//5 Surface Forces & Liquid Films (nswers to Exercise Problems) Wuge H. Briscoe wuge.briscoe@bris.ac.uk URL: wugebrisco7.wix.com/briscoegroup Exercise : van der Waals forces & liquid films When octane is placed in a quart vessel, it wets the walls of the vessel, as shown in the figure below. Quart ir Octane Medium Hamaker constant ( - J) (for two identical media interacting across air) Octane.5 Quart 6. PTFE.8
//5 Exercise : van der Waals forces & liquid films ) The van der Waals interaction energy per unit area at a particular height h (with a corresponding film thickness D) is Use the Hamaker constant values for two identical media interacting across air (vacuum) given in the table, estimate the Hamaker constant, aoq, for air and quart interacting across octane. Comment on the physical significance of the sign of aoq you have estimated. (Eq. 8) aoq aa oo qq oo.5 6..5.7 J The negative sign means the vdw interaction is positive, that is, it is repulsive thus the film formation! Exercise : van der Waals forces & liquid films ) The gravitational potential energy per unit area of the film at a height, h, above the liquid surface is given by where =.7 g cm - is the density of liquid octane and g = 9.8 m s - is the acceleration due to gravity. Sketch both W vdw and W gra, and then the total energy, W total = W vdw + W gra, as a function of film thickness D at h = cm, using the aoq value you have calculated in Part ). 8x -6 h = cm 6 W tot aoq Wvdw Wgrav D ghd W (J) W_tot W_grav If h, which way would W tot shift (towards bigger or smaller D)? W_vdw 5 Film Thickness D (nm)
//5 Exercise : van der Waals forces & liquid films ) The equilibrium film thickness at a height h corresponds to the minimised total energy W total. Derive an expression of film thickness D as a function of the height h. Then evaluate the equilibrium film thickness at h = cm and m. W W W aoq D tot vdw grav ghd Minimise W tot with respect to D, we obtain W D tot 6D gh aoq Rearrange this, we obtain aoq D 6 gh / 7.5 h( cm) / nm t h = m, D =.75 nm t h = cm, D = 7.5 nm. (but experimental value is smaller due to the retardation effect) 5 Exercise : van der Waals forces & liquid films ) If the vessel was made of PTFE (Teflon ), using the Hamaker constant value given in the table above, discuss if an octane film would form on the vessel wall above the liquid level. W tot aoq Wvdw Wgrav D ghd W (J) - - -6 h = cm W_grav W_tot W_vdw aot aa oo TT oo.5.8.5.6 J (Eq. 8) -8 -x -6 Film Thickness D (nm) 5 The positive sign means the vdw interaction is negative, that is, it is attractive thus no film formation! 6
//5 Exercise : DLVO theory; the empirical Schult-Hardy rule (critical The DLVO theory states that the colloidal stability can be accounted for by considering the sum of the electric double layer interaction and the van der Waals interaction. For two spherical particles of radius R at a surface separation D apart with a constant surface potential y in an electrolyte of bulk concentration r, the total DLVO interaction energy can be expressed as (assuming weak overlap the double layers): 6 Γ e 6 where is the Hamaker constant, is related to the surface potential y Γ and k - is the Debye length, / with i the valency of ion i. 7 Exercise : DLVO theory; the empirical Schult-Hardy rule (critical ) The curves in the figure below show how W DLVO is affected by increasing the electrolyte (salt) concentration, in particular how the repulsive energy barrier is lowered from a to d. Explain qualitatively how the salt concentration can affect W EDL and W vdw respectively. Using the above equation, can you also explain the formation of river deltas on which ancient civilisations (ancient Egyptians, Chinese and Babylonians) settled? W DLVO 8
