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

1

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

2

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

3

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. 57-59), 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:

pVUH += (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

edddd wVpqU +−= [1]

edddd wpVqH ++= [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

4

Vdd qU = (2.15)

pdd qH = (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:

5

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 temperature-time 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:

THC ∆∆−= /r [3]

where ∆T is the temperature change obtained from the experiment by plotting temperature readings against time (Fig. 1).

time, t, [s]

temperature, T, [K

]

T1

T0

∆T

Fig.1. Graphical determination of values of ∆T = T1 – T0 We can write the relation:

6

iiiii HMmHnH mrmrr )/( ∆⋅=∆⋅=∆ [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:

THMmC iii ∆∆⋅−= /)/( mr [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:

mnmdismtot 2 HHH ∆+∆=∆ [6]

7

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

T

qS revd

d = [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

T

qS rev=∆ [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

i

frev ln

V

VnRTq = [3]

So the entropy change, ∆S, accompanying this process is

8

i

flnV

VnRS =∆ [4]

2. Phase transition As the change in enthalpy is equal to the heat supplied at a constant pressure:

pqdHd = or pqH =∆ , p=const. (2.24)

the entropy change at a constant pressure is given by

T

HS

∆=∆ , p=const. [5]

Analogous expression for the entropy of the phase transition ∆trsS is :

trs

trstrs

T

HS

∆=∆ (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

p

pT

HC

∂

∂= (2.27)

Using Eqn (2.24) we can transform Eqn (2.27) into

TCq p dd rev = , 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

( ) ( ) ∫+=f

i

dif

T

T

p

T

TCTSTS (4.19)

When Cp is independent of temperature, the above integral can be solved analytically:

9

( ) ( ) ( )i

fiif ln

df

i

T

TCTS

T

TCTSTS p

T

T

p +=+= ∫ (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,

3aTCp = [7]

Substituting it into Eqn (4.19) we obtain

)(dd)(313

31

0

2

0

3

TCaTTTaTT

aTTS p

TT==== ∫∫ [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:

...d)(d)(

)()(trs2

trs2

trs1

trs113

1 trs2

trs1

trs1

1

+∆

++∆

++= ∫∫ T

H

T

TTC

T

H

T

TTCTCTS

T

T

pT

T

pp [9]

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)

10

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

2.09 14.43 36.44 62.55 87.03 111.0 131.4 139.4 165.3 179.6 192.8 237.6

160.0 170.0 180.0 190.0 200.0

247.3 256.5 265.1 273.0 280.3

Solution

1. Debye extrapolation According to Eqn [8]

KJ/mol 70.009.2)10()10(31

m,31

m, ⋅=⋅== pp CS

2. Integration

a) Fitting a polynomial to the experimental data

11

T / K

0 50 100 150 200 250

Cp,m / J K

-1 m

ol-1

-50

0

50

100

150

200

250

300

Coefficients:b0=-32.78628b1= 2.729234b2=-6.743575e-3b3= 4.391396e-6

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:

33

2210m, )( TbTbTbbTCp +++=

with the coefficient given in Fig. 1.

b) Analytical integration of the polynomial from 10 K to 200 K.

( )

[ ]

( ) ( ) ( )

KJ/mol 51.297

10200102001020010

200ln

ln

d/

d)10()200(

333

1222

110

200

10

333

1222

110

200

10

23210

200

10

33

2210

m,m,

⋅=

−+−+−+=

+++=

+++=

+++=−

=

=

=

∫

∫

bbbb

TbTbTbTb

TTbTbbTb

TT

TbTbTbbSS

T

T

Tpp

12

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 Cp/J mole–1 K–1

12.97 14.14 15.12 15.57 16.66 16.80 16.94 18.13 18.32 18.45

4.602 6.360 6.694 7.489 9.749 9.121 9.414 11.171 11.339 11.673

19.34 20.26 20.33 20.85 21.84 22.24 22.24

12.845 14.644 14.728 15.062 17.573 17.866 18.410

Transition at 23.66 K

25.02 25.61 25.61 26.75 28.00

22.68 23.31 22.89 24.06 25.31

1 W.F.Giauque and H.L.Johnstone, J.Am.Chem.Soc., 51(1929)2300

13

28.08 29.88 30.63 31.08 33.05 33.33 34.41 35.57 35.77 37.59 37.85 38.47 39.99 40.18 40.67 42.21

