
22
Name …………………………………………… ID ………… Sec …..
Experiment 4: ACID-BASE TITRATION CURVES
REPORT SHEET
1: NaOH–HCl Titration
i) VNaOH at equivalence
point = _______ mL
ii) pH at equivalence
point = _____
iii) Determine whether the
equivalence point is
neutral, acidic or
basic. _________
iv) Suitable indicator(s):
mL NaOH
pH
0.0
5.0
10.0
15.0
17.5
20.0
21.0
22.0
23.0
23.5
24.0
24.5
mL NaOH
pH
25.0
25.5
26.0
26.5
27.0
27.5
28.0
29.0
30.0
32.0
35.0
___________________
2: NH3 –HCl Titration
i) VHCl at equivalence
point = _______ mL
ii) pH at equivalence
point = _____
iii) Determine whether the
equivalence point is
neutral, acidic or
basic. _________
iv) Suitable indicator(s):
____________________
mL HCl
pH
0.0
5.0
10.0
15.0
16.0
17.0
18.0
19.0
20.0
21.0
22.0
23.0
23.5
mL HCl
24.0
24.5
25.0
25.5
26.0
26.5
27.0
27.5
28.0
29.0
30.0
32.0
35.0
3: HC2H3O2 –NaOH Titration [refer to Experiment 3]
i) VNaOH at equivalence point = _______ mL
ii) pH at equivalence point = _______
iii) Determine whether the equivalence point is neutral, acidic, or basic _________
iv) Suitable indicator(s): _________________________
23
pH
Experiment 5:
HYDROLYSIS OF SALTS
AND THE ACTION OF A BUFFER SOLUTION
OBJECTIVE
To learn about the concept of hydrolysis and to gain familiarity with the behavior
of buffer solutions
A- HYDROLYSIS OF SALTS
DISCUSSION
We expect solutions of substances such as HCl and HNO2 to be acidic and
solutions of NaOH and NH3 to be basic. However, we may be somewhat surprised
at first to discover that aqueous solutions of some salts such as NaNO2 and
KC2H3O2 are basic, whereas others such as NH4Cl and FeCl3 are acidic. Recall
that salts are the products formed in neutralization reactions of acids and bases.
For example, when NaOH and HNO2 (nitrous acid) react, the salts NaNO2 is
formed:
NaOH (aq) + HNO2 (aq) → NaNO2 (aq) + H2O (l)
[1]
Nearly all salts are strong electrolytes and exist as ions in aqueous solutions.
Many ions react with water to produce acidic or basic solutions. The reactions of
ions with water are frequently called hydrolysis reactions. We will see that anions
such as CN‾ and C2H3O2‾ that are the conjugate bases of the weak acids HCN and
HC2H3O2, respectively, react with water to form OH‾ ions. Cations such as NH 4+
and Fe3+ come from weak bases and react with water to form H+ ions.
Hydrolysis of Anions: Basic Salts
Anions of weak acids react with proton sources. When placed in water these
anions react to some extent with water to accept protons and generate OH – ions
and thus cause the solution pH to be greater than 7. Recall that proton acceptors
are BrØnsted-Lowry bases. Thus, the anions of weak acids are basic in two senses:
24
they are proton acceptors, and their aqueous solutions have pH’s above 7. The
nitrite ion, for example, reacts with water to increase the concentration of OH–
ions:
NO2‾ (aq) + H2O (l) ⇌ HNO2 (aq) + OH‾ (aq)
[2]
This reaction of nitrite ion is similar to that of weak bases such as NH 3 with water:
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH‾ (aq)
[3]
Thus, both NH3 and NO2‾ are bases and as such have a basicity or basedissociation constant, Kb, associated with their corresponding equilibria.
According to the BrØnsted-Lowry theory, the nitrite ion is the conjugate base of
nitrous acid. Let’s consider the conjugate acid-base pair HNO2 and NO2‾ and their
behavior in water:
HNO2 ⇌ H+ + NO2‾
Ka = [H+] [NO2‾] / [HNO2]
[4]
NO2– (aq) + H2O (l) ⇌ HNO2 (aq) + OH‾ (aq)
Kb = [HNO2] [OH‾] / [NO2‾]
[5]
Multiplication of these dissociation constants yields:
Ka  Kb = [H+] [OH‾] = Kw = 1.0  10‾14
[6]
Where Kw is the ion-product constant of water.
