Soft Condensed Matter, 3rd Year

Topics to Cover

The subject matter is defined by the lecture course Soft Condensed Matter. The main topics are:

(i) Dispersion interactions between large uncharged particles.

(ii) Electrical interactions between large charged particles (electrical double layer interactions).

(iii) The factors determining colloid stability, i.e. the combination of (i) and (ii) above.

(iv) The self-assembly of amphiphilic molecules in water (micelle formation, etc.).

(v) The surface behaviour of amphiphilic molecules, including both insoluble monolayers and soluble monolayers in equilibrium with the sub-phase (applications of the Gibbs equation). See introduction to the Gibbs equation.



Intermolecular and Surface Forces by Israelichvili covers the intermolecular forces very well and is also good for colloidal particles. For some topics you may find the book Molecular Driving Forces by by Dill & Bromberg useful. There are also good books by Evans and Wennerstrom: The Colloidal Domain, and Basic Principles of Colloid Science by Everett (if you can get hold of it.


Many of the hints that you need are included in a


This asks multiple choice questions (you may need a calculator) and the quiz finishes when you have answered 15 questions correctly. The range of questions extends beyond the present topic but the additional questions should be useful revision. It would be sensible to make sure that your performance on the quiz is good before you do the problems.

Dispersion Forces between Large Particles

1. Helium wets BaF 2 . If a plate of BaF2 is immersed in liquid helium the thickness of the wetting film is determined by the balance of van der Waals and gravitational forces at any point in the wetting helium film. The film thickness therefore decreases with height (see figure alongside).

For an area dA of film the gravitational energy is ρghldA and, at equilibrium, this must equal the van der Waals attraction VdA , where V is given in terms of the Hamaker constant by

wetting of a surface by a liquid

Show that the thickness of the film varies with height according to the equation

The following data were obtained for such a film
h/m 3.0 8.0 17.0 30.0 70.0
l/nm 3.9 2.8 2.2 1.8 1.4

Interpret these results using the equation you have derived. Determine the value of m.

2. Two silica particles of radius R separated by a distance D are shown below.

interaction between two spheres

Their interaction energy is given by

where A11 is the Hamaker constant.

Calculate the energy of interaction of two 0.5 μm radius silica spheres at a separation distance of 100 nm in a vacuum and compare this value with the thermal energy of the particles at room temperature (A11 = 6 10−20 J for silica).

Show that the Hamaker constant, A11, for the interaction between the particles in a vacuum must be replaced by the Hamaker constant, A131, for interaction between the particles in water where

A131 = A11 + A33 - 2A13 = A11 + A33 - 2(A11A33)1/2.

Given that A33 for water is 8 10−21 J calculate the van der Waals interaction between the same particles in water.

3. When octane is placed in a quartz vessel, the octane wets the walls of the vessel, as shown in the sketch alongside.

The energy per unit area, UVdeW, of a film of octane of thickness, D, due to van der Waals interactions is given by

UVdeW = -A/12πD2

where the Hamaker constant A = -710-21 J.

The gravitational potential energy per unit area, UG, of the film at a height, h, above the liquid surface is given by

UG = ρghD

where ρ is the density of the liquid and g = 9.81 m s2 is the acceleration due to gravity.

wetting of the walls of a glass vessel

(i) Sketch the form of (i) UVdeW, (ii) UG and (iii) UVdeW + UG as a function of D.

(ii) Evaluate the equilibrium thickness of the film at h = 1 cm. (Take ρ to be 703 kg m3.)

4. The Hamaker constant for water interacting with itself across a vacuum is AWW = 3.710-20 J while for a typical hydrocarbon oil, AOO = 5.110-20 J.

(i) Estimate the Hamaker constant, AWO, for water interacting with oil across a vacuum.

(ii) Determine the sign of the Hamaker constant for a film of oil on water in air.
(You may use without proof the following combining relation for medium 1 interacting with medium 2 across medium 3: A132 = A12 + A33 - A13 - A23).

(iii) Hence predict whether oil will spread on water.

Electrical Double Layer Interactions

5. A direct measurement of the force between cylindrical mica surfaces in aqueous solution of electrolytes led to values for the energy of interaction of the surfaces as a function of separation as follows:
KNO3 (10−4M) U/mJ m−20.1000.0730.0530.038
KNO3 (10−3M) U/mJ m−20.0900.0330.0120.005
Ca(NO3)2 (10−4M) U/mJ m−20.0160.0090.0050.003

Account for the variation of the interaction with separation and explain why the interaction is different in the different electrolyte solutions.

6. Dispersions of polystyrene in water can be made extremely monodisperse. By reducing the electrolyte concentration to very low levels the repulsion between the particles can be given a sufficiently long range that the particles order into what is called a colloidal crystal. In such a crystal the particles are arranged in a true crystalline lattice even though they are mainly separated by solution. The figure on the right shows a neutron diffraction pattern of such a colloidal crystal. It is recorded on a detector that is 1 m square situated 40 m from the sample. The colloidal crystal is cubic so the lowest angle diffraction spots (six of them symmetrically arranged about the centre of the detector) correspond to diffraction from the (100) planes of the crystal. (There is an introduction to the structure of cubic crystals and diffraction, which forms part of the Surfaces and Solids tutorial.)