26.86 27.67 29.04 28.99 31.46 32.34 33.81 34.56 35.52 37.99 38.16 41.00 41.00 41.51 42.51 44.89

Transition at 43.76 K

45.90 47.76 48.11 48.97 50.55 51.68 52.12

46.11 46.32 46.07 45.98 46.07 46.15 46.28

Melting point at 54.39 K

14

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 ∆trsH/J mole–1

23.66 43.76 54.39 90.13

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)

15

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

16

3. Distribution of acetic acid between two immiscible solvents

Exercise No: 15

Theory Atkins: The thermodynamic description of mixtures + 7.1(a)-(b) (pp. 163-167), 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

21 µ=µ [1]

The chemical potentials can be described as functions of activities a1, a2 of the solute in the two phases, respectively:

1011 ln aRT+µ=µ and 2

022 ln aRT+µ=µ [2]

and 01µ , 0

2µ are the standard chemical potentials. Substituting Eqn [2] into

[1] we obtain

01

02

2

1ln µ−µ=a

aRT [3]

At a constant temperature the standard chemical potentials, 01µ and 0

2µ , are

constant, so we can write

17

a2

1 const ka

a≡= [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

c2

1 const kc

c≡= [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:

w0

wc

cc

ck

−= [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.

18

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:

2xKT =∆ [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:

2211

22

11

2

//

/

MmMm

MmK

nn

nKT

+=

+=∆ [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:

19

1

212 1m

mM

T

KM

−

∆= [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.

Time

Temperature

-4

-2

0

2

4

solvent

solution

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.

20

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]:

( )OH3

2104 AAT −⋅=∆ − [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.

21

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

))pH(/()( s ∆∆=β c [1]

22

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] VHA [cm3] pH0 pH1 ∆(pH) = |pH1 – pH0| ∆cs β

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

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.

23

Calculations 1. Calculate the change in the salt concentration, ∆cs, on addition of HCl to

the buffer solution, from the following equation:

))/(c( HCltHClHCls VVVc ⋅⋅=∆ [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.

24

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

[HIn]

]OH[]In[ 3In

+−

≈K [2]

Using the definition of the pH

]OH[loglogpH 3OH3

+−≈−= +a (9.27)

and after some mathematical transformations, Eqn [2] takes the form

pHp]In[

[HIn]log In −≈

−K (9.51)

25

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:

]cid[a

]salt[logppH a += K [3]

26

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

NaOH~]salt[ V [4]

and the concentration of the acid is proportional to the volume of the non-neutralized acid

NaOH20~]acid[ V− [5]

Now, Eqn [3] takes the form:

NaOH

NaOH

-20log76.4pH

V

V+= [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–]

−

−

−

−=

InHIn

In0[HIn][HIn]

AA

AA [9]

[In-] = [HIn]0 – [HIn] [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

27

graph the value of the pH at which the curves intersect. This value corresponds to pKIn.

Table 1 Absorbance of solutions A and B

Absorbance, A Wavelength, λ [nm] Solution A Solution B

400 420 440

460 480 500 520 540 560 580 600 620 640 660 680 700

λanalytical = AHIn =

AIn– = Table 2 Influence of pH on the concentration of acid and base forms of the indicator

VNaOH [cm3]

pH A [HIn] [In–]

2 4 6 8

28

10 12 14 16 18

[HIn]0 =

29

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

-BA νν+

in

terms of the equilibrium:

(solid)BA-νν+

(aq)B(aq)A-zz

−+ ν+ν+

[1]

The thermodynamic equilibrium constant of the above reaction is given by Eqn [2]:

−ν+ν

−

−

+

+

νν

=BA

BA zz

a

aaKa [2]

where −+ zz BA, aa and

−ν+ν BAa stand for the equilibrium activities of the

species in the saturated salt solution. Since the activity of the precipitate,

−ν+ν BAa , can be assumed constant, one can write:

( )ν±

νν

=

=

−

−

+

+ aaaKK aaL zz BA, [3]

where KL ,a is called the solubility constant or the solubility product, a± - the

mean activity, and ν = ν+ +ν– .

