Download Laboratory Exercises in Physical Chemistry

Laboratory Exercises
in Physical Chemistry
A.Jakubowska, S.Lamperski, G.Nowicka, W.Nowicki
Editor in chief: Stanisław Lamperski
Adam Mickiewicz University
Faculty of Chemistry
Poznań 2005
Contents
Introduction ....................................................................................................................................... 2
Enthalpy of neutralisation ............................................................................................................... 3
Calculation of the molar entropy of oxygen at the boiling point ............................................. 7
Distribution of acetic acid between two immiscible solvents ................................................. 16
Cryoscopic determination of the molecular mass ..................................................................... 18
Dependence of the buffer capacity on the composition of a mixture of a weak acid
and its salt with a strong base....................................................................................................... 21
Spectrophotometric determination of the equilibrium constant of an acid-base
indicator............................................................................................................................................ 24
Determination of the thermodynamic solubility constant of lead (II) chloride ................... 29
Determination of the silver electrode potential ......................................................................... 33
Determination of the Gibbs energy, enthalpy and entropy changes in the
electrochemical cell reaction from the temperature dependence of the cell potential........ 36
Determination of the molar conductivity of a strong electrolyte ........................................... 40
Kinetics of sucrose inversion......................................................................................................... 44
Determination of the rate constant and the activation energy of the ester
hydrolysis reaction ......................................................................................................................... 47
Determination of the critical micelle concentration of ionic surfactants ............................... 52
Determination of the diffusion coefficient.................................................................................. 55
Adsorption from solution. Determination of the specific surface area of activated
carbon................................................................................................................................................ 58
1
Introduction
Welcome to the laboratory course in Physical Chemistry. This brochure will
explain to you how to perform particular exercises. Descriptions of each experiment
are proceeded by relevant theoretical introduction. The theory is needed to
understand the idea of the experiment. You can get the theory from the handbook
“Physical Chemistry” written by P.W.Atkins (Oxford University Press, sixth
edition). Study it before reading the description of the exercise. In the brochure we
refer to the equations presented in the Atkins’ handbook. Atkins’ equation numbers,
preceded by the number of chapter, are given in the parentheses (e.g. Eqn (25.12)
means equation 12 from the chapter 25), while the numbers of our equations are
given in the square brackets (e.g. Eqn [7]). If possible, we use the same symbols and
notation as Atkins did.
A note “Exercise No: ##” given just below the title of the exercise will help you to
find your exercise in the laboratory room.
Many exercises require the application of the least-squares fitting method. The
necessary programme (nmk.exe) is installed in the computers located in the
laboratory room. Your tutor will explain to you how to run and use this programme.
Alternatively you can use the Excel programme.
We would also like to thank Mrs Barbara Stoińska for her technical assistance.
Poznań, 2005
Authors
2
1. Enthalpy of neutralisation
Exercise No: 2
Theory
Atkins: 2.2 (Introduction) + 2.2(a) (pp. 48-49), Work and heat + 2.3+2.3(a)2.3(e), 2.4+2.4(a) (pp. 51-56), 2.5+2.5(a)-2.5(b) (without Example 2.2) (pp. 5759), 5.7(c) (pp. 69-70)
The enthalpy, H, like the internal energy, U, is a thermodynamic state
function. The enthalpy is related to the internal energy by the equation:
H = U + pV
(2.23)
where p and V are the pressure and volume, respectively.
The internal energy or the enthalpy is used in mathematical representation
of the first law of thermodynamics. So, according to the first law of
thermodynamics, the enthalpy and internal energy changes (denoted as: dH
and dU, respectively) accompanying an elementary process occurring in
closed system are equal to
d U = d q − p d V + d we
[1]
d H = d q + V d p + d we
[2]
where dq is the heat and dwe is the work other than that due to a volume
change (the so-called the “non-volumetric work”, called also the additional
work). The expression of (– pdV) represents the work of expansion or
contraction (called the “volumetric work”).
For any elementary process taking place at a constant volume or at a
constant pressure, if the non-volumetric work is not done, Eqn [1] and Eqn
[2] become
3
dU = d qV
(2.15)
d H = d qp
(2.24a)
Then, in this case, the total heat of a chemical reaction or other process is
independent of the path of the process but depends only on the initial and
final states of the system under consideration. This statement is known as
Hess’s law, which is a consequence of the application of the first law of
thermodynamics to chemical reactions.
In this experiment we determine the neutralisation enthalpy, ∆nH, for the
reaction between a strong acid (H2SO4) and a strong base (KOH)
H2SO4 + 2KOH → K2SO4 + 2 H2O
The experiment consists of three parts:
1) determination of the calorimeter constant, C;
2) determination of the dissolution enthalpy of H2SO4, ∆disH;
3) determination of the total heat effect, ∆totH, accompanying the addition
of H2SO4 to the KOH solution.
Experimental
In this experiment a thermos bottle is used as a calorimeter. The thermos
bottle is closed with a lid.
Determination of the calorimeter constant, C:
1) fill the thermos bottle with 300 ml of distilled water, put a stirrer into it
and take temperature-time readings at 10-second intervals until the
difference between the successive readings becomes close to zero (T0);
2) weigh 10 g of KCl (a substance with a known dissolution enthalpy), add
it into the thermos bottle, begin stirring and take temperature-time
readings at 10-second intervals until the difference between successive
readings becomes close to zero (T1).
Determination of the dissolution enthalpy of H2SO4:
1) fill the thermos bottle with 300 ml of distilled water, put a stirrer into it
and take temperature-time readings at 10-second intervals until the
difference between successive readings becomes close to zero (T0);
2) begin stirring and add exactly 5 ml of concentrated H2SO4 (density, ρ =
1.84 g·cm-3) from a cylinder. Take temperature-time readings in the way
described above (T1).
Determination of the total heat effect, ∆totH, accompanying addition of H2SO4 to
KOH solution:
4
1) fill a beaker with 300 ml of distilled water, dissolve a quantity of KOH
necessary for neutralisation of 5 ml of concentrated H2SO4 and lead the
solution to the temperature of distilled water used earlier (see above in
determination of T0);
2) fill the thermos bottle with KOH solution, put a stirrer into it, begin
stirring, add exactly 5 ml of concentrate H2SO4 and take temperaturetime readings in the way described above (T1).
Calculations
The calorimeter with the reacting substances can be regarded as an isolated
system for which ∆H = 0. Therefore we can write
C·∆T (for calorimeter) + ∆rH (for reactants) = 0.
Consequently, the calorimeter constant can be determined from the
equation:
C = −∆ r H / ∆T
[3]
where ∆T is the temperature change obtained from the experiment by
plotting temperature readings against time (Fig. 1).
temperature, T, [K]
T1
∆T
T0
time, t, [s]
Fig.1. Graphical determination of values of ∆T = T1 – T0
We can write the relation:
5
∆ r H = n i ⋅ ∆ r H m i = ( mi / M i ) ⋅ ∆ r H m i
[4]
where: ∆rH – the enthalpy of the reaction in kJ,
ni - the number of moles of the i-th reactant,
∆rHm,i – the molar enthalpy of the reaction of the i-th substance
expressed in kJ·mol-1
mi – the weight of the i-th reactant in grams,
Mi – the molecular weight of the i-th reactant in g·mol-1
We can combine Eqn [3] with Eqn [4] to obtain the expression:
C = − (mi / M i ) ⋅ ∆ r H m i / ∆T
[5]
1) Determine the calorimeter constant, C, from Eqn [5]. Use the molar
enthalpy of dissolution of KCl: ∆rHm,KCl = 18.33 kJ·mol-1.
2) Determine the molar enthalpy of dissolution of H2SO4, ∆disHm, and next the
total molar heat effect, ∆totHm, accompanying addition of H2SO4 to KOH
solution from Eqn [5] assuming that the i-th reactant is H2SO4.
3) Estimate the molar neutralisation enthalpy, ∆nHm, of the reaction H2SO4
with KOH for 1 mol of H2O from the equation:
∆ tot H m = ∆ dis H m + 2 ∆ n H m
6
[6]
2. Calculation of the molar entropy of oxygen at the
boiling point
Theoretical exercise
Theory
Atkins: 4.2+4.2(a)-(b) (pp. 99-104), 4.3+4.3(a)-(c) (without Example 4.3) (pp.