We note that the stronger the acid is, the larger its Ka, and the weaker its
conjugate base, the smaller its Kb. Likewise, the weaker the acid (the smaller the
Ka), the stronger the conjugate base (the larger the Kb).
Anions derived from strong acids, such as Cl‾ from HCl, do not react with water
to affect the pH. Nor do Br‾, I‾, NO3‾, SO42‾, and ClO4‾ affect the pH, for the same
reason. They are spectator ions in the acid-base sense and can be described as
neutral ions. Similarly, cations from strong bases, such as Na + from NaOH or K+
from KOH, do not react with water to affect the pH. Hydrolysis of an anion occurs
only when it can form a molecule or ion that is a weak electrolyte in reaction with
water. Strong acids and bases do not exist as molecules in dilute water solutions.
25
Hydrolysis of Cations: Acidic Salts
Cations that are derived from weak bases react with water to increase the
hydrogen-ion concentration; they form acidic solutions. The ammonium ion is
derived from the weak base NH3 and reacts with water as follows:
NH4+ (aq) ⇌ NH3 (aq) + H+ (aq)
[7]
This reaction is completely analogous to the dissociation of any other weak acid,
such as HC2H3O2 or HNO2. The acid dissociation constant of NH4+ in [7] is related
to the Kb of NH3, which is the conjugate base of NH4+:
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH‾ (aq)
[3]′
Cations of the alkali metals (Group 1A) and the larger alkaline earth ions, Ca2+,
Sr2+, and Ba2+, do not react with water, because they come from strong bases. Thus
these ions have no influence on the pH of aqueous solutions. Consequently, they
are described as being neutral in the acid-base sense. The cations of most other
metals do hydrolyze to produce acidic solutions. Metal cations are coordinated
with water molecules, and it is the hydrated ion that serves as the proton donor.
The following equation illustrate this behavior for the hexaaqua iron (III) ion:
Fe(H2O)63+ (aq) + H2O (l) ⇌ Fe(H2O)5(OH)2+ (aq) + H3O+ (aq)
[8]
We frequently omit the coordinated water molecules from such equations. For
example, Equation [8] may be written as
Fe3+ (aq) + H2O (l) ⇌ Fe(OH)2+ (aq) + H+ (aq)
[9]
Additional hydrolysis reactions can occur to form Fe(OH) 2+ and even to the
precipitation of Fe(OH)3. The equilibria for such cations are often complex, and
not all species have been identified. However, equations such as [8] and [9] serve
to illustrate the acidic character of dipositive and tripositive ions and account for
most of the H+ in these solutions.
26
Summary of Hydrolysis Behavior of Salts
Whether a solution of a salt will be acidic, neutral or basic can be predicted on the
basis of the strengths of the acid and base from which the salts was formed.
1. Salt of a strong acid and a strong base: Examples: NaCl, KBr, and
Ba(NO3)2. Neither the cation nor anion hydrolyzes, and the solution has a
pH of 7.
2. Salt of a strong acid and a weak base: Examples: NH4Br, ZnCl2, and
Al(NO3)3. The cation hydrolyzes, forming H+ ions, and the solution has a
pH less than 7.
3. Salt of a weak acid and a strong base: Examples: NaNO2, KC2H3O2, and
Ca(OCl)2. The anion hydrolyzes, forming OH‾ ions, and the solution has a
pH greater than 7.
4. Salt of a weak acid and a weak base: Examples: NH4F, NH4C2H3O2, and
Zn(NO2)2. Both ions hydrolyze. The pH of solution is determined by the relative
extent to which each ion hydrolyzes.
In this experiment, we will test the pH of water and of several aqueous salt
solutions to determine whether these solutions are acidic, basic, or neutral. In each
case, the salt solution will be 0.1 M. Knowing the concentration of the salt solution
and the measured pH of each solution allows us to calculate Ka or Kb for the ion
that hydrolyzes.
PROCEDURE
1- Boil approximately 500 mL of distilled water for about 10 min to expel
dissolved carbon dioxide. Allow the water to cool to room temperature.
2- Determine the pH of unboiled and boiled distilled water.
3- Determine the pH of the following solutions that are 0.1 M: NaCl,
NaC2H3O2, NH4Cl, ZnCl2, KAl(SO4)2, and Na2CO3. Use about 25 mL of
each of these solutions. Rinse the beaker with boiled distilled water when
you go from one solution to the next.
