CE70006 - Colloids and Interface Science

Question 1

What is a colloid?

Answer 1

A material which is an intermediate between a single molecule and large objects that are dominated by gravitational forces.

A dispersed material in which one of its dimensions is in the range 1-1000 nm.

Thus we can have 1,2 and 3 dimensional colloids.

“A colloid is a mixture in which one substance consisting of microscopically dispersed insoluble particles is suspended evenly throughout another substance.”

Question 2

Give examples of 1, 2, and 3 dimensional colloids

Answer 2

1D: plates e.g. clays

2D: Needles e.g. asbestos

3D: Particles e.g. titania or latex

Question 3

What is the name for the following dispersed phase - dispersion media systems?

1. Gas-Liquid
2. Liquid-Liquid
3. Solid-Liquid
4. Gas-Solid
5. Liquid-Solid
6. Solid-Solid
7. Liquid-Gas
8. Solid-Gas

Answer 3

1. Foam
2. Emulsion
3. Dispersion
4. Solid foam
5. Solid emulsion (ice cream)
6. Solid dispersion
7. Aerosol
8. Solid aerosol

Question 4

How are colloids prepared?

Answer 4

1. By the breaking up of large lumps of material via grinding, milling, etc.
   This will give an average particle size of no smaller than ~1um.
2. By the aggregation of small molecules. This is hard to achieve and costly to control, but can obtain better products.
   No real limit on particle size.

Question 5

Do colloids have high or low surface areas?

Answer 5

Very high - it is the property of their surfaces that determines their properties.

Total particle volume represented by Φ.

Question 6

What is the Van der Waal equation?

Answer 6

(P + an^2/V^2)*(V-nb) = RT

Which considers non-deal gas behaviour

P = pressure
R = universal gas constant
T = absolute temperature
V = molar volume
b = gas constant
a = gas constant
V = molar volume
n = moles

b is subtracted from V to account for the finite size of the molecules.

a/V^2 was added to pressure to account for the attractive intermolecular forces.

Question 7

What are Van der Waals forces?

Answer 7

Distance-dependent interactions between atoms or molecules.

The force results from a transient shift in electron density.

Question 8

Describe the following Van der Waals ineractions:

Keesom
Debye
London

Answer 8

Keesom: permant-permanent dipole interaction

Debye: permanent-induced dipole interactions

London: fluctuation-induced dipole (or dispersion) interaction [always present]

Van der Waals forces are long range (0.2 - 10 nm)

Question 9

What are the (4) main features of dispersion forces?

Answer 9

1. They are long ranged (acting over several atomic diameters)
2. Net forces may be attractive or repulsive, but only repulsive between different materials in a medium
3. Dispersion forces can align molecules
4. Dispersion interaction of two bodies is affected by the presence of other bodies nearby

Question 10

What is the equation to consider the dispersion interactions between atoms?

Answer 10

w(r) = -3a^2I/(4(4piε0)^2r^6)

Where:
a - atomic polarisability
I - ionisation potential
ε0 - permittivity of free space
r - distance between the two atoms

Question 11

How do Van der Waals interactions behave with macroscopic bodies?

Answer 11

The interactions between molecules is of the form w(r) = -C/r^n.
Here, ‘C’ is a constant and ‘n’ is a parameter determining the nature of the interaction.

Assuming additivity, the interaction of a molecule with a surface is the sum of its interactions with all molecules in the body.

W(D) = - πCρ / (6*D^3)

The equation W(D) = - πCρ / (6*D^3) calculates the interaction energy (W) between a molecule and a surface, where ‘D’ represents the distance from the surface, ‘C’ is a constant from the molecule-surface interaction equation, and ‘ρ’ is the number density of molecules in the body.

This equation assumes that the interaction between a molecule and a surface is the sum of all interactions between that molecule and all the molecules within the body, following an additive principle.

Question 12

If the thermal (k*T) and Van der Waals energy are the same, what does this mean?

Answer 12

It would be the boiling or condensation point.

Van der Waals forces start to dominate over thermal energy, leading to a change in the physical state of matter.

Question 13

What is the Van der Waals interaction equation between:
1. A sphere and a plate
2. Two equally sized spheres
3. Parallel surfaces

Answer 13

1. W(D) = -AR/(6D)
2. W(D) = -AR/(12D)
3. W(D) = -A/(12D^2)

Question 14

What is the problem with the Hamaker approach?

Answer 14

It is not suitable when atoms are further away (e.g. 100 nm)

An alternative approach is the Lifshitz theory, which treats matter as a continuum. It is considered to be more accurate but in effect it is an alternative way to calculate the Hamaker constant. The basic relation remains the same.

Question 15

What happens if the Hamaker constant is negative?