//5 Exercise : DLVO theory; the empirical Schult-Hardy rule (critical The Debye length - W DLVO e i i i rkt / (Eq. 6) The Debye length k - is the characteristic decay length in the electric double layer for i, x and the electric double layer force. Thus k - indicates how diffused (or compact) the double layer is. Electrolyte concentration, k - t 5 o C, with c in M,. / c nm (for :, e.g. NaCl).76 / c nm (for : or :, e.g. CaCl).5 / c nm (for :, e.g. MgSO ) (Eq.6) Increasing the salt concentration will compress the EDL, and reduce the EDL repulsion. It has little effect on the vdw interaction. ( a subtle issue ) 9 Exercise : DLVO theory; the empirical Schult-Hardy rule (critical ) Rapid coagulation between the colloidal particles is triggered at the critical coagulation electrolyte concentration ( ) ccc where the repulsive energy maximum is just falling to ero (Curve d in the figure above). t this condition, the total interaction energy approaches ero (W DLVO = ) and the repulsive energy maximum is overcome (i.e. it is an inflection point, dw DLVO /dd = ). Show that the maximum energy (i.e. the repulsive energy barrier) of the DLVO interaction occurs at D m = -. W 6kT R R D) WEDL Wvdw exp( D) 6D DLVO( From the st condition, W DLVO =, 6kT R R exp( Dm ) 6Dm From the nd condition, dw DLVO /dd =, kt R R D 6 exp( Dm ) D 6D m m Substituting the result from the st condition, R R or Dm or D m Energy maximum at 6D 6D -. m m W DLVO 5
//5 Exercise : DLVO theory; the empirical Schult-Hardy rule (critical ) Show that at this condition of rapid coagulation, the following relation holds for an : symmetric electrolyte: t ccc, we have D m Substituting this back to the result from the st condition, W DLVO =, / 8πkT e 6 / T / / or From the definition of the Debye length (Eq.6), we also have for a : electrolyte W DLVO / T Substitution of this gives T 5 6 D m D Exercise : DLVO theory; the empirical Schult-Hardy rule (critical ) For surface potential > mv, derive the Schult-Hardy rule, i.e. From ) above we have Recall Eq. 68 T 5 6 tanh e ( mv ) tanh (Eq. 68) kt t high surface potentials > mv, =. In this limit, we have T 5 6 const. That is, we obtain the empirical Schult-Hardy rule, ccc 6 6
//5 Exercise : DLVO theory; the empirical Schult-Hardy rule (critical 5) If the critical coagulation concentration (ccc) for a : electrolyte is mm, what is the ccc for a : electrolyte and a : electrolyte? Rationale your answers by explaining how the valency of the electrolyte affects the DLVO interactions. From the Schult-Hardy rule, ccc 6 If the ccc is mm for a : electrolyte, i.e. =, the ccc value for a : electrolyte ( = ) is / 6 =.56 mm; and for a : electrolyte ( = ) it is / 6 =. mm. That is, multivalent ions are much more effective in causing coagulation between colloidal particles. Why? suppressing the Debye length, and thus the EDL repulsion Exercise : DLVO theory; the empirical Schult-Hardy rule (critical 6) For low surface potentials, /T. Show that for constant surface potentials, the Schult-Hardy rule is modified to Return to the full expression for the Schult-Hardy rule, T 5 6 For low surface potentials, /T. The above relation becomes T is a constant, if we assume the surface potential is constant. Thus we obtain the modified Schult-Hardy rule, ccc 7
//5 Exercise : DLVO theory; the empirical Schult-Hardy rule (critical 7) Discuss the validity of the assumptions regarding the surface potential in real colloidal dispersion systems. In deriving the Schult-Hardy rule, we have assumed either high surface potentials or constant low surface potentials. This is often not the case in real systems. Usually, the surface potential falls due to adsorption of ions. This effect is more pronounced for ions with higher valency, i.e. multivalent ions have a higher tendency to adsorb to the surface to reduce the surface potential, Then for the low surface potential case, the full Schult-Hardy rule is ccc 6 This recovers the Schult-Hardy rule. 5 Exercise : DLVO theory; the empirical Schult-Hardy rule (critical 8) Whilst colloidal particles acquire a surface charge readily in an aqueous medium, it is more difficult for them to do so in a non-polar solvent. Discuss how colloids might be stabilised against coagulation in non-polar solvents. QQ w( r ) (Eq. ) r r dielectric constant of medium (~ 78 for water; ~ for nonpolar solvents) Charging mechanisms in water would not work in non-polar solvents Ioniable or dissociable surface groups (e.g. COOH, -OH) dsorption of charged species (e.g. ionic surfactants or polyelectrolytes) Surface lattice charge due to isomorphic substitution (e.g. many natural minerals such as mica) In non-polar solvents, colloidal stability may be achieved by polymers (adsorbed or brushes; e.g. ancient Egyptian ink), charging facilitated by inverse micelles, polymers bearing charges and large counterions. 6 8