106-108), 4.6 (pp. 113-118)
In thermodynamics, the entropy, S, is defined by the expression
dS =
d q rev
T
[1]
where qrev is the heat exchanged between the system and its surrounding
and T is temperature of the process. For the finite isothermal process Eqn [1]
takes the form
∆S =
q rev
T
[2]
Let us consider the change in the entropy of three fundamental
processes.
1. Isothermal expansion
The heat of the isothermal reversible expansion of perfect gas from the initial
volume Vi to the final volume Vf is
q rev = nRT ln
Vf
Vi
So the entropy change, ∆S, accompanying this process is
7
[3]
∆S = nR ln
Vf
Vi
[4]
2. Phase transition
As the change in enthalpy is equal to the heat supplied at a constant
pressure:
dH = dq p or ∆H = q p , p=const.
(2.24)
the entropy change at a constant pressure is given by
∆S =
∆H
, p=const.
T
[5]
Analogous expression for the entropy of the phase transition ∆trsS is :
∆ trsS =
∆ trs H
Ttrs
(4.16)
where ∆trsH and Ttrs are the enthalpy and the temperature of the phase
transition, respectively.
3. Heating at a constant pressure
We start from the definition of a constant pressure heat capacity
 ∂H 
Cp = 

 ∂T  p
(2.27)
Using Eqn (2.24) we can transform Eqn (2.27) into
d q rev = C p d T , p=const.
[6]
When introducing this expression into Eqn [1] and after integrating it from
the initial temperature Ti to the final Tf we obtain
Tf
S(Tf ) = S(Ti ) +
∫
Ti
Cp dT
T
(4.19)
When Cp is independent of temperature, the above integral can be solved
analytically:
8
Tf
∫
S(Tf ) = S(Ti ) + C p
Ti
T
dT
= S(Ti ) + C p ln f
T
Ti
(4.20)
However, in the general case, when Cp depends on temperature, we use Eqn
(4.19) and it is convenient to perform integration numerically.
The entropy of a system at the temperature T related to the entropy at
T=0 can be evaluated from Eqn (4.19) and if in the temperature range of
interest the phase transition exists, also Eqn (4.16) must be used. If we want
to apply Eqn (4.19) to the real system, some modifications are needed. The
first one concerns the entropy at T=0. We must make use of the third law of
thermodynamics according to which the entropy of a system at T=0 equals
zero, S(0)=0. The second modification concerns the heat capacity at
temperatures close to 0 K, where it is extremely difficult to measure Cp.
Here, the Debye extrapolation is usually applied. According to the Debye
theory [Eqn (11.11)] at low temperatures the heat capacity is proportional to
T3,
C p = aT 3
[7]
Substituting it into Eqn (4.19) we obtain
T
∫0
S(T ) =
aT 3
dT = a
T
T
∫0 T
2
dT = 13 aT 3 = 13 C p (T )
[8]
If phase transitions occur in the investigated range of temperatures, also the
corresponding entropies of phase transitions [Eqn(2.16)] should be included.
Finally we have:
S(T ) = 13 C p (T1 ) +
Ttrs1
∫T
1
C p (T ) dT
T
+
∆ trs1 H
+
Ttrs1
Ttrs2
C p (T ) dT
trs1
T
∫T
+
∆ trs2 H
+ ... [9]
Ttrs2
The integrals in Eqn [9] can be evaluated by fitting a polynomial to the
experimental heat capacity data and integrating this polynomial analytically.
Example (Atkins, exercise 4.18)
Calculate the molar entropy of anhydrous potassium hexacyanoferrate (II) at
T = 200 K using the following heat capacity data:
Table 1
Heat capacity of anhydrous potassium hexacyanoferrate (II)
9
T/K
Cp,m / J K-1 mol-1
10.0
20.0
30.0
40.0
50.0
60.0
70.0
80.0
90.0
100.0
110.0
150.0
160.0
170.0
180.0
190.0
200.0
2.09
14.43
36.44
62.55
87.03
111.0
131.4
139.4
165.3
179.6
192.8
237.6
247.3
256.5
265.1
273.0
280.3
Solution
1. Debye extrapolation
According to Eqn [8]
S p , m (10) = 13 C p , m (10 ) = 13 ⋅ 2.09 = 0.70 J/mol ⋅ K
2. Integration
a) Fitting a polynomial to the experimental data
10
300
-1
Cp,m / J K mol
-1
250
200
150
Coefficients:
b0=-32.78628
b1= 2.729234
b2=-6.743575e-3
b3= 4.391396e-6
100
50
0
-50
0
50
100
150
200
250
T/K
Fig. 1. The third degree polynomial fitted to the capacity-temperature data
of anhydrous potassium hexacyanoferrate (II)
The polynomial fitted to the capacity data has the form:
C p , m (T ) = b0 + b1T + b2T 2 + b3T 3
with the coefficient given in Fig. 1.
b) Analytical integration of the polynomial from 10 K to 200 K.
S p , m ( 200 ) − S p , m (10) =
=
b0 + b1T + b2T 2 + b3T 3
dT
T = 10
T
∫
200
200
2
∫T =10 (b0 /T + b1 + b2T + b3T )d T
[
= b0 ln T + b1T + 21 b2T 2 + 13 b3T 3
]
200
T = 10
200
+ b1 (200 − 10 ) + 21 b2 (200 − 10 )2 + 13 b3 (200 − 10 )3
10
= 297.51 J/mol ⋅ K
= b0 ln
11
The total molar entropy at T=200 K amounts 0.70+297.51 = 298.21 J/mol·K.
Exercise
Calculate the molar entropy of oxygen at the boiling point (90.13 K)
using the data of the heat capacity (Table 2) and the enthalpy of phase
transition (Table 3) determined by Giauque and Johnstone1. Fit the
polynomial to the heat capacity data separately for each temperature
interval. Try different degrees of the polynomials to get the best fitting.
Collect the results in a table similar to Table 4.
Table 2
Heat capacity of oxygen
T/K
12.97
14.14
15.12
15.57
16.66
16.80
16.94
18.13
18.32
18.45
19.34
20.26
20.33
20.85
21.84
22.24
22.24
Transition at 23.66 K
25.02
25.61
25.61
26.75
28.00
1
Cp/J mole–1 K–1
4.602
6.360
6.694
7.489
9.749
9.121
9.414
11.171
11.339
11.673
12.845
14.644
14.728
15.062
17.573
17.866
18.410
22.68
23.31
22.89
24.06
25.31
W.F.Giauque and H.L.Johnstone, J.Am.Chem.Soc., 51(1929)2300
12
28.08
26.86
29.88
27.67
30.63
29.04
31.08
28.99
33.05
31.46
33.33
32.34
34.41
33.81
35.57
34.56
35.77
35.52
37.59
37.99
37.85
38.16
38.47
41.00
39.99
41.00
40.18
41.51
40.67
42.51
42.21
44.89
Transition at 43.76 K
45.90
46.11
47.76
46.32
48.11
46.07
48.97
45.98
50.55
46.07
51.68
46.15
52.12
46.28
Melting point at 54.39 K
13
60.97
61.48
65.57
65.92
68.77
69.12
70.67
71.38
73.31
74.95
75.86
77.58
78.68
81.13
82.31
82.96
84.79
86.43
86.61
86.97
87.32
90.13
53.18
53.18
53.18
53.18
53.26
53.35
53.43
53.47
53.60
53.76
53.56
53.72
53.68
53.89
53.81
53.89
54.10
54.02
54.18
54.06
54.02
54.35
Table 3
Enthalpy of the phase transition of oxygen
Ttrs/K
23.66
43.76
54.39
90.13
∆trsH/J mole–1
93.81
743.08
444.76
6814.90
Table 4
Collection of the results
S/J mole–1 K–1
0 – 11.75 K (Debye extrapolation)
11.75 – 23.66 K (integration of polynomial)
14
Phase transition at 23.66 K
23.66 – 43.76 K (integration of polynomial)
Phase transition at 43.76 K
43.76 – 54.39 K (integration of polynomial)
Fusion at 54.39 K
54.39 – 90.13 K (integration of polynomial)
Vaporization at 90.13 K
Entropy of oxygen at the boiling point
15
3. Distribution of acetic acid between two
immiscible solvents
Exercise No: 15
Theory
Atkins: The thermodynamic description of mixtures + 7.1(a)-(b) (pp. 163167), 7.3+7.3(a) (pp. 171-173), Activities + 7.6 (pp. 182-183), 6.4 (pp. 145-146)
Consider a two-phase system composed of two immiscible solvents, 1
and 2. When we add the third component, which is soluble in both solvents,
this component (solute) will be distributed between the two phases. At
equilibrium the chemical potentials of the solute in both phases are equal
µ1 = µ2
[1]
The chemical potentials can be described as functions of activities a1, a2 of the
solute in the two phases, respectively:
µ 1 = µ 01 + RT ln a1
and
µ 2 = µ 02 + RT ln a2
[2]
and µ 01 , µ 02 are the standard chemical potentials. Substituting Eqn [2] into
[1] we obtain
RT ln
a1
= µ 02 − µ 01
a2
[3]
At a constant temperature the standard chemical potentials, µ 01 and µ 02 , are
constant, so we can write
16
a1
= const ≡ ka
a2
[4]
which is the exact form of the so-called distribution law. The ratio, equal to
the constant in Eqn [4] is commonly referred to as the distribution ratio. For
the ideal or nearly ideal solutions, the activities can be replaced by the of the
mole fractions x1 and x2. Further, for the dilute solutions the concentrations
c1 and c2 can be used in place of mole fractions and thus Eqn [4] takes the
form
c1
= const ≡ k c
c2
[5]
This is the mathematical form of the distribution law. It says that: at a
constant temperature the dissolved substance, irrespective of its total amount,
distributes itself between two immiscible or slightly miscible liquids at a constant
concentration ratio.