(i) Using Bragg's law of diffraction (λ = 2dsinθ where 2θ is the angle of diffraction) calculate the lattice constant of the crystal. (The wavelength of the monochromatic neutron beam is 1 nm and the angle between the spots in the pattern (2θ) is 0.26o

neutron diffraction pattern from a colloidal crystal of polystyrene spheres in water

(ii) Given that the Debye length is 10 nm in aqueous 0.001 mol dm -3 electrolyte (1:1) and that the separation of the particles should be approximately the same as the Debye length, estimate the concentration of 1:1 electrolyte in the colloidal crystal.

Double Layer Repulsion and Van der Waals Attraction Combined

7. (a) Discuss the factors that determine the shape of the interaction potential shown alongside for two spherical particles in a stable colloidal dispersion in an aqueous NaCl solution. What would be the effect of (i) an increase in the NaCl concentration, (ii) the addition of a divalent electrolyte, (iii) the effect of a change in the solvent to 90 % water, 10 % methanol, the NaCl remaining at the same molarity, and (iv) the addition of a polymer to the solution.

8. The forces between two sapphire (Al2 O3) surfaces in 10-3 M NaCl solutions are shown as a function of pH in the diagram below. Give a qualitative explanation of the changes that occur.

9. Foam films are easily generated by drainage from a supporting frame. They consist of a layer of solution sandwiched between two layers of amphiphilic molecules ("soap"). In the vertical geometry shown on the right they are initially very thick but the solution drains rapidly and the film thins with time and eventually becomes too thin to reflect any significant amount of light and appears colourless. This is called a "black" film (click here for an applet illustrating the physical origin of the colours). Because of a balance between the van der Waals attractive force and the double layer repulsive forces from the electrolyte solution in the film a metastable black film can often be generated. For the series of films shown such a metastable film of thickness 10 nm could be obtained when the 1:1 electrolyte concentration was 0.1 mol dm-3 .

(a) The van der Waals' interaction is given by

sequential pictures of the drainage of a single foam film

U(energy per unit area) = -A131 /12πd2

where A131, the Hamaker constant for the interaction of two slabs of air across a thin film of water can be taken to be approximately 6 x 10-20 J.
Show that the pressure generated by the van der Waals' force is

P= A131/6πd 3

and calculate its value when d is 10 nm.

(b) The repulsive pressure generated by the electrical double layer repulsion is given by

where c is the bulk electrolyte concentration, Φ is the electrical potential at the surface of each face of the film, and 1/κ is the Debye length. Calculate the attractive pressure when Φ = 60 mV, taking T = 300 K , κ = 1 nm (Debye length for a 0.1 mol dm-3 solution), and e = 1.6 x 10-19 and show that it approximately balances the repulsion at this distance.


10. The CMC and micellar aggregation number of sodium dodecyl sulphate (SDS) varies with added NaCl as shown in the table below.
[NaCl]/mol dm−300.030.100.30
CMC/mol dm−30.00810.00310.00150.0007
M (no: of molecules)587193123

(a) Taking the volume of the hydrocarbon chain of a single SDS molecule to be 0.325 nm3, calculate, for the concentration of 0.10 mol dm−3, (i) the volume of the hydrocarbon core, (ii) the effective radius of the hydrocarbon core, assuming spherical geometry, and (iii) the cross-sectional area per chain at the aggregate surface, assuming that it is located 0.3 nm outside the hydrocarbon core.

(b) What is the main physical origin of the factor causing the changes in the CMC and in (iii) above?

11. State the relation between the concentration, c, of a non-ionic surfactant in solution in water and the surface tension, γ, of the solution (Gibbs equation).

The variation of surface tension with concentration, dγ/dc, of a number of non-ionic surfactants of general formula C 12 H25 (OC 2 H4 ) m OH (designated C 12 E m ) at their critical micelle concentrations, CMC, and at 298 K is given in the table below.

Compound (dγ/dc)/N m-1 mol -1 dm3 CMC x 105/mol dm-3
C12E3 -381 3
C12E4 -187 5
C12E5 -136 6
C12E6 -88 8.5
C12E8 -84 8
C12E12 -46 12.5

Use the Gibbs equation to calculate the surface excess of each surfactant at its CMC and hence determine the area occupied by each molecule at the CMC. Comment on the values you obtain.

Given that the volume of a C12H25 chain is 0.325 nm3 and, taking the value of the area per molecule to be the same as the head group area in a micelle, estimate the radius and aggregation number of micelles of C12E8 and C12E12 in aqueous solution (assume that they are spherical).

Given that the maximum extension of a C12H25 chain is 1.65 nm, how would you expect the aggregation of C12E5 and C12E4 in aqueous solution to differ from that of C12E8 and C12E12?

12. At 300 K measurements of the surface tension of solutions of the surfactant C12H25(OC2H4)2OH in water as a function of its concentration are given below. At concentrations higher than 5.6 x 10−5 mol dm−3 the surface tension remains approximately constant at 29.5 mN m−1.

106 x c/mol dm−3 1.80 3.20 5.60 10.0 18.0 32.0 56.0 100.0
γ/mN m-1 64 59 52 45 38 31 29.5 29.0

(a) Plot a graph of γ against lnc. Account qualitatively for the shape of this graph and estimate the critical micelle concentration (CMC).

(b) The variation of the surface tension below the CMC is well accounted for by the equation

/dlnc = − 13720c/(1 + 106c)

where the units of γ and c are N m-−1 and M respectively.

Use the Gibbs equation to calculate the surface excess (Γ) of the surfactant at a concentration of 3.0x10-5 M.

(c) A molecule in a saturated monolayer of the closely related surfactant C12H25(OC 2H4) 6 OH occupies 0.52 nm2 . Comment on the reasons for the difference in this value for the two molecules.