30

Substituting the ion activities, ai , by the product of the ion

concentrations, ci , and the activity coefficients, γi , of ions one can obtain:

γ

νν

=

γ

γ=

−

−−

+

++ KKccK cLaL ,BBAA

, zzzz [4]

where:

−

−

+

+

νν

=

zz BA, ccK aL [5]

and

( ) ν±

νν

γ γ=

γ

γ=

−

−

+

+ zz BAK [6]

The mean activity coefficient of the ions, γ± , can be estimated from the extended Debye-Hückel law:

c

c

IdB

IzzA

+−=γ

−+

±1

log [7]

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:

i

i

ic czI ∑= 221 [7]

Combining Eqn [4] with Eqn [6] one can obtain:

±γν−= logloglog ,, aLcL KK [9]

Substituting Eqn [7] into Eqn [9] leads to:

c

c

aLcLIdB

IzzAKK

+ν+=

−+

1loglog ,, [10]

Thus, the dependence of cLK ,log on c

c

IdB

I

+1 should be rectilinear with

the intercept equal to aLK ,log .

31

Experimental The objective of the experiment is determination of the solubility constant,

aLK , 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

of the ratio: V

E

∆

∆, where ∆E is the change in the cell potential brought about

by the addition of the ∆V portion of the titrant. Then calculate the qa value from Eqn [11]:

2

1

2 E

Eqa

∆

∆= [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]:

aqVa ∆= [12]

If ∆E1 is previous to ∆E2 add the correction a to the titrant volume Vmax i.e.

32

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 cLK , values

from Eqn [5]. Plot the dependence of aLK ,log on c

c

IdB

I

+1 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 aLK , from the intercept.

33

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: Ag+(aq) + e

– → Ag(s) [2a]

L: AgCl(s) + e– → Ag(s) +Cl

–(aq) [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:

−+

=

ClAg

AgCl

aa

aQ [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:

+

=

Ag

1

aQ [5a]

34

or

±γ

=+Ag

1

cQ [5b]

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:

±−+ γ++−= +Aglj

00 ln)Cl Ag,/AgCl()Ag/Ag( cF

RTEEEE [6]

After rearranging [6] we obtain the expression for the potential of the silver electrode:

±+− γ+=−+ +Ag

0lj

0 ln)Ag/Ag()Cl Ag,/AgCl( cF

RTEEEE [7]

or

++γ+= ±++

Ag

0 lnln)Ag/Ag()Ag/Ag( cF

RT

F

RTEE [8]

where

lj0 )Cl,Ag/AgCl()Ag/Ag( EEEE −+= −+ [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.

35

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

M/Ag+c +Ag

ln)/( cFRT 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.

36

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.

37

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,

EFnw −=maxe, [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

EFnG −=∆r (10.43)

The Gibbs energy change accompanying the process at T = const is related to the enthalpy and entropy changes by the equation:

STHG rrr ∆−∆=∆ (10.61)

Further, since

p

rr

∂

∆∂=∆−

T

GS [2]

It can be shown that differentiation of Eqn (10.43) with respect to temperature at a constant pressure leads to the result

dT)/d(r EFnS =∆ (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.

38

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 = A T + B [4]

39

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.

40

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

c

κ=Λm

]m[mol

]mS[3

1

−

−

[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

c1000

mκ

=Λ]dm[mol]m[dm

]mS[333

1

−−

−

⋅ [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

210mm cΚ−Λ=Λ [3]

In this equation Κ is a constant while 0mΛ is the limiting molar conductivity.

The word “limiting” means that we consider the molar conductivity in the

41

limit of zero electrolyte concentration, at which the ion–ion interactions can be neglected.

The conductivity, κ, is directly related to the conductance, G:

][Sl

AG

κ= [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:

R

G1

= [5]

we have

RA

l=κ [6]

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

A

lp = [7]

to obtain

R

p=κ [8]

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.

42

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.

