Worked solutions for: ‘Applied Colloid & Surface Chemistry’ by R.M. Pashley and M.E. Karaman (J.Wiley, 2004) Chapter 2 1. Answer: Use Laplace equation in the form (eqn 2.8): ∆P = 2γ cos θ . Gives pressure r of 2 bars. Addition of surfactants will reduce the interfacial tension and hence the pressure. The use of hydrophobic porous materials to prevent water flow is used in clothing and in concentrating orange juice. 2. Answer: the Laplace equation for spherical soap films is ∆P = 4γ . Hence we can r calculate the pressure difference between each bubble and the atmosphere and therefore between the bubbles. The pressure difference between the two bubbles can then be used in the Laplace equation to obtain the radius (1.5cm) of the (spherical) soap film formed between the bubbles. 3. Answer: Use the capillary rise equation (2.9). Make sure that you convert the parameters to SI units. The calculated surface tension value is: 0.62Nm-1. 4. Answer: Apply the full Laplace equation to calculate the pressure difference across the water interface, relative to atmospheric pressure. The difference in these two pressures must equal the hydrostatic pressure difference in the water (ie hρg). This gives a height difference of 4.1mm. 1 5. Answer: Use the form of the Laplace pressure equation for capillaries ( ∆P = 2γ cos θ r ). The Laplace pressure for a stable meniscus is 1.5 bars (convert to SI) and hence the capillary radius must be 0.33 microns (given that we know that the contact angle of water on Teflon is 1100). 6. Answer: The particles of sand can be held together by the Laplace pressure generated in a meniscus (see diagram below) of wetting water. By comparison, for full immersion or complete drying no meniscus can form. 7. Answer: Most powdered building materials (eg sand and clay) are sufficiently hydrophilic for water to wet and form a meniscus which will hold the particles together (see diagram below). 8. Answer: First we must realize that two forces act on the lower particle due to the meniscus. Both of these act to hold the particles together. These are the pressure generated by the Laplace pressure and the surface tension forces acting along the threephase line. (Also, it should be assumed that the weight of the lower particle is negligible relative to the meniscus forces.) The Laplace pressure across the curved interface can be directly calculated using the known radii R1 and R2. Care, however, because the radii are in different directions, that is R2 is positive (in the sense of contributing and attractive pressure between the particles) and R1 is negative (it acts to push the particles apart). Thus the Laplace pressure is 1 1 therefore given by the equation: ∆P = γ ( − ) . This pressure acts to hold the R2 R1 particles together and acts over an area of πR12 . Hence the total force, due to the Laplace pressure, acting on the particle can be calculated. The second term is a surface tension force acting along the TPL of the lower particle. This surface tension acts in the direction of the interface at the point of contact with the particle. Simple geometry tells us that the angle the droplet surface attaches to the droplet relative to the vertical direction is given by cosΦ = 0.4 (ie Φ = 66.40). Thus, the vertical component of the force acting along the TPL on the lower particle can be easily calculated as: FTPL = 0.4 * γ * 2 * π * R1 . Combining these two forces gives a total force acting on the lower particle of 0.18mN. 2 9. Answer: This problem follows the same principle as the previous question, that is we must calculate the combined forces due to the Laplace pressure and the vertical component of the surface tension force acting along the three-phase-line (TPL). Once again the two principal radii of the curved meniscus are in opposite directions and therefore their signs are different. Thus the Laplace pressure is given by: 1 1 ∆P = γ ( − ) , where the pressure acts across the area πR12 to hold the particle to R2 R1 the substrate. The Laplace pressure term contributes a binding force of 5.6E-5N. The second term is given by the vertical component of the surface tension around the TPL: FTPL = Sinφ * γ * 2 * π * R1 . This