4- Add three drops of each of the following indicators: methyl red,
bromothymol blue, and phenolphthalein to each solution (one indicator per
solution). Record the colors.
From the pH values that you determined, calculate the hydrogen- and
hydroxide- ion concentrations for each solution (pH = – log10 [H+], and [H+] 
[OH‾] = 110‾14). Complete the tables on the report sheets and calculate the Ka or
Kb as appropriate.
27
B- THE ACTION OF A BUFFER SOLUTION
DISCUSSION
The pH of solution is often accomplished by the use of buffers. A buffer
solution has the important property of resisting large changes in pH upon the
addition of small amounts of strong acids or bases. A buffer solution must have
two components– one that will react with H+, and the other that will react with
OH‾. The two components of a buffer solution are usually a weak acid and its
conjugate base, such as HC2H3O2–C2H3O2‾ or NH4+–NH3. Thus buffers are often
prepared by mixing a week acid or a week base with a salt of that acid or base. For
example, the HC2H3O2–C2H3O2‾ buffer can be prepared by adding NaC2H3O2 to a
solution of HC2H3O2; the NH4+–NH3 can be prepared by adding NH4Cl to a
solution of NH3. By appropriate choice of components and their concentrations,
buffer solutions of virtually any pH can be made.
To examine how a buffer works, consider, for example, the HC2H3O2–C2H3O2‾
buffer. If OH‾ ions are added, they react with the acid component of the buffer:
OH‾ (aq) + HC2H3O2 (aq) → C2H3O2‾ (aq) + H2O (l)
[10]
If H+ are added, they react with the base component of the buffer:
H+ (aq) + C2H3O2‾ (aq) → HC2H3O2 (aq)
[11]
Buffers resist changes in pH most effectively when the concentrations of the
conjugate acid-base pair, HC2H3O2 and C2H3O2‾ in the above example, are about
the same.
Here you will demonstrate the action of a buffer solution by comparing the pH
changes after the addition of small measured amounts of HCl and NaOH to a
buffer solution and to pure water.
28
PROCEDURE
123-
4-
5678-
9-
10-
Fill a burette with 0.1 M HCl and another with 0.1 M NaOH.
Using a measuring cylinder, put 25 mL of a buffer solution (of pH = 7)
in a 50 mL beaker.
Rinse the pH meter electrode with distilled water from a wash bottle,
and put it into the beaker. Make sure that the glass bulb is completely
immersed and record the pH.
Place the beaker under the burette containing NaOH and, making sure
the alkali does not fall directly on the electrode, add 1 drop of 0.1 M
NaOH. Stir gently to ensure thorough mixing and record the pH.
Add more NaOH to make the total volume added 1.0 mL and record the
pH as before.
Add more NaOH to make the total volume added 5.0 mL and record the
pH.
Rinse the electrode in distilled water and stand it in a flask of distilled
water.
Take another 25 mL portion of the buffer and record its pH. Record the
pH on the addition of 1 drop, 1.0 mL, and 5.0 mL of 0.1 M HCl in the
same way you did for NaOH. Again, rinse the electrode carefully and
stand it in distilled water.
Put 25 mL of distilled water in a 50 mL beaker and, keeping its
exposure time to the air as short as possible, record its pH. If the water
is absolutely pure its pH will be 7.0, but it is very difficult to achieve
this. If the pH is less than 6.0, wash the beaker and the electrode more
carefully and try again.
When you have a pH between 6.0 and 7.0 for the ‘pure’ water, record
the pH changes on the addition of 0.1 M NaOH and 0.1 M HCl
(separately) just as you did for the buffer solution. Take special care to
wash the electrode when you change from using alkali to acid.
29
Name …………………………………………… ID ………… Sec …..