43

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, 0mΛ , 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 [mole dm-3]

c1/2

[mole dm-3]1/2

Rp

[Ω]

<Rp>

[Ω] κ

[S m–1] Λm

[S m2 mol–1]

0.01 0.0081 0.0064 0.0049 0.0036 0.0025 0.0016

0.1 0.09 0.08 0.07 0.06 0.05 0.04

RKCl = p = 0mΛ =

44

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 T

λα][ is defined as:

cdT α

=α λ][

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.

45

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:

ss ckt

c=−=ν

d

d [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 0ss cc = at t = 0, where

0sc is the initial sucrose concentration, we obtain:

0lnln ss ctkc +−= [2]

Since sc is proportional to αt – α∞ and0sc to α0 – α∞ , Eqn [2] can be written

as:

)(ln)(ln 0 ∞∞ α−α+−=α−α tkt [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.

46

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.

47

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:

[ ][ ]BAkv = [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):

[ ] [ ] [ ]

tttv

d

Cd

d

Bd

d

Ad=−=−= [3]

For equal concentrations of the reactants, [A]=[B], and using Eqn [3], the rate law of the reaction [1] can be written

[ ]

[ ]2Ad

Adk

t−= [4]

Its integrated form is

48

[ ] [ ] 0A

1

A

1−=kt (25.12b)

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

0

0

0 ]A[

]A[]A[

]A[

1 −=t

kT [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:

( ) ( )0

0MCOOCH

0MOH

M0

0COOCH

0OH

[A]

[A]-[A]

[A]

[A]

[A]

[A]-[A]

[A]

[A]

3

3

+−+−

+−−

+++=

=++=

GGGG

GGGGt

[7]

It is convenient to rewrite Eqn [7] in the form

0

0

00

[A]

[A]-[A]

[A]

[A]∞+= GGGt [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];

49

∞

∞

−

−=

GG

GG

t

00

[A]

[A] [9]

which can be used to express the rate constant [6] in terms of conductance. As

∞−

−=−=

−

GG

GG

t

t000 1[A]

[A]

[A]

[A][A] [10]

Eqn [6] can take the form

∞−

−=

GG

GG

tk

t

tT

0

0[A]

1 [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

+−

+= ∞t

GG

kGG t

Tt

0

0[A]

1 [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:

0[A]

1

bkT = [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

2

1

lnln

lnln

2

1

TR

EAk

TR

EAk

aT

aT

−=

−=

[14]

and solve it for the activation energy, Ea:

−=

1

2ln12

21a

T

T

k

kR

TT

TTE [15]

50

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 least-

squares 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 = [K] G0 = [S]

t [s] Gt [S] (G0 – Gt)/t [S · s-1]

180

51

360 540 … 1800

52

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

53

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]:

0

00

s

sss

VV

Vcc

+= [1]

where 0sV is the volume of the added surfactant stock solution of

concentration 0sc .

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]

12

21CMC

aa

bbcs

−

−= [2]

Cs

0.00 0.01 0.02

G

0

4

8

12

cmc

54

Fig. 1. Conductance, G, versus surfactant concentration, cs. The Figure illustrates a method of the CMC determination.

55

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:

tx

cAD

t

n

∂

∂−=

d

d [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

56

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:

π

=tD

Acnt 02 [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]:

rtt Vnc /= [3]

where Vr is the volume of the solvent in the reservoir. Combining Eqns [2] and [3] we obtain:

π

=tD

V

A

c

c

r

t 2

0

[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 '0G of the dilute

solution, whose concentration is further denoted as '0c . Boil the remaining 45

cm3 of the initial solution of concentration 0c , add l g of agar crystals and

4 Hadgraft,J. Int. J. Pharmaceut. 1979, 2, 177-194

57

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 '0G is proportional to '

0c .

Taking into account that 2

10

'0

V

Vcc = and 2)2/(dA π= , where d is the

cylinder diameter, Eqn [4] can be transformed into:

tDVV

Vd

G

G

r

t π=1

22

0 2 [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.

58

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:

cK

cKxx m

+=1

[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:

mm xcKxx

1111+= [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

59

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:

am sNxS = [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 least-squares 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.