This form of the distribution law does cannot be applied to the case when
the solute associates or dissociates in one or in both phases.
Experimental
The experimental goal is to determine the distribution ratio of the acetic
acid between water and butanol. To do it pour into four clean and dry
bottles 10 cm3 of butanol and than 10 cm3 of the aqueous solution of acetic
acid of concentrations: 0.025, 0.035, 0.05, 0.075 and 0.10 mol/dm3. Leave the
bottles, after tight stopping, for attaining the equilibrium distribution,
shaking the contents every 5 minutes. In the meantime determine, by the
titration with the 0.01 M NaOH solution, the exact concentration co of acetic
acid in the stock solutions. When the equilibrium distribution is attained,
which takes about 1 hour, separate the two liquid phases from each bottle,
using separating funnels. The lower phase is the aqueous solution.
Determine by titration with the 0.01 M NaOH solution, the concentration cw
of the acetic acid in the aqueous phases. Calculate the distribution ratio kc for
each sample from the equation:
kc =
cw
c0 − cw
[6]
Note, that in the above determination of kc the dissociation of acetic acid in
water is neglected and the absence of its association in butanol is assumed.
17
4. Cryoscopic determination of molecular mass
Exercise No: 17
Theory
Atkins: 7.5(a)-(c) (pp. 177-179)
The freezing-point depression of a solution that results from addition of
a small amount of a solute is called a colligative property. For dilute
solutions it depends on the number of the solute particles, but not on any of
the properties of the solute particles. These particles could be small
molecules, macromolecules or ionic species. Only the number of these
particles in a given amount of the solvent affects the freezing point
depression as well as the other colligative properties (vapour-pressure
lowering, boiling-point elevation, osmotic pressure).
The dependence between the freezing-point depression, ∆T, and the
solute concentration can be expressed, for dilute solutions, by the equation:
∆T = K x 2
[1]
where K is the so-called freezing-point depression constant (also called the
cryoscopic constant) and x2 is the mole fraction of a solute. Since, for dilute
solutions the mole fraction x2 is equal to n2/(n1+n2), where n1 and n2 are the
numbers of moles of the solvent and the solute, respectively, Eqn [l] can be
written as:
∆T = K
n2
m2 / M 2
=K
n1 + n1
m1 / M 1 + m2 / M 2
[2]
where m1, m2 and M1, M2 denote masses and molar masses of the solvent and
the solute, respectively. Eqn [2], after simple transformations, gives:
18
m
 K

M2 = 
− 1  M1 2
∆
T
m1


[3]
and can be used for the determination of the molecular mass of substances.
Experimental
In the experiment the freezing point of a solution containing a known
weight of an “unknown” solute in a known weight of water is determined
from the cooling curves. The shapes of the cooling curves that may be
obtained are presented in Figure l. In the experiment a precise platinum
resistance thermometer is employed.
Prepare, by weighing, the solution of an “unknown” substance (about
0.5 g) in water (about 5 g). The weights of the solvent and the solute should
be known exactly. Tubes as well as the thermometer and the stirrer should
be clean and dry.
Prepare in a beaker an ice-salt cooling mixture by mixing about one part
by volume of NaCl with about four parts by volume of finely crushed ice.
Temperature
4
2
solvent
0
-2
solution
-4
Time
Fig. 1. Schematic cooling curves: (a) and (c) show the cooling curves for pure
solvent; (b) and (d) show the cooling curves for a solution.
19
Insert the thermometer and the stirrer into the test tube containing the
solvent. Mount the thermometer in such a way that the sensing element is
concentric with the tube so that the stirrer can easily pass around it. Place
the tube into a cooling bath of temperature –5 to –10°C and immediately
start the stirring of the solution. The success of this experiment much
depends on the technique of stirring. The motion of the stirrer should carry
it from the bottom of the tube up to the surface of the liquid. The stirring
should be continuous throughout the run and should be at the rate of about
1 stroke per second. The outer bath should be stirred a few times per minute.
Follow carefully the readings of the millivoltometer. Find, as shown in Fig.
1, the readings AH2O (in mV) corresponding to the freezing point of water.
Repeat the experiment for the prepared solution and find the readings A (in
mV) of the milivoltmeter corresponding to the freezing point of the solution.
The freezing-point depression, ∆T, of the examined solution can be
found from Eqn [4]:
(
∆T = 4 ⋅ 10 −3 A − AH 2 O
)
[4]
Substituting the obtained ∆T value, masses of solute and water used for the
preparation of the solution, molecular mass of water and the cryoscopic
constant of water, K = 103.1 K, into Eqn [3] calculate the molecular mass of
the “unknown” substance.
20
5. Dependence of the buffer capacity on the
composition of the mixture of a weak acid
and its salt with a strong base
Exercise No: 21
Theory
Atkins: 9.5+9.5(a)-(d) (pp. 229-235), 10.8+10.8(a) (pp. 267-268)
Aqueous mixtures of a weak acid and its salt with a strong base are
buffer solutions. Such solutions exhibit constant concentration of hydrogen
ions, which remains practically constant in time of dilution or after addition
of a small amount of strong acids or bases. This property is known as the
buffer action. The buffer action of a solution of a weak acid, HA, and its
highly ionised salt (conjugate base), A–, in the presence of hydrogen or
hydroxyl ions added is explained by the reactions:
–
H3O+ + A H2O + HA
–
–
OH + HA H2O + A
In this experiment we study the buffer capacity, β, against the composition
of the acetate buffer solutions. The buffer capacity is defined as the number
–
of moles of H3O+ or OH ions that must be added to 1 dm3 of the buffer
solutions for the unit change in pH. Practically, the buffer capacity is
expressed as the ratio of the change in the salt concentration, ∆cs, and the
change in pH, ∆(pH), on addition of a very small amount of a strong acid or
base to the buffer solution
β = ( ∆c s ) /( ∆(pH))
21
[1]
The buffer capacity reaches its maximum, βmax, in the solution containing
equivalent amounts of a weak acid and its salt and then pH = pKa [Eqn
(9.45)].
Experimental
1. Prepare 13 buffer solutions in a 50-mL beakers from the stock solutions
(0.02 mol·dm-3 CH3COOH and 0.02 mol·dm-3 CH3COONa) according to
Table 1.
Table 1
No. Vs [cm3]
1
2
3
4
5
6
7
8
9
10
11
12
13
3,5
5,0
6,5
8,5
10,5
12,5
14,5
16,5
18,5
20,0
21,5
23,0
23,5
VHA [cm3]
pH0
pH1
∆(pH) = |pH1 – pH0| ∆cs
β
21,5
20,0
18,5
16,5
14,5
12,5
10,5
8,5
6,5
5,0
3,5
2,0
1,5
2. Put a stirrer into the first beaker containing the buffer solution studied
and begin stirring, immerse the glass electrode into the solution and
measure pH of the buffer (pH0). In the same way measure the pH
(denoted as pH0) of the consecutive buffer solutions.
3. Pipette 0.5 cm3 of 0.1 mol·dm-3 HCl into the each beaker containing the
buffer solution.