Experiment 5: HYDROLYSIS OF SALTS AND THE ACTION
OF A BUFFER SOLUTION
REPORT SHEET
A. Hydrolysis of Salts
Table I
Solution
(0.1 M)
Ion expected
to hydrolyze
Spectator
ion(s)
CO32‾
Na+
Al3+
K+, SO42‾
NaCl
Na2CO3
NaC2H3O2
NH4Cl
ZnCl2
KAl(SO4)2
30
Table II
Indicator Color
Solution
pH
[H+]
[OH‾]
H2O (unboiled)
H2O (boiled)
NaCl
Na2CO3
NaC2H3O2
NH4Cl
ZnCl2
KAl(SO4)2
31
Methyl red
Bromothymol
blue
Phenolphthalein
Table III
Solution
(0.1 M)
Net-ionic equation for
hydrolysis
Expression for equilibrium
constant (Ka or Kb)
C2H3O2‾ (aq) + H2O  HC2H3O2 (aq) + OH‾ (aq)
Kb = [HC2H3O2] [OH‾] / [C2H3O2‾]
Zn2+ (aq) + H2O  Zn(OH)+ (aq) + H+ (aq)
Ka = [Zn(OH)+] [H+] / [Zn2+]
Na2CO3
NaC2H3O2
NH4Cl
ZnCl2
KAl(SO4)2
32
Value of
Ka or Kb
B- The Action of a Buffer Solution
Table IV
pH on addition of
0.1 M NaOH to
Volume (total)
/ mL
buffer
pure water
pH on addition of
0.1 M HCl to
buffer
pure water
0.0
1 drop
1.0 mL
5.0 mL
Questions
i) Compare the changes in pH upon addition of NaOH to water and to buffer
solution:
ii) Compare the changes in pH upon addition of HCl to water and to buffer
solution:
33
Experiment 6:
DETERMINATION OF THE SOLUBILITY
PRODUCT CONSTANT FOR A SPARINGLY
SOLUBLE SALT
OBJECTIVE
To become familiar with equilibria involving sparingly soluble substances by
determining the value of the solubility product constant for a sparingly soluble
salt.
DISCUSSION
Inorganic substances may be broadly classified into three different categories:
acids, bases, and salts. According to the BrØnsted-Lowry theory, acids are proton
donors, and bases are proton acceptors. When an acid reacts with a base in
aqueous solution, the products are a salt and water. With a few exceptions, nearly
all common salts are strong electrolytes. The solubilities of salts span a broad
spectrum, ranging from slightly or sparingly soluble to very soluble. This
experiment is concerned with heterogeneous equilibria of slightly soluble salts.
For a true equilibrium to exist between a solid and a solution, the solution must be
saturated. Calcium hydroxide is slightly soluble salt, and in a saturated solution
this equilibrium may be represented as follows:
Ca(OH)2 (s) ⇌ Ca2+ (aq) + 2OH‾ (aq)
[1]
The equilibrium constant for Equation [1] is
Ksp = [Ca2+][OH‾]2
[2]
Ksp is called the solubility product constant. At a given temperature the value of
Ksp is constant. The solubility product for a sparingly soluble salt can be easily
calculated by determining the solubility of the substances in water. Suppose, for
example, we determined that 7.410‾2 g of Ca(OH)2 dissolves in 100 mL of water.
The molar solubility of this solution (that is, the molarity of the solution) is 0.010
M.
34
We see from Equation [1] that for each mole of Ca(OH) 2 that dissolves, one mole
of Ca2+ and two moles of OH‾ are formed. It follows, therefore, that
solubility of Ca(OH)2 in moles/liter = [Ca2+] = ½ [OH‾] = 0.010 M
and
Ksp = [Ca2+][OH‾]2 = [0.010][20.010]2 = 4.010‾6
In a saturated solution the product of the molar concentrations of Ca 2+ and OH‾
cannot exceed 4.010‾6. If the ion product [Ca2+][OH‾]2 exceeds 4.010‾6,
precipitation of Ca(OH)2 would occur until this product is reduced to the value of
Ksp. Or if a solution of NaOH is added to a solution of CaCl2, Ca(OH)2 would
precipitate if the ion product [Ca2+][OH‾]2 is greater than Ksp.
To determine the solubility product constant for a sparingly soluble substance,
we need only to determine the concentration of one of the ions, because the
concentration of the other ion is related to the first one’s concentration by a simple
stoichiometric relationship. Any method that we could use to accurately determine
the concentration would be suitable. In this experiment, you will determine the
solubility product constant Ksp for Ca(OH)2. You will determine the concentration
of hydroxide ion by titration with hydrochloric acid, according to:
H+ (aq) + OH– (aq) → H2O (l)
[3]
The concentration of Ca2+ can be determined by noting that at equilibrium [ Ca2+] =
½ [OH‾].
35
PROCEDURE
1- You will be provided with a bottle containing about 2 g of powdered
calcium hydroxide in 100 mL of distilled water. The bottle has been shaken
well and set aside for more than a day. Minimize shaking the bottle as
possible.