gives a contribution of 2.15E-6N. Summing these two forces gives the minimum force F (ie 58µN) required to detach the particle. Chapter 3 1. Answer: First of all it should be realized that from the Gibbs’ adsorption isotherm the maximum adsorption density (ie minimum area per adsorbed molecule) occurs at the maximum slope on the graph (ie in both cases that is the linear slope of the graph). Further, it should be realized that for a 3:1 electrolyte (ie CoRCl3) 4 molecular species adsorb at the air-water surface per surfactant molecule. Also, for measurements in concentrated salt (150mM NaCl) the bulk concentration of only one species (the ionized surfactant monomer is changing with surfactant concentration. Take care with the units. The Gibbs adsorption isotherm appropriate for analysis of the water isotherm is: 1 ⎛ ∂γ ⎞ Γ=− ⎜ ⎟ , because 4 molecular species adsorb. It should be realized that in 4 RT ⎝ ∂ ln C ⎠ obtaining data from the graph we do not need to convert molarity concentration units to ⎛C ⎞ SI (ie moles/m3) because ∂ ln C = ln⎜⎜ 2 ⎟⎟ and hence the units cancel. Using the water ⎝ C1 ⎠ data on the linear region of the graph, the adsorption density is calculated as: 1.2 µmoles/m2. This value can then be converted to the corresponding area per molecule: 1.4E-18m2/molecule or a minimum molecular area of 1.4nm2. The form of the Gibbs equation for the lower graph in high electrolyte is 1 ⎛ ∂γ ⎞ Γ=− ⎜ ⎟ , since only one species changes bulk concentration (increases and RT ⎝ ∂ ln C ⎠ hence increases its chemical potential) as the surfactant concentration is increased. Using 3 this equation the graph gives a maximum adsorption density of 1.8 µmoles/m2, which corresponds to a minimum area of 0.94nm2. These results are interesting because they show that the Gibbs model gives reasonable results, even in this complex system. The area per molecule should decrease with increased electrolyte as the ionic head-groups become screened and can pack closer together. 2. Answer: A: The horizontal line corresponds to zero adsorption. B: The curved region corresponding to increasing adsorption densities, occurring over a fairly narrow concentration range. C: The linear region corresponds to constant adsorption density. Note that the interfacial tension still falls in this region. D: The horizontal region here does not correspond to zero adsorption but is caused by the surfactant concentration remaining constant, as all the added molecules self assemble into aggregates such as micelles. This occurs at concentration above the critical micelle concentration (cmc) value. Chapter 4 3. Answer: Equation 4.8 gives the standard free energy of transfer of surfactant molecules from an aqueous environment, as free monomers, to molecules in a micelle, as a function of its cmc value. Thus, the difference in free energy of transfer between surfactants with the same head group but different chain lengths must be due to the free energy associated with the transfer of the extra methylene groups from aqueous to liquid hydrocarbon environment, in the center of a micelle. Thus, the average free energy of transfer is 2.5E-21J or about 0.6kT per group. 4 Chapter 6 1. Answer: Using the Debye length equation 6.11 we obtain the values for NaCl: 30nm, 3nm and 1nm. For MgSO4 the corresponding values are 15nm, 1.5nm and 0.5nm. 2. Answer: Use either the approximate equation for the decay in potential next to a flat surface (eqn 6.14) or the exact equation (eqn 6.13) to calculate the potentials and then at several points use the Boltzmann distribution function (eqn 6.6) to calculate the individual ion concentrations. 4. Answer: Use equation 6.47: ∏R ≈ 2εoDκ2ψo2exp(-κd), which gives the swelling pressure using the superposition principle. First calculate the Debye length (κ -1) of the solution, which is 7.88nm . The permittivity of free space, ε0 , is equal to 8.854E-12 C2J-1m-1 and the dielectric constant of water is 80. Substitution into the equation gives a swelling pressure of 3250Nm-2. 