4. Measure the pH (denoted as pH1) of the consecutive buffer solutions in a
way described above (p.2.).
5. Collect the experimental results of the pH0 and the pH1 in Table 1.
22
Calculations
1. Calculate the change in the salt concentration, ∆cs, on addition of HCl to
the buffer solution, from the following equation:
∆c s = (VHCl ⋅ cHCl )/(Vt ⋅ VHCl )
[2]
where Vt is the total volume of the buffer solution: Vt = VHA + Vs.
2. Calculate the buffer capacity, β, of the each buffer solution studied
using Eqn [1].
3. Plot β versus the pH0, read βmax and pH corresponding to βmax.
23
6. Spectrophotometric determination of the
equilibrium constant of an acid-base indicator
Exercise No: 26
Theory
Atkins: 9.5(e) (p. 235), 16.2 (pp. 458-459)
Acid-base indicators are usually weak organic acids. In an aqueous
solution they exists as acids (HIn) and their conjugate bases (In–). The
equilibrium between these form is given by the expression:
HIn (aq) + H2O (l) In– (aq) + H3O+ (aq)
[1]
The two forms of the indicator differ in colour. As the concentration of the
indicator is very small, we can assume that the activity of water is equal to 1
and we can replace the activities by the molar concentrations. Then the
equilibrium constant of [1] is given by the formula
K In ≈
[In − ] [H 3 O + ]
[HIn]
[2]
Using the definition of the pH
pH = − log aH
3O
+
≈ − log [H 3 O + ]
(9.27)
and after some mathematical transformations, Eqn [2] takes the form
log
[HIn]
≈ pK In − pH
[In − ]
24
(9.51)
It follows from this equation that at pH = pKIn the concentrations of HIn and
In– are equal. Thus the aim of this exercise to measure the dependence of
[HIn] and [In–] against pH using the spectrophotometric method and to find
the value of the pH at which [HIn] = [In–]. The experiment consists of two
parts:
1) measurement of the analytical wavelength
2) determination of pKIn.
1. Measurement of the analytical wavelength, λanalytical
1.1. Prepare the following solutions of the indicator:
A: 5 cm3 of 0.0001 M solution of the indicator and 5 cm3 of 0.1 M HCl dilute
with water in the volumetric flask to 50 cm3;
B: 5 cm3 of 0.0001 M solution of the indicator and 5 cm3 of 0.1 M NaOH
dilute with water in the volumetric flask to 50 cm3.
The indicator is mainly in its acid form (HIn) in solution A, and in the basic
form (In–) in solution B.
1.2. Measure the absorbance of both solutions at the wavelengths given in
Table 1. Note down the results in your report. Prepare a graph showing the
dependence of the absorbance, A, of solutions A and B on the wavelength, λ.
Find the wavelength at which the difference in the absorbance between A
and B is the largest. This wavelength, λanalytical, is called the analytical
wavelength.
2. Determination of pKIn
2.1. Prepare 9 solutions of the indicator in a buffer at pH close to pKIn. The
buffer will ensure the stability of pH and on the basis of its composition you
will be able to calculate the pH of the solution.
Prepare the following solutions:
Add to each of the 9 volumetric flasks 5 cm3 of 0.0001 M solution of the
indicator, 20 cm3 of 0.1 M acetic acid and subsequently 2, 4, 6, …, 18 cm3 of
0.1 M NaOH. Dilute all these solutions with water to 50 cm3. You have
prepared in all the flasks the buffer composed of a weak acid (acetic acid)
and its salt (sodium acetate). The pH of such a buffer is given by:
pH = p K a + log
25
[salt ]
[a cid]
[3]
where Ka is the acidity constant. Concentration of the salt in the buffer
prepared is proportional to the volume of 0.1 M NaOH, VNaOH, added to
each volumetric flask
[salt ] ~ VNaOH
[4]
and the concentration of the acid is proportional to the volume of the nonneutralized acid
[acid ] ~ 20 − VNaOH
[5]
Now, Eqn [3] takes the form:
pH = 4.76 + log
VNaOH
20 - VNaOH
[6]
as pKa of the acetic acid equals 4.76. Calculate pH for each of the solutions
and note down the results in a table similar to Table 2.
2.2. Measure the absorbance of each solution at the analytical wavelength.
Note down the results in a table.
The overall absorbance, A, of a solution composed of the acid and base
forms of the indicator can be expressed as the sum of absorbances, AHIn, AIn–,
of pure forms multiplied by the corresponding mole fractions
A = AHIn ([HIn]/[HIn]0) + AIn– ([In–]/[HIn]0)
[7]
where [HIn]0 is the total concentration of the indicator:
[HIn]0 = [HIn] + [In–]
[8]
This concentration is exactly the same concentration of the indicator you
have prepared in the volumetric flasks. Calculate this concentration.
Solving the set of Eqns [7] and [8] we obtain expressions for [HIn] and
[In–]
[HIn] = [HIn]0
A − AIn −
AHIn − AIn −
[In-] = [HIn]0 – [HIn]
[9]
[10]
Calculate these concentrations, note them down in a table and prepare a
graph showing the dependence of [HIn] and [In–] on pH. Read out from this
26
graph the value of the pH at which the curves intersect. This value
corresponds to pKIn.
Table 1
Absorbance of solutions A and B
Wavelength, λ
[nm]
400
420
440
460
480
500
520
540
560
580
600
620
640
660
680
700
Absorbance, A
Solution A
Solution B
λanalytical =
AHIn =
AIn– =
Table 2
Influence of pH on the concentration of acid and base forms of the indicator
VNaOH
[cm3]
2
4
6
8
pH
A
27
[HIn]
[In–]
10
12
14
16
18
[HIn]0 =
28
7. Determination of the thermodynamic solubility
constant of lead (II) chloride
Exercise No: 19
Theory
Atkins: 10.2 (without Justification) + 10.2 (pp. 248-253), 10.7 (without
Example) + 10.6 (p. 266)
One can discuss the solubility of a sparingly water soluble salt A ν + B ν - in
terms of the equilibrium:
+
-
A ν + B ν - (solid) ν + A z (aq) + ν − B z (aq)
[1]
The thermodynamic equilibrium constant of the above reaction is given
by Eqn [2]:
ν
 a
 +  a 
z+
z−
Ka =  A   B 
aA ν B ν
+
where a
Az
+
,a
Bz
−
and aA ν
+
B ν−
ν−
[2]
−
stand for the equilibrium activities of the
species in the saturated salt solution. Since the activity of the precipitate,
aA ν B ν , can be assumed constant, one can write:
+
−
K L , a = K a  a z+ 
 A 
ν+
 a 
−
 Bz 
ν−
= (a± )ν
[3]
where KL ,a is called the solubility constant or the solubility product, a± - the
mean activity, and ν = ν+ +ν– .
29
Substituting the ion activities, ai , by the product of the ion
concentrations, ci , and the activity coefficients, γi , of ions one can obtain:
K L , a =  c z+ γ z+ 
 A A 
ν+
 c γ

−
− 
 Bz Bz 
ν−
= KL, c K γ
[4]
where:
K L , a =  c z+ 
 A 
ν+
 c 
−
 Bz 
ν−
[5]
and
K γ =  γ z+ 
 A 
ν+
 γ

− 
 Bz 
ν−
= (γ ± ) ν
[6]
The mean activity coefficient of the ions, γ± , can be estimated from the
extended Debye-Hückel law:
log γ ± = −
A z+ z−
Ic
[7]
1 + B d Ic
where A and B are constant for a given solvent and temperature, zi is the
charge number of an ion i, d-the effective diameter of ions and Ic is the ionic
strength of the solution defined as:
Ic =
∑ zi2 ci
1
2
[7]
i
Combining Eqn [4] with Eqn [6] one can obtain:
log K L , c = log K L , a − ν log γ ±
[9]
Substituting Eqn [7] into Eqn [9] leads to:
log K L , c = log K L , a + ν
Thus, the dependence of log K L , c on
A z+ z−
1 + B d Ic
Ic
1 + B d Ic
the intercept equal to log K L , a .
30
Ic
[10]
should be rectilinear with
Experimental
The objective of the experiment is determination of the solubility constant,
K L , a of lead(II) chloride by analysis of chloride ion concentrations in the
aqueous solutions of potassium nitrate of various concentrations saturated
with PbCI2. KNO3 influences the ionic strength of the solution but it does not
have a common ion with PbCI2. The chloride ion concentration is
determined by the potentiometric titration with the silver nitrate standard
solution in the presence of the ion selective electrode.