2- Rinse and fill the burette with standardized hydrochloric acid (0.100 M).
3- Filter the contents of the bottle, allowing the first 5 mL to run to waste and
collecting the rest in a dry conical flask. The first few mL are rejected
because they are less concentrated in solute than the rest. The filter paper
absorbs solute until it attains equilibrium with the solution. Yet another
equilibrium!
To minimize absorption of carbon dioxide, steps 4 and 5 should be done
quickly (with due care!)
4- Rinse the pipette with the calcium hydroxide solution and transfer 25.0 mL
to a conical flask (this needs not be dry).
5- Add two drops of phenolphthalein to the flask and titrate the solution until
the pink color just disappears. Record your burette readings in your report
sheet.
6- Repeat steps 4 and 5 for 2 or 3 solutions of calcium hydroxide.
7- Record the temperature.
8- Calculate the concentrations of OH‾ and Ca2+ ions, and Ksp for Ca(OH)2.
36
Name …………………………………………… ID ………… Sec …..
Experiment 6: DETERMINATION OF THE SOLUBILITY PRODUCT CONSTANT
FOR A SPARINGLY SOLUBLE SALT
REPORT SHEET
Ca(OH)2–HCl Titration
Trial
Burette readings
(0.100 M HCl)
/ mL
1
2
3
4
Final
Initial
Volume / mL
Average volume
/ mL
Ksp for Ca(OH)2
Moles of OH‾ = moles H+ = _________________________________ mol
[OH‾] = ______________________________ M
[Ca2+] = ________________________ M
Solubility of Ca(OH)2 = _______________ M
Ksp for Ca(OH)2 = [Ca2+][OH‾]2 = ___________________________________
at _________ oC.
37
Experiment 7:
RATE OF CHEMICAL REACTION: THE
IODINATION OF ACETONE
OBJECTIVE
To measure the effect of concentration upon the rate of the reaction of iodine with
acetone; to determine the order of the reaction with respect to reactant
concentrations; to obtain the rate law for the chemical reaction; and to calculate
the activation energy for the r
eaction.
DISCUSSION
Experiments show that rates of homogeneous chemical reactions in solution
depend upon:
1- The nature of the reactants
2- The concentration of the reactants
3- The temperature
4- The presence of catalysts
Before a reaction can occur the reactants must come into direct contact via
collisions of the reacting particles. However, even then, the reacting particles (ions
or molecules) must collide with sufficient energy to result in a reaction. With these
considerations in mind, we can qualitatively explain how the various factors
influence the rates of reactions.
For a given reaction, the rate typically increases with an increase in the
concentration of any reactant. The relation between rate and concentration is a
simple one in many cases, and for the reaction
aA + bB → Products
the rate can usually be expressed by the equation:
Rate = k [A]m [B]n
38
[1]
Where m and n are usually, but not always, integers, 0, 1, 2 or possibly 3; [A] and
[B] are the concentration of A and B (ordinarily in mol/L); and k is a constant,
called the rate constant of the reaction. The numbers m and n are called the
orders of the reaction with respect to A and B, respectively. If m is 1 the reaction
is said to be first order with respect to reactant A. If n is 2 the reaction is second
order with respect to reactant B. The overall order is the sum of m and n. In this
example the reaction would be third order overall. Equation [1] is called the rate
law for the reaction.
The rate of a reaction is also significantly dependent on the temperature at which
the reaction occurs. An increase in temperature increases the rate, an often-cited
approximate rule being that a 10oC rise in temperature will double the rate.
As with the concentration, there is a quantitative relation between reaction rate
and temperature. This relation is based on the idea that to react, the reactant
species must have a certain minimum amount of energy present at the time the
reactants collide in the reaction step; this amount of energy, which is typically
furnished by the kinetic energy of motion of the species present, is called the
activation energy for the reaction. The equation relating the rate constant k to the
absolute temperature T and the activation energy Ea can be written as:
E
ln k = ln A – a
[2]
RT
Where R is the gas constant (8.314 J/mol.K) and A is a constant, or nearly so, as
temperature is varied. The constant A is called the frequency factor and is related
to the frequency of collisions and the probability that the molecules are suitably
oriented for reaction. By measuring k at different temperatures, we can determine
graphically the activation energy for a reaction.