//5 Electrical double layer in a nonpolar liquid In pure water In decane Charing mechanism: OT inverse micelles W.H. Briscoe & R.G. Horn, Direct measurement of surface forces due to charging of solids immersed in a nonpolar liquid, Langmuir 8, 95-56 () W.H. Briscoe and P. ttard, Counterion-only Electric Double Layer: Constrained Entropy pproach, J. Chem. Phys. 7, 55-56 () 7 Exercise : Polymer brush mediated surface forces ) Describe ways in which polymers can either stabilise or destabilise colloidal dispersions. When present, polymers dominate inter-surface forces vast field of rich behaviour Forces depend on MW Conformation Surface density Surface layer thickness Concentration Molecular architecture Chemical nature mphiphilicity R.G.Horn (99) J.. Ceramic Soc. 8 9
//5 Exercise : Polymer brush mediated surface forces (cf. Briscoe, chp6, in Colloid Science, Wiley (); Klein, Briscoe et al., chp, Polymer brushes and surface forces, in Polymer dhesion, Friction, and Lubrication, Wiley (). N-mer per chain blobs L s polymer brush can be characterised by the following parameters: N the number of monomers in the chain a the volume of a monomer L the brush thickness s the spacing between brush 9 Exercise : Polymer brush mediated surface forces.) What is volume fraction of the monomer in the brush? Na Ls N-mer per chain blobs.) For random flight polymers in a semi-dilute regime, the osmotic pressure due to monomers in a brush is Π Give the expression for the osmotic pressure due to monomer crowding in terms of N, a, L and s. kt N a kt a s L osm L s N number of monomers in the chain a monomer volume L brush thickness s spacing betw. brush Dunlop IE, Briscoe WH et al. Macrom. Chem. Phys. 5, 5 () Titmuss S, Briscoe WH et al. J. Chem. Phys., 8 9 ()
//5 Exercise : Polymer brush mediated surface forces.) The elastic energy of stretching a polymer chain is What is elastic force experienced by a single polymer chain in the brush? Then give the elastic pressure el (i.e. force per unit area) that resists an increase in the brush thickness L. The elastic force is F W L kt L Na el el The negative sign because it is stretching. The osmotic pressure per chain is F el el s kt Na s Dunlop IE, Briscoe WH et al. Macrom. Chem. Phys. 5, 5 () Titmuss S, Briscoe WH et al. J. Chem. Phys., 8 9 () L L N-mer per chain s blobs N number of monomers in the chain a monomer volume L brush thickness s spacing betw. brush Exercise : Polymer brush mediated surface forces.) By balancing the osmotic and elastic pressures on the brush, determine how L scales with N, a and s. osm el N-mer per chain blobs N a kt s kt L L Na s Rearranging this gives the brush thickness L L / / 5 / L Ns a kt N Compare L with the Flory radius R F or radius of gyration R g ; both ~ N /5 in a good solvent. Polymer brushes are highly stretched compared to their coiled conformation in solution. s N number of monomers in the chain a monomer volume L brush thickness s spacing betw. brush.5) rgue qualitatively why polymer brushes give rise to a repulsion between two surfaces coated with brushes. (Entropy: monomer crowding)
//5 Experimental methods Surface Force pparatus (SF) Interactions at soft bio-nano interfaces F n wuge.briscoe@bris.ac.uk FECO To spectrometer D F s White light G.5 PMM+DTB Nature (6); Soft Matter (); CIS (, 5); Science (9); CS Nano (); Soft Matter (); Langmuir (5) Fundamental forces in nature Intermolecular forces van der Waals (adhesion) Electric double layer Surface forces Dynamic forces Hydrodynamic forces Friction, etc. Equilibrium forces DLVO forces Non-DLVO forces Structrual force Hydration force Hydrophobic force Capillary force Surfactant mediated Polymer-mediated Bridging Depletion Steric repulsion Polymer brushes Polyelectrolytes Biological interactions Biolubrication Future Membrane fusion challenges Cell adhesion Nanofluids mediated forces Nanotoxicity Nanostructured surfaces Hydrodynamic forces (e.g. surface deformations)