Pipette into dry beakers 10 cm3 of the supplied saturated solutions of
PbCI2 in water and in aqueous KNO3 solutions of concentrations: 0.05, 0.10,
0.20, 0.30, 0.40 and 0.50 mole/dm3. Take care not to introduce the precipitate
into a pipette. Place the beaker filled with the solution on the magnetic
stirrer, immerse the reference and ion selective electrodes beforehand
connected to the potentiometer and start adding the standard solution of
AgNO3 from the burette. Follow carefully changes in the readings of the
potentiometer upon the addition of titrant and note down a titrant volume at
which the changes become distinct. Repeat the titration of the same solution
(and also the same volume) quickly introducing a certain titrant volume that
will not cause the distinct changes in the potential of the ion selective
electrode, yet. Afterwards, continue the titration by introducing small
volumes of the titrant and read simultaneously the cell potential, E, on the
potentiometer. End the titration when the changes in E start to decrease.
Determine the titrant volume, VEP, corresponding to the equivalence
point, EP. To this end the Hahn’s method is advised. Find the highest value
∆E
, where ∆E is the change in the cell potential brought about
of the ratio:
∆V
by the addition of the ∆V portion of the titrant. Then calculate the qa value
from Eqn [11]:
qa =
∆E1
2 ∆E2
[11]
where ∆E1 and ∆E2 denote changes in the cell potential that are on both sides
of the highest change ∆Emax and ∆E1>∆E2. Assuming that near EP the ∆V
portions of the titrant are constant, calculate a correction a (in cm3) from Eqn
[12]:
a = ∆V q a
[12]
If ∆E1 is previous to ∆E2 add the correction a to the titrant volume Vmax i.e.
31
the total titrant volume corresponding to ∆Emax. In the opposite case subtract
a from Vmax.
On the basis of the obtained results calculate the values of the ionic
strength of the solutions (from Eqn [8]) and the corresponding K L , c values
from Eqn [5]. Plot the dependence of log K L , a on
Ic
1 + B d Ic
assuming
B=3.287 [dm3/2 mole–1/2 nm–1] and d=0.3 [nm]. Show in this plot the
experimental points together with the straight line fitted by the least-squares
method. Find the value of K L , a from the intercept.
32
8. Determination of the potential of the silver electrode
Exercise No: 23
Theory
Atkins: Electrochemical cells + 10.3+10.4+10.5+10.6 (pp. 253-266)
Consider the following electrochemical cell:
g(s)|AgCl(s)|KCl(aq,saturated) AgNO3(aq, c M)+KF(aq, 0.1–c M)|Ag(s) [1]
The two reduction half-reactions are
R:
L:
Ag+(aq) + e– → Ag(s)
–
–
AgCl(s) + e → Ag(s) +Cl (aq)
[2a]
[2b]
Subtracting from the right-hand half-reaction the left-hand one, we obtain
the reaction corresponding to the cell considered:
Ag+(aq) + Cl– (aq) → AgCl(s)
[3]
with the reaction quotient, Q, equal:
Q=
aAgCl
aAg + aCl −
[4]
We can assume that the pure AgCl has unit activity and that the activity of
Cl– ions in the saturated solution also equals 1. Thus, the expression for the
reaction quotient reduces to:
Q=
1
aAg +
33
[5a]
or
Q=
1
[5b]
c Ag + γ ±
Note that the right-hand cell compartment contains the electrolyte of a
constant ionic strength: AgNO3 of concentration c M and KF of concentration
(0.1–c) M. Thus, the mean activity coefficient, γ±, is independent of c.
We use in our laboratory a commercial silver/silver chloride electrode with
a salt bridge mounted in it. The salt bridge reduces significantly but not
completely the liquid-junction potential, Elj,. Therefore, the precise
expression for the overall cell potential must include Elj:
E = E0 (Ag + / Ag ) − E0 (AgCl / Ag, Cl − ) + Elj +
RT
ln c Ag + γ ±
F
[6]
After rearranging [6] we obtain the expression for the potential of the silver
electrode:
E + E0 (AgCl / Ag, Cl − ) − Elj = E0 (Ag + / Ag ) +
RT
ln c Ag + γ ±
F
[7]
or
E(Ag + / Ag ) = E0 (Ag + / Ag ) +
RT
RT
ln γ ± +
ln cAg +
F
F
[8]
where
E(Ag + / Ag ) = E + E0 (AgCl / Ag , Cl − ) − Elj
[9]
Experimental
1. Use the supported stock solutions (0.1 M KF and 0.09 M KF + 0.01 M
AgNO3) to prepare in the volumetric flasks the solutions of AgNO3 of
concentrations given in Table 1 (prepare the fifth solution from the third
one).
2. Wash up the electrodes in the distilled water, dry them with the blotting
paper and dip in the solution examined.
3. Connect the electrodes to the digital voltmeter and read the cell
potential, E, when the potential readings stabilises . If the electrodes are
connected correctly, then E is positive.
34
4. Repeat the measurements for the remaining solutions. Note down the
results in a table.
Calculations
1. Calculate the potential of the silver electrode, E(Ag+/Ag), from Eqn [9]
using the measured values of the cell potential, E, and assuming
E0(AgCl/Ag,Cl–) = 0.2042 V, E1j = 0.008 V.
2. Fit a straight line to the experimental points: E(Ag+/Ag) vs. (RT/F)
lncAg+ (Eqn [8]) using the least-squares method. Calculate the mean
activity coefficient, γ±, from the intercept (E0(Ag+/Ag) + (RT/F)lnγ±)
assuming that E0(Ag+/Ag) = 0.799 V.
3. Show the dependence of E(Ag+/Ag) on (RT/F)lncAg+ on a graph
presenting both the experimental points and the least-squares result.
Table 1
Collection of the results
c Ag + / M
( RT / F ) ln c Ag +
E/V
E(Ag+/Ag)/V
0.01
0.0025
0.001
0.0004
0.0001
2
Note that the exact value of E0(AgCl/Ag,Cl–) equals 0.220 V, the difference is due to
our assumption that activity of chloride ions in the saturated KCl is equal 1.
35
9. Determination of the Gibbs energy, enthalpy
and entropy changes in the electrochemical
cell reaction from the temperature dependence
of the cell potential
Exercise No: 24
Theory
Atkins: 10.9 (pp. 268-270)
A galvanic cell consists of two half-cells combined in such a way that
after their connection with a metallic conductor the electric charges start
flowing across this circuit. A half-cell is a system of two or more conducting
phases being in contact, one of them is an electronic (metallic) conductor (the
so-called electrode) and the other is an electrolytic conductor (electrolyte). At
the surface of separation between the electrode and the electrolyte there is a
difference in the electric potential and electrons or ions flow across the
interfacial surface. Then, the electrochemical reactions of oxidation and
reduction occur. In a galvanic cell the reaction of oxidation takes place at the
negative half-cell (anode) and the reaction of reduction occurs at the positive
half-cell (cathode)
anode reaction [-]
Red1 → Ox1 + n e–
cathode reaction [+]
Ox2 + n e– → Red2
where n is the stoichiometric coefficient of electrons, e–, participating in a
half-cell reaction.
When a cell is open (no current flows in a cell), a galvanic cell stays in a state
of electrochemical equilibrium. Under these conditions, the zero-current cell
potential, E, is equal to the difference of potentials between half-cells.
36
When a galvanic cell is closed, it means when its electrodes are connected by
a metallic conductor, the flow of electrons and ions takes place and the
electric current flows in the cell (the spontaneous anode and cathode
reactions occur inside it). In this case, the galvanic cell is not in a state of
electrochemical equilibrium. However, during the flow of infinitesimally
small currents across the circuit, the system is always virtually in
equilibrium (the quasistatic process) and the galvanic cell does the
maximum electrical work, we,max,
we, max = − n F E
[1]
where n is stoichiometric coefficient of electrons in an equation of a half-cell
reaction and F is Faraday’s constant.