Catalysts, in some cases, are believed to increase reaction rates by bringing
particles into close juxtaposition in the correct geometrical arrangement for
reaction to occur. In other instances, catalysts offer an alternative route to the
reaction, one that requires less energetic collisions between reactant particles. At
the end of the reaction, the catalyst can be recovered chemically unchanged.
In this experiment we will study the kinetic of the reaction between iodine and
acetone:
O
O

H3C
C
H
H3C
CH3 (aq) + I2 (aq) 
C
CH2I (aq) + I‾ (aq)
[3]
This reaction, in the vicinity of room temperature, proceeds at a moderate,
relatively easy measured rate. The rate of the reaction is found to depend on the
39
concentration of hydrogen ion in the solution as well as presumably on the
concentration of the two reactants. By Equation [1], the rate law for this reaction
is:
Rate = k [acetone]m [H+]n [I2]p
[4]
Where m, n and p are the orders of the reaction with respect to acetone, hydrogen
ion and iodine, respectively, and k is the rate constant for the reaction.
The rate of this reaction can be expressed as the small change in concentration
of I2, Δ[I2], that occurs, divided by the time interval required Δt for the change:
Rate =
 [ I 2 ]
t
[5]
The minus sign is to make the rate positive (Δ[I2] is negative). Ordinarily, since
rates varies as the concentrations of the reactants according to Equation [4], in a
rate study it would be necessary to measure, directly or indirectly, the
concentration of each reactant as function of time; the rate would typically vary
markedly with time, decreasing to very low values as the concentration of at least
one reactant becomes very low. This makes reaction rates studies relatively
difficult to carry out and introduces mathematical complexities that are difficult
for beginning students to understand.
The iodination of acetone is a rather typical reaction, in that it can be easily
investigated experimentally. First of all, iodine has color, so that one can readily
follow changes in iodine concentration visually. A second and very important
characteristic of this reaction is that it turns out to be zero order in I2
concentration. This means (see Equation [4]) that the rate of the reaction does not
depend on [I2] at all no matter what the value of [I2] is, as long as it is not itself
zero.
Because the rate of the reaction does not depend on [I2], we can study the rate
by simply making I2 the limiting reactant present in a large excess of acetone and
H+ ion. We then measure the time required for a known initial concentration of I2
to be completely used up. If both acetone and H+ are present at much higher
concentration than that of I2, their concentrations will not change appreciably
during the course of the reaction, and the rate will remain, by Equation [4],
effectively constant until all the iodine is gone, at which time the reaction will
stop. Under such circumstances, if it takes t seconds for the color of a solution
having an initial concentration of I2 equals [I2]o to disappear, the rate of the
reaction, by Equation 5, would be:
Rate =
 [ I 2 ]  ([I 2 ]  [I 2 ]o ) [I 2 ]o
=
=
t
t  to
t
40
[6]
Although the rate of the reaction is constant during its course under the conditions
we have set up, we can vary it by changing the initial concentrations of acetone
and H+ ion. If, for example, we should, in a reaction Mixture II, double the initial
concentration of acetone over that in Mixture I, keeping [H +] and [I2] at the same
values in both mixtures, then the rate of Mixture II would, according to equation
[4], be different from that of mixture I:
Rate II = k [2A]m [H+]n [I2]0
[7a]
Rate I = k [A]m [H+]n [I2]0
[7b]
Dividing the first equation by the second, we obtain simply:
Rate I
= 2m
Rate II
[8]
Having measured both Rate II and Rate I by equation [6], we can find their ratio,
which must be equal to 2m. We can then solve for m in Equation [8] either by
inspection or using logarithms and so find the order of the reaction with respect to
acetone.
By similar procedure we can measure the order of the reaction with respect to
+
H ion concentration and also confirm the fact that the reaction is zero order with
respect to I2. Having found the order with respect to each reactant, we can then
calculate k, the rate constant of the reaction.
The last part of this experiment is to study the rate of this reaction at different
temperatures to find its activation energy. The general procedure here would be to
study the rate of reaction in one of the mixtures at room temperature and at two
other temperatures, one above and one below room temperature. Knowing the
rates, and hence the k’s, at the three temperatures, you can then find Ea by plotting
ln k versus 1/T. The slope of the resultant straight line, by Equation [2], must be –
Ea/R.
The determination of the orders m and n, the confirmation of the fact that p, the
order with respect to I2, equals zero, the evaluation of the rate constant k for the
reaction at room temperature, and the evaluation of the activation energy Ea for the
reaction comprise your assignment in this experiment.