5. Answer: The hydrostatic pressure on a column of water 1cm high is given by the fundamental equation of hydrostatics: P = hρg. Hence the hydrostatic suction pressure acting to drain the wetting film is 98 Nm-2. If the air/water interface has no charge this means that the film can be considered as one half of two flat surfaces with –60mV on each surface but separated by twice the thickness of the film. (Since dψ = 0 at the middz plane, which therefore has no charge, from equation 6.16). Equation 6.47: ∏R ≈ 2εoDκ2ψo2exp(-κd), can then be used to calculate the answer, since all the parameters are known. Note that ‘d’ is twice the thickness of the water film and that we must assume that the Debye length of water is (typically) about 50nm. These values lead to a calculated film thickness of 75nm. 5 6. Answer: The chemical free energy of adsorption (-ve) must be equal to the electrostatic work done (+ve) in bringing a single charge from bulk solution to the charged surface. Thus: ∆µ chem = − z i q e ψ 0 , where z i q e is the charge on the ion. Which means that the chemical potential of adsorption is –1.28E-20 J per ion or about -3kT . 8. Answer: Using the superposition principle we can obtain the electrostatic potential at the mid-plane and then use the Boltzmann distribution to calculate the Na+ ion concentration. Thus, the mid-plane potential is given by: ψm ≈ 2ψd/2 ≈ 2ψoexp(–κd/2) , which in this case is equal to –24mV. ⎡ Ziqψ(x)⎤ The corresponding Boltzmann distribution equation is Ci(x)=Ci(B) exp⎢– kT ⎥ ⎣ ⎦ ⎛ qe ψ m ⎞ M Bulk ⎟ where q e is the positive value of the which becomes: C Na = C Na exp⎜⎜ − ⎟ kT ⎝ ⎠ electronic charge. This gives a Na+ ion concentration at the mid-plane of 2.5mM. 9. Answer: The electrostatic decay away from a flat, charged surface is given approximately by the equation: ψ(x) ≅ ψo exp(-κx) . Since the Debye length in this case is 30nm, if the potential 5nm away is –10mV the surface potential must be –11.8mV. The Boltzmann equation gives the concentration of each ion from the equation: ⎡ Ziqψ(x)⎤ Ci(x)=Ci(B) exp⎢– kT ⎥ . Note that in using this equation the signs of Zi and Ψ(x) ⎦ ⎣ are important. Hence, the Na+ ion concentration 5nm from the surface is 0.15mM and the concentration of Cl- ions is 0.068mM. Since the total ion concentration 5nm away is 0.218mM, the osmotic pressure given by ⎡ T ⎤ the equation: ∏os=kT⎢⎣Cm – 2C(B)⎥⎦ equals 45 Nm-2 . (Careful with concentration units – use SI.) 6 Chapter 7 2. Answer: The magnitude of the refractive index measured at visible frequencies is a measure of the strength of its van der Waals interaction or Hamaker constant because a high value corresponds to significant UV absorption – the main contributor to the vdw interaction. The three possible interactions are: 1:3:1, 1:3:2 and 2:3:2. In the first case (1:3:1) we will have the strongest attraction because the particles have well correlated frequencies (since the materials are the same) and a strong UV contribution. In the second case (1:3:2) the vdw interaction with the solvent will be strong and the replacement of solvent with an approaching particle will lead to a weaker vdw interaction. Hence in this case the vdw force will be repulsive and will act to separate particles 1 and 2. In the final case (2:3:2) the correlation between the same materials (ie 2) will dominate over the interaction with the solvent and the vdw force will be attractive. 3. Answer: The interaction will be the same. The Hamaker constants for the cases A121 are equal to those for A212 for all materials. 4. Answer: The rapid coagulation condition shown above is useful because under these conditions both Vs = 0 and dVs/dH = 0 for the same separation distance H = H*. Thus: 2πaε 0 Dψ 02 exp(−κH ) = aA121 12 H and dVs aA = 0 = (−κ )2πaε 0 Dψ 02 exp(−κH ) + 1212 dH 12 H 7 where on direct substitution it is found that the special distance H* = κ-1 (ie the Debye length). The critical surface potential can then be directly obtained by substitution of H* into either equation above. The critical coagulation value is then found to be 29mV. 5. Answer: The interaction energy for close separation distances for liquids which only have vdw interactions must be equal to the work done to create two new surfaces per unit area, ie 2γ, where γ is the surface tension of the liquid. Thus, 2γ = V121 and therefore γ = 55mJm-2. 8