In a thermodynamically reversible process, at a constant temperature (T)
and pressure (p), the maximum work other than that due to a volume
change is equal to the Gibbs energy change, we,max = ∆rG (the Gibbs energy,
G, is a thermodynamic state function). So, when the thermodynamically
reversible (quasistatic) electrochemical reaction takes place in the galvanic
cell, it is consequently possible to write
∆rG = − n F E
(10.43)
The Gibbs energy change accompanying the process at T = const is related to
the enthalpy and entropy changes by the equation:
∆ r G = ∆ r H − T∆ r S
(10.61)
 ∂∆ G 
− ∆ rS =  r 
 ∂T  p
[2]
Further, since
It can be shown that differentiation of Eqn (10.43) with respect to
temperature at a constant pressure leads to the result
∆ r S = n F (dE / dT)
(10.60)
In this experiment we measure the potential of the Clarc cell at different
temperatures , which allows determinations of the Gibbs energy, ∆rG,
enthalpy, ∆rH, and entropy, ∆rS, changes for the cell reaction.
The Clarc cell is shown in Fig. 1.
37
Fig.1. The Clarc cell
This cell is denoted
[-] 7% Zn in Hg|ZnSO4⋅7H2O (solid)|saturated solution of ZnSO4|Hg2SO4 (solid)|Hg [+]
where the vertical bars denote phase boundaries.
The overall chemical reaction taking place in the cell is
Zn + Hg2SO4 ZnSO4 + 2Hg
[3]
Experimental
The Clarc cell is placed in the thermostat bath. The cell potential, E, is
measured using the high-resolution digital voltmeter with large input
impedance. Then the cell work has nearly thermodynamically reversible
character.
1. Determine E as a function of temperature at 5-degree intervals in the
range from 293 to 313 K: for each temperature the Clarc cell should be
thermostated for a minimum of 10 min. before the measurement of E.
The digital voltmeter must be switched off between the measurements.
2. After ending the measurements switch off the digital voltmeter and
disassemble the measurement circuit.
Calculations
1. Plot E versus the absolute temperature, T. The equation relating E to T is
E=AT+B
38
[4]
where A is the temperature coefficient of the cell potential, dE/dT.
2. Use linear regression to determine A.
3. Determine ∆rG (in kJ·mol–1), ∆rS (in kJ·K–1mol–1) and ∆rH (in kJ·mol–1) for
the Clarc cell reaction using the following equations:
Eqn (10.43) for ∆rG for each temperature,
Eqn (10.60) for ∆rS valid for the whole temperature interval,
Eqn (10.61) for ∆rH for each temperature.
39
10. Determination of the molar conductivity of
a strong electrolyte
Exercise No: 5
Theory
Atkins: 24.7 (pp. 737-740)
The molar conductivity, Λm, is defined as
Λm =
κ [ S m −1 ]
c [mol m − 3 ]
[1]
where κ is the conductivity and c the molar electrolyte concentration. To
ensure consistence with the SI units the concentration in this equation must
be given in moles per cubic meter. However, in the laboratory practice we
usually express concentration in moles per cubic decimetre (litre). Eqn [1]
adapted to these units takes the form
Λm =
[ S m −1 ]
κ
1000 c [dm 3 m − 3 ] ⋅ [mol dm − 3 ]
[2]
where c is the electrolyte concentration in mol/dm3, now.
The Kohlrausch’s law says that the molar conductivity, Λm, of a strong
electrolyte depends linearly on the square root of the concentration, c
Λ m = Λ0m − Κ c 1 2
[3]
In this equation Κ is a constant while Λ0m is the limiting molar conductivity.
The word “limiting” means that we consider the molar conductivity in the
40
limit of zero electrolyte concentration, at which the ion–ion interactions can
be neglected.
The conductivity, κ, is directly related to the conductance, G:
G=
κA
[S ]
l
[4]
Here l is the distance between the conductivity cell electrodes and A is the
electrode cross-section area. Knowing that the conductance is the reciprocal
of the resistance, R:
1
R
[5]
l
RA
[6]
G=
we have
κ=
For a given conductivity cell the parameters l and A are constant, so we can
simplify Eqn [6] by introducing the constant, p, of the conductivity cell
p=
l
A
[7]
κ=
p
R
[8]
to obtain
When the value of the constant p is known, determination of the
conductivity κ and the molar conductivity, Λm, reduces to the measurement
of the resistance, R.
The shape of the presently used conductivity cell is different from that
shown in Fig. 24.15 (Atkins). It is similar to a probe and you just dip it into
the solution examined. Therefore, we call it the dipping conductivity
electrode. When you want to use it remember to wash it up in the distilled
water before each measurement, then shake the cell down to remove the
excess of water and immerse for several times in the examined solution to
remove air bubbles which can be present in the electrode compartment. The
solution studied must be placed in the thermostat for minimum 15 minutes
before the measurement. The thermostat temperature should be set at 25°C.
41
Use the Wheatstone bridge to measure the resistance. The scheme of the
Wheatstone bridge is shown in Fig. 1. The most important elements of the
bridge are: the decade resistor Rp and the oscilloscope. Measure the
resistance of the 0.01 KCl electrolyte. To do it you must find such a
resistance on the decade resistor which gives the smallest signal (sinusoidal
vibrations) on the oscilloscope screen. The bridge in the laboratory is
constructed in such a way that the resistance, Rp, settled on the decade
resistor corresponds exactly to the resistance R of the examined electrolyte.
Repeat the measurement for three times, calculate the mean value of the
resistance and use it to determine the constant, p, of the conductivity
electrode from Eqn [8]. The conductivity of 0.01 M KCl solution, κ, is 0.14114
S·m–1.
Fig. 1. The Wheatstone bridge
Prepare in the volumetric flasks solutions of NaCI according to the
concentrations given in Table 1. Measure the resistance, Rp, of each
electrolyte in the same way as you did with the KCl solution. Using the
determined earlier value for p calculate κ (Eqn [7]) and Λm (Eqn [2]). Collect
all the results in a table similar to Table 1.
42
The aqueous solution of NaCI is a strong electrolyte which fulfils the
Kohlrausch’s law described by Eqn [3]. Fit a straight line to your Λm vs. c1/2
data using the linear least-squares method. The equation of the straight line
is y = a+bx where a corresponds to the limiting molar conductivity, Λ0m , of
the electrolyte studied while x and b to c1/2 and K in Eqn [3], respectively.
Plot the molar conductivity results (experimental data and the fitted line) as
a function of c1/2.
Table 1
Collection of results
c
c1/2
-3
[mole dm ] [mole dm-3]1/2
0.01
0.1
0.0081
0.09
0.08
0.0064
0.0049
0.07
0.0036
0.06
0.0025
0.05
0.04
0.0016
Rp
[Ω]
RKCl =
p=
Λ0m =
43
<Rp>
[Ω]
κ
[S m–1]
Λm
[S m2 mol–1]
11. Kinetics of sucrose inversion
Exercise No: 12
Theory
Atkins: 25.2+25.3 (pp. 763-771)
In the presence of hydrogen ions, sucrose in aqueous solution hydrolyses
to fructose and glucose according to the reaction:
Sucrose as well as the products of its hydrolysis are optically active
substances, that is they rotate the plane of polarisation of the plane-polarised
light. The specific rotations3 of sucrose and inverted sucrose are
considerably different, as shown in the diagram:
3
Specific rotation [ α]Tλ is defined as:
[α ]Tλ =
α
dc
where α is the rotation observed in degrees if the light of the wavelength λ passes
through d decimeters of the solution of the active substance of concentration c
expressed in g/cm3. T denotes the experimental temperature.
44
Thus, the observation of the optical activity behaviour of the reaction
mixture with time is a very convenient and easy way to follow the reaction
of sucrose hydrolysis. Note, that the instantaneous sucrose concentration, cs,
in the reaction mixture is proportional to the difference between the angle of
optical rotation αt measured at time t of the reaction and the angle α∞
measured after the reaction is complete.
Since in the examined reaction of the sucrose hydrolysis, H+ ions act as
the catalyst and the changes in water concentration can be neglected, the rate
ν of the reaction can be described by:
ν=−
d cs
= k cs
dt
[1]
where k is the specific reaction rate, or simply a rate constant, applicable to
the specified conditions (temperature and, in the present case, hydrogen-ion
concentration) which are constant throughout the course of the reaction.
If Eqn [1] is integrated with the initial condition c s = c s0 at t = 0, where
c s0 is the initial sucrose concentration, we obtain:
ln c s = − k t + ln c s0
[2]
Since c s is proportional to αt – α∞ and c s0 to α0 – α∞ , Eqn [2] can be written
as:
ln ( α t − α ∞ ) = − k t + ln ( α 0 − α ∞ )
[3]
where α0 is the rotation angle of the sucrose solution at time t = 0.