PROCEDURE
1- Select two regular test tubes; when filled with distilled water, they should
appear to have identical color when you view them down the tubes against
a white background.
41
2- Starting with Mixture 1 in the report sheet, measure out, with a graduated
cylinder, 10.0 mL of the 4.0 M acetone solution and pour it into a clean 125
mL Erlenmeyer flask. Then measure out 10.0 mL of 1.0 M HCl and add
that to the acetone in the flask. Add 20.0 mL distilled water to the flask.
Drain the graduated cylinder, shaking out any excess water, and then use
the cylinder to measure out 10.0 mL of 0.0050 M I2 solution. Be careful not
to spell the iodine solution on your hands or clothes.
3- Noting the time on your stopwatch to 1 second, pour the iodine solution
into the Erlenmeyer flask and quickly swirl the flask to mix the reagents
thoroughly. The reaction mixture will appear yellow because of the
presence of the iodine, and the color will fade slowly as the iodine reacts
with the acetone.
4- Fill one of the test tubes ¾ full with the reaction mixture, and fill the other
test tube to the same depth with distilled water. Look down the test tubes
toward a well-lit piece of white paper, and note the time the color of the
iodine just disappears. Measure the temperature of the mixture in the test
tubes. Record the time and the temperature in your report sheet.
5- Repeat the experiment, using as a reference the reacted solution instead of
distilled water. The amount of time required in the two runs should agree
within 20 seconds.
6- Repeat steps 2-5 for other mixtures listed in your report sheet, making sure
the temperate is kept within about a degree of that in the initial run.
7- Calculate the rate of the reaction by dividing the initial concentration of I2
in the reaction mixture by the elapsed time. Since the reaction is zero order
in I2, and since both acetone and H+ ion are present in great excess, the rate
is constant throughout the reaction and the concentration of both acetone
and H+ remain essentially at their initial values in the reaction mixtures.
8- Determine the value of m and n by calculating the order of reaction with
respect to acetone and H+, respectively, and show that the order of the
reaction with respect to I2 is zero (that is, p = 0). Determine the overall
order of reaction.
9- Having found the order of the reaction for each species on which the rate
depends, evaluate k, the rate constant of the reaction, from the rate and
concentration data in each of the mixtures you studied. If the temperatures
at which the reactions were run are all equal to within a degree or two, k
should be about the same for each mixture. Calculate the average k.
10- Select one of the reaction mixtures you have already used which gave a
convenient time, and use that mixture to measure the rate of reaction at
about 0oC and at about 40oC. From the two rates you found, plus the rate at
room temperature, calculate the energy of activation for the reaction, using
Equation [2].
42
Name …………………………………………… ID ………… Sec …..
Experiment 7: RATE OF CHEMICAL REACTION: THE IODINATION
OF ACETONE
REPORT SHEET
I- Reaction Rate Data
Volume / mL
Mixture
Time
Temp.
st
4.0 M
acetone
1.0 M
HCl
0.0050
M I2
H2O
1
10
10
10
20
2
20
10
10
10
3
10
20
10
10
4
10
10
5
25
Mixture [acetone]
[H+]
[I2]o
0.80 M
0.20 M
0.0010 M
1
2
3
4
43
1
run
nd
2
run
Average
time / s
Rate = [I2]o / ave. time
(mol/L.s)
/ oC
II- Reaction Orders (show your work)
1) with respect to acetone:
m = _________ ≈ _______
2) with respect to H+:
n = _________ ≈ _______
3) with respect to I2:
p = _________ ≈ _______
Overall order of reaction: ________
44
III- Rate Constant
Mixture
1
2
3
4
Average
ln k
1 / T(K)
k
units
Show calculation for k for an exemplary mixture
IV- Activation Energy
Reaction mixture used _______
Temp. / oC
room temp. (a)
Aver. time
/s
Rate /
(mole/L.s)
(b)
(c)
k
(d)
(a), (b) & (c): from part I
(d): average k from part III
Slope of plot of ln k vs. 1/T =
units
Ea =
units
45
Experiment 8:
Oxidation-Reduction Titration: Determination of Oxalate
Objective
To gain some familiarity with redox chemistry through analysis of an oxalate
sample.