Experimental
1. Connect to the mains the polarimeter Polamat, the thermostat (set up the
temperature on 35 – 38°C) and the water bath (set up 60 – 70°C).
2. Prepare 100 cm3 of the sucrose aqueous solution of concentration 0.24
g/cm3.
45
3. Put into a clean and dry beaker 25 cm3 of the sucrose solution prepared
and 25 cm3 of HCI aqueous solution of the concentration 2 mole/dm3.
Rinse and next fill the polarimeter cell with the reaction mixture. Place
the cell into the thermostated polarimeter chamber. After the desired
temperature is reached (which takes about 15 minutes) start the timing.
Note down the polarimeter readings together with the corresponding
reaction time t.
4. In order to determine α∞ pour the remaining reaction mixture into a
clean and dry stoppered vessel, put the vessel into the water bath of a
temperature between 60–70°C and heat it for about one hour. Next, cool
the reaction mixture, refill, after rinsing, the polarimeter cuvette with the
mixture and place it in the polarimeter chamber. After the temperature
of the reaction is reached, read the rotation angle of the solution which
can be assumed as equal to α∞.
5. Plot the dependence of ln(αt – α∞) on t indicating the experimental points
as well as the straight line fitted with the linear least-squares method.
6. Find the rate constant, k, of the reaction examined from the slope of the
ln(αt – α∞) versus t dependence.
46
12. Determination of the rate constant and the
activation energy of the ester hydrolysis reaction
Exercise No: 13
Theory
Atkins: 25.5 (pp. 775-777)
Before reading this exercise repeat the theoretical introduction to the previous
exercise.
The rate, v, of the second order reaction
A+B→C
[1]
is proportional to the concentration of each reactant raised to the first power:
v = k [A ][B]
[2]
The coefficient k is the rate constant of the reaction and the expression [2] is
called the rate law of the reaction. To define the rate constant we can use the
concentration of the reactants or the product(s):
v=−
d [A ]
d [B] d [C ]
=−
=
dt
dt
dt
[3]
For equal concentrations of the reactants, [A]=[B], and using Eqn [3], the rate
law of the reaction [1] can be written
d [A ]
= − k [A ] 2
dt
Its integrated form is
47
[4]
kt =
1
−
1
(25.12b)
[A] [A] 0
where [A]0 denotes the concentration of A at t=0.
The example of the second-order reaction is the very early stage of ester
hydrolysis
CH3COOC2H5 + OH– → CH3COO– + C2H5OH
[5]
When the initial concentrations of reactants are equal, [ester]0 = [OH–]0 =
[A]0, the integrated rate law is given by Eqn (25.12b). After simple
transformations this equation takes the form
kT =
1 [A ]0 − [A ]
t [A ]0
[A ]0
[6]
where the subscript T indicates the temperature of the reaction.
According to the reaction [5] the ions OH– and CH3COO– give rise to the
overall conductance. As the conductance of OH–, GOH–, is several times
higher than tha of CH3COO–, GCH3COO–, we can monitor the progress of the
reaction by measuring the conductance of the solution. The overall
conductance at the time t is a sum of the conductance of both kinds of ions
multiplied by the corresponding mole fractions and the constant
conductance of cations, GM+, which are also present in the reacting solution:
[A]0 - [A]
[A]
+ GCH COO −
+ GM + =
3
[A]0
[A]0
[A]0 - [A]
[A]
+ GM +
+ GCH COO − + GM +
3
[A]0
[A]0
Gt = GOH −
(
= GOH −
)
)
(
[7]
It is convenient to rewrite Eqn [7] in the form
Gt = G 0
[A]0 - [A]
[A]
+ G∞
[A]0
[A]0
[8]
where G0 indicates the conductance of the solution at t = 0 while G∞ the
conductance when the reaction would theoretically proceed to the end
(according to the reaction [5]). Simple arithmetic transformations of Eqn [8]
lead to the expression for [A]0/[A];
48
[A]0 G0 − G∞
=
[A] Gt − G∞
[9]
which can be used to express the rate constant [6] in terms of conductance.
As
[A]0 − [A] [A]0
G − Gt
=
−1= 0
[A]
[A]
Gt − G ∞
[10]
Eqn [6] can take the form
kT =
1 G0 − Gt
t [A]0 Gt − G∞
[11]
The conductances G0 and Gt, are easy to measure during the experiment. The
problem is with G∞ as the reaction [5] does not proceed to the end but stops
when reaches the equilibrium. We can overcome this problem noting that
Eqn [11] can be transformed into the form
Gt = G ∞ +
1
G0 − Gt
+
kT [A]0
t
[12]
Now, we can fit a straight line, y = a+bx, to the Gt vs. (G0 – Gt)/t data and
calculate the rate constant from the slope, b:
kT =
1
b [A]0
[13]
The temperature dependence of the rate constant is given by the Arrhenius
equation (Eqn (25.24)). When the rate constants are known for two different
temperatures T1 and T2, one can create a set of equations
ln kT1 = ln A −
Ea
R T1
ln kT2 = ln A −
Ea
R T2
[14]
and solve it for the activation energy, Ea:
Ea =
 kT
T1 T2
R ln  2
 kT
T2 − T1
 1
49




[15]
The activation energy is the minimum kinetic energy that the reactants must
have in order to form products (Atkins, p. 777). The higher the activation
energy, the stronger the influence of temperature on the rate constant and
the rate of reaction.
Experimental
1. Switch the thermostat on and set up the temperature at approximately
30°C.
2. Pipette 40 cm3 of 0.02 M CH3COOC2H5 solution into one beaker and the
same amount of 0.02 M NaOH into another one. Place both beakers into
the thermostat bath.
3. Switch on the conductometer. Wash up the conductivity electrode in the
distilled water, shake the electrode down to remove the excess of water,
and place the electrode in one of the solutions (to warm it up).
4. Read the temperature after approximately 15 minutes of heating the
solutions.
5. Mix both solutions together and stir the resulting solution properly.
6. Read G0 immediately after mixing solutions. Do not forget to dip the
electrode for several times in the examined solution before each
conductance measurements. Read out the conductance, Gt, at intervals of
3 minutes in the time of 30 minutes. Note down the readings in a table
similar to Table 1.
7. Repeat the measurements at 50°C using newly made CH3COOC2H5 and
NaOH solutions.
Calculations
1. Fit a straight line, y = a+bx, to the Gt vs. (G0 – Gt)/t data using the leastsquares method. Calculate the rate constants, kT1 and kT2, from the slope
of the straight line, b, using Eqn [13]. Note that the concentration [A]0
after mixing the solutions amounts to 0 01 M.
2. Calculate the activation energy from Eqn [15].
Table 1
Collection of results
T=
t [s]
[K]
G0 =
Gt [S]
[S]
(G0 – Gt)/t [S · s-1]
180
50
360
540
…
1800
51
13. Determination of the critical micelle
concentration of ionic surfactants
Exercise No: 9
Theory
Atkins: 23.9 (pp. 702-706)
The goal of the experiment is to determine the critical micelle
concentration (CMC) of two ionic surfactants: sodium dodecyl sulfate,
CH3(CH2)11OSO3Na, MW 288.38 D, abbreviated further as SDS and cetyl
trimethyl ammonium bromide, [CH3(CH2)15N+(CH3)3]Br–, MW 364,46 D,
abbreviated as CTAB.
CMC is the threshold concentration of surfactant at which micellization,
i.e. aggregation of the surfactant molecules, begins. Usually, CMC is
determined from the discontinuity of some physical properties of the
surfactant solution on varying concentration. In the case of ionic surfactants
the CMC can be easily found from the dependence of the solution
conductance on the surfactant concentration. The tendency of the
conductance to rise with the increasing surfactant concentration in the
solution decreases at CMC owing to the lower mobility of larger micelles.
Experimental
Students are provided with the aqueous stock solutions of 8.00⋅10–2
mole/dm3 SDS and 6.00⋅10–3 mole/dm3 CTAB. The procedure of the CMC
determination proceeds as follows. To a beaker, placed on the magnetic
stirrer and containing a volume V = 100 cm3 of water, add small portions (1
cm3) of the surfactant stock solution from a burette. After introduction of
each portion of the surfactant stock solution into the beaker wait 1 minute
52
and then measure the conductance, G, of the resulting solution. The total
volume of the surfactant solution added should amount 25 cm3. Calculate
the surfactant concentration cs, from Eqn [1]:
cs =
c s0 Vs0
[1]
V + Vs0
where Vs0 is the volume of the added surfactant stock solution of
concentration c s0 .
Calculations and Results
The plot of G versus cs should contain two rectilinear regions as shown in
Figure 1. By rectilinear regression analysis, find the best slopes a1, a2 and
intercepts b1, b2 for each linear region. The CMC value of a given surfactant
can be found from the intersection of the two straight lines. Calculate this
point from Eqn [2]
c sCMC =
b1 − b 2
a 2 − a1
[2]
G
12
8
4
0
0.00
cmc
0.01
Cs
53
0.02
Fig. 1. Conductance, G, versus surfactant concentration, cs. The Figure
illustrates a method of the CMC determination.
54
14. Determination of the diffusion coefficient
Exercise No: 6
Theory
Atkins: Diffusion + 24.10(a)-(c) + 24.11(a) (pp. 746-751)
When the concentration of a dissolved substance in various points of a
solution is different, we observe the migration of the substance from the
regions of higher concentration to those of lower concentration. This
migration is called the diffusion. In one dimensional diffusion, the rate of the
diffusion, dn/dt, of dn moles of dissolved substance that passes through the
area A in the time dt is proportional to the concentration gradient, dc/dx:
dn
 ∂c 
= − AD 
dt
 ∂x  t
[1]
The constant D is called the diffusion coefficient. The SI units of D are meter
squared per second (m2/s).
A major problem in diffusion coefficient measurement is to make sure
that the solute movement is solely a result of diffusion, i.e., random
molecular motion, and not by bulk liquid movement, e.g., by convection.
Bulk movement is prevented if the liquid is held in a gel. A gel is a
comparatively rigid structure composed of a three-dimensional open
network of polymer chains, the voids filled with liquid. Small molecules of
solvent or solute can move about in a gel as freely as in an ordinary liquid.
This can easily be shown by measuring the conductivity of an ionic solute in
a cooling solution of agar in water. The conductivity falls as the temperature
falls, but there is no discontinuity as the mobile liquid turns into a firm gel.
A convenient experimental set for the diffusion coefficient measurement
consists of a cylinder of agar gel formed inside a glass cylinder. The gel
55
contains a concentrated solution of the solute under study. The glass
cylinder (reservoir) is held vertical with the lower open end immersed in a
beaker with a large volume of the stirred solvent. At the moment of
immersion the solvent contains no solute, and the solute concentration from
then on increases with time. Diffusion through the gel is in one dimension
only, down the vertical axis. A mathematical analysis of this system, given
by Hadgraft4, shows that if less than 5% or so of the total amount of the
solute in the gel has diffused out, then the number of the solute moles nt
diffused out at the time t is:
nt = 2 c 0 A
Dt
π
[2]
where c0 is the initial solute concentration in the gel, D denotes the diffusion
coefficient of the solute and A - the cross-section of the cylinder. By
monitoring the solute concentration in the beaker it is possible to determine
the solute diffusion coefficient. Note, that the solute concentration ct in the
beaker after time t is small but finite, and it can be calculated from Eqn [3]:
c t = nt /Vr
[3]
where Vr is the volume of the solvent in the reservoir. Combining Eqns [2]
and [3] we obtain:
ct 2 A Dt
=
c 0 Vr
π
[4]
Experimental
The goal of the experiment is to determine the diffusion coefficient of
potassium chloride (KCl). Changes in KCl concentration in the beaker can be
monitored conductometrically.
Dissolve about 0.4 g KCl in 50 cm3 of water in a beaker. Dilute with
distilled water 5 cm3 (V1) of the prepared solution (of concentration c0) to 250
cm3 (V2) into a volumetric flask. Measure the conductance G0' of the dilute
solution, whose concentration is further denoted as c 0' . Boil the remaining 45
cm3 of the initial solution of concentration c 0 , add l g of agar crystals and
4
Hadgraft,J. Int. J. Pharmaceut. 1979, 2, 177-194
56
dissolve them on stirring. Measure precisely the inner diameter, d, of the
glass cylinder, pour the hot solution into this cylinder and leave it to cool.
When the room temperature of the gel is reached, which takes about 45
minutes, clamp the cylinder vertically at such a height that the lower end
will be approximately 2 cm above the bottom of the 250-cm3 beaker
containing 150 cm3 (Vr) of water with a dipping conductivity cell immersed
beforehand. Note down the water conductance, GH2O. The water should be
vigorously stirred by a magnetic stirrer. At the moment of the immersion of
the cylinder filled with gel into the reservoir start a stopwatch. In suitable
time intervals (1, 2, 5, 10, 20, 30, 40, 55, 70 and 90 min) measure the
conductance Gt of the solution in the beaker.
Results
At low electrolyte concentrations there is no significant error in
assuming that conductance is linearly proportional to concentration. Thus,
we can assume that Gt is proportional to ct, and G0' is proportional to c 0' .
V
Taking into account that c 0' = c 0 1 and A = π ( d / 2 )2 , where d is the
V2
cylinder diameter, Eqn [4] can be transformed into:
d 2 V2
Gt
=
G0 2 V1 Vr
πD t
[5]
Using Eqn [5] calculate the values of D for different t. Calculate the diffusion
coefficient of KCl as the arithmetic mean of the obtained D values. Plot Gt
against
t . Submit the obtained data and results.
57
15. Adsorption from solution. Determination of
the specific surface area of activated carbon
Exercise No: 11
Theory
Atkins: The extend of adsorption + 28.3+28.4 (pp. 857-863)
Many solids are able to adsorb solutes from solution, and the
relationship between the adsorbed amount and the concentration of the
solute (adsorbate) in the solution at equilibrium is often described by the
Langmuir adsorption isotherm. For the adsorption of solutes from solution
the isotherm takes the form:
x=
xm K c
1+ Kc
[1]
where x represents the amount of the solute adsorbed (in moles or grams per
unit mass of adsorbent), c is the equilibrium concentration of the solute, i.e.,
the solute concentration in the solution that is in equilibrium with the
adsorbent, xm is the limiting amount of adsorbate that can be taken up by
unit mass of adsorbent, and K is a constant. Both K and xm are constant for
the particular system being studied and for a given temperature. Eqn [1] can
be rearranged into:
1
1 1
1
=
+
x xm K c x m
[2]
Thus a plot of 1/x against 1/c should be linear with a gradient l/xm K and
intercept on the 1/x axis of l/xm. By fitting the parameters of Eqn [2], using
e.g. the least-squares method, to experimental data (x and c) one can find the
58
xm value. Using this xm value (in moles per a unit mass of the adsorbent) one
can determine the approximate specific area S of the adsorbent:
S = xm N s a
[3]
where N is the Avogadro constant and sa is the area occupied by one
molecule of the adsorbate.
Experimental
The adsorbent used in the experiment is a commercial activated carbon.
The adsorbate is the acetic acid
Prepare from the aqueous stock solution of acetic acid of the
concentration 0.20 mole/dm3 five CH3COOH solutions of a volume 100 cm3
and the following concentrations: 0.15, 0.12, 0.09, 0.06 and 0.03 mole/dm3.
Determine, by the titration with NaOH in the presence of phenophtaleine,
the exact concentrations of the prepared solutions (including the stock
solution).
Adsorption isotherm is obtained as follows. Accurately weighed samples
of activated carbon of about 1 g are placed in six separate conical flasks
containing 50 cm3 of CH3COOH solutions of varying concentrations. The
flasks are then stoppered and left for one hour to reach equilibrium (in the
meantime they should be often shaken). At the end of this time the acetic
acid solutions are separated from the activated carbon by filtration and the
CH3COOH equilibrium concentrations are determined by the titration with
the standard NaOH solution.
On the basis of the obtained results plot the isotherm (x versus c) of the
CH3COOH adsorption on the activated carbon. Find the limiting amount of
moles of acetic acid that can be taken up by the unit mass of activated
carbon, xm, by fitting Eqn [2] to the experimental data, using the leastsquares method. Present the l/x against l/c dependence graphically drawing
the calculated regression straight line between the experimental points.
Then, assuming the area occupied by one molecule of acetic acid sa = 21·10–20
A2, calculate from Eqn [3] the specific surface area of activated carbon.
59