Apparatus
50-mL buret
250-mL conical flask (3)
thermometer
glass stirring rods
balance
weighing bottle
hot plate
ring stand and buret clamp
Chemicals
Na2C2O4 (primary standard) (1g)
~ 0.02 M KMnO4 (100 mL)
1.00 M H2SO4 (650 mL)
unknown oxalate sample (1g)
Introduction
Potassium permanganate reacts with oxalate ions in acidic solution as follows:
5C2O42– (aq) + 2MnO4– (aq) + 16H+ (aq) → 10CO2 (g) + 8H2O (l) + 2Mn2+ (aq)
It is necessary to heat the solution gently because this reaction proceeds slowly at
room temperature. The permanganate ion is intensely purple, whereas the
manganese(II) ion is nearly colorless The first slight excess of permanganate
imparts a pink color to the solution, signaling that all the oxalate has been
consumed.
In this experiment, you will standardize a KMnO4 solution-that is, you will
determine its molarity -by titrating it against a very pure sample of sodium oxalate,
Na2C2O4. You will then use your standardized KMnO4 to determine the
percentage of oxalate ion, C2O42–, in an unknown sample. The basis of the
determination is that the reagents react in a molar ratio of 5:2
mol C2O42– = (5/2) × mol MnO4–
Measuring the volume of KMnO4 that reacts with a known mass of sodium
oxalate, Na2C2O4, allows us to calculate the molarity of KMnO4 solution.
46
Procedure
A) Standardization of KMnO4 Solution
Weigh out from a weighing bottle, to the nearest 0.1 mg, triplicate portions of
about 0.08 g each of pure Na2C2O4 into 250-mL conical flasks. Run each trial as
follows: Add about 80 mL of 1.0 M H2SO4; then stir the solution (not with the
thermometer), and warm the mixture until the oxalate has dissolved and the
temperature has been brought to 80 to 90 oC. Titrate with KMnO4, stirring
constantly, while keeping the solution above 70 oC all the times. Add the
permanganate dropwise when near the end point, allowing each drop to decolorize
before adding the next. The end point is reached when the faintest visible shade of
pink remains even after the solution has been allowed to stand for 15 s.
Note: A 0.02 M KMnO4 solution is so deeply colored that the bottom of the
meniscus is difficult to read. Thus, it is necessary to read the top of the surface of
the solution.
B) The Determination of Oxalate in Unknown
Into three separate 250-mL conical flasks, weigh out from a weighing bottle, to the
nearest 0.1 mg, triplicate potions of about 0.12 g of unknown. Add about 80
mL of 1.0 M H2SO4. Titrate with standardized KMnO4 as described above in part
A.
Waste disposal instructions
Dispose all the oxalate-and metal-containing solutions in appropriate containers.
47
Name …………………………………………… ID ………… Sec …..
Experiment 8: OXIDATION-REDUCTION TITRATION: DETERMINATION
OF OXALATE
REPORT SHEET
A) Standardization of KMnO4 Solution
Mass of Na2C2O4
Trial 1
Trail 2
Trial 3
Weighing bottle initial mass
___________________
____________________
____________________
Weighing bottle final mass
___________________
____________________
____________________
Mass of Na2C2O4
____________________
____________________
____________________
Final reading
____________________
____________________
____________________
Initial reading
____________________
___________________
____________________
Volume of KMnO4
____________________
____________________
____________________
Moles of Na2C2O4
____________________
____________________
____________________
Moles of KMnO4
____________________
____________________
____________________
Molarity of KMnO4
____________________
____________________
____________________
Titration volume of KMnO4
Calculations
Average molarity
____________________________________
Standard deviation (show calculations)
____________________________________
48
B) Analysis of Oxalate Unknown
Mass of sample
Trial 1
Trail 2
Trial 3
Weighing bottle initial mass
___________________
____________________
____________________
Weighing bottle final mass
___________________
____________________
____________________
___________________
____________________
____________________
Final reading
____________________
____________________
____________________
Initial reading
____________________
___________________
____________________
Volume of KMnO4
____________________
____________________
____________________
Mass of sample
_
Titration volume of KMnO4
Calculations
Moles of KMnO4
____________________
____________________
____________________
Moles of oxalate, C2O42–
____________________
____________________
____________________
Mass of oxalate, C2O42–
____________________
____________________
____________________
Percent oxalate, C2O42–
____________________
____________________
____________________
Average percent oxalate
____________________________________
Standard deviation (show calculations)
____________________________________
49