Answer 15

The Van der Waals force will be positive, i.e. repulsive

Question 16

What is the affect of solvent on the Hamaker constant, A?

Answer 16

A.effective = A.final - A.initial

Meaning the effective Hamaker constant of a system following separation of a solvent is equal to the difference in the system A before and after separation.

Positive A.eff leads to attractive forces
Negative A.eff leads to repulsive forces

Question 17

What is the Hamaker constant of a vacuum?

Answer 17

0

Air is also 0

Question 18

What are the 3 sources of a charged surface?

Answer 18

1. Dissolution of soluble ions, mainly metal ions, e.g. in clays
2. Adsorption of ions, generally anions, to surfaces e.g. iodide in silver iodide
3. Adsorption of surfactants or polyelectrolyte

With a charged surface, for electrical neutrality to be met, there must be ions of opposite charge (counter ions) present somewhere in the solution.
From an enthalpic view, it is favourable for the ions to be as close to the charged surface as possible.
From an entropic view, it is favourable for the ions to be as far as possible.
In reality, something in between forms - an electrical double layer.

Question 19

What is the Hamaker constant?

Answer 19

The Hamaker constant is a coefficient accounting for the van der Waals interaction between two materials.
It has a strong correlation with various physical phenomena, such as liquid wettability, adhesion, friction, adsorption, colloidal stability, polymer flow, and deformation.

Question 20

How does the decay of interactions vary between atoms and macroscopic bodies?

Answer 20

w(r) ∝ 1/r^6 for atoms
w(D) ∝ 1/D for macroscopic bodies

So the interaction with 1/D decays much slower than the 1/r^6 dependence of the intermolecular pair potential.

Question 21

What are the units of the Hamaker constant, A?

Answer 21

Joules

Question 22

How does Debye length vary with concentration?

Answer 22

As concentration increases, Debye length decreases - rapidly at first and then very gradually (on a nm scale).

3-3 electrolytes (i.e. substances with a 3+ 3- charge like AlCl) start with a shorter Debye length, k, than a 1-1 electrolyte like NaCl.

Question 23

What are the two origins of repulsive potential between 2 charged surfaces?

Answer 23

1. Electrostatic interaction from the emanating electric field, exerting a force, Fel, on the ions.
2. Osmotic pressure difference between the double layer and the bulk (where the conc. of ions close to a surface is much larger than that of the bulk).

Question 24

What is zeta potential?

Answer 24

Zeta potential (ζ) is the electric potential at the shear plane of a particle suspended in a liquid. It is measured in volts (V).

The zeta potential arises due to the presence of charges on the surface of colloidal particles or other solid surfaces in a liquid. These charges may come from dissociated ions, adsorbed ions, or other surface functional groups.

Zeta potential provides information about the stability of colloidal dispersions. Particles with higher zeta potentials tend to repel each other more strongly, leading to increased stability. On the other hand, lower zeta potentials may result in particle aggregation or flocculation.

Question 25

What are the 2 main ways to define the dimensions of a polymer chain?

Answer 25

1. Root mean square end to end distance (r^2)^0.5
2. Root mean square distance of the elements of the chain from its centre of gravity, also known as the radius of gyration (s^2)^0.5

(r^2)^0.5 = (6s^2)^0.5

Question 26

What is the size of a polymer molecule proportional to?

Answer 26

The square root of its molecular weight.

(r^2)^0.5 = lx^0.5
Where:
r - root mean square end to end distance
l - length of links
x - number of links

Question 27

What’s the molecular weight of a polymer?

Answer 27

They don’t have a unique Mw. Instead, they have a distribution (since there can be varying chain lengths).

Number average (Mn) and weight average (Mw) molecular weights can be calculated.

Always, Mw > Mn, and Mw : Mn is often used as a measure of molecular weight distribution.

Note:
Mn = Σni Mi / Σni

Mw = Σwi Mi / Σwi

Question 28

What is the assumption for ideal polymer solutions?

Answer 28

All intermolecular interactions are the same, so the enthalpy of mixing is 0.

Question 29

How do polymers behave on a surface?

Answer 29

Polymers adsorbed onto surfaces generally do not lie flat on the surface, but extend away (forming loops and tails).

This means that when polymers adsorb to particle surfaces, they may interfere with each other before Van der Waals forces can play a part and aggregate the particles.

A polymer when it is adsorbed to a surface generally does not form a compact layer, but a
fuzzy, relatively thick monolayer. The polymer layer itself may only contain around 5% polymer, the rest is solvent.

Question 30

What happens when 2 particles coated with polymers interact?

Answer 30

The two polymer layers overlap and compress. This has two effects one osmotic, enthalpic, and one entropic.

The osmotic effect is that the concentration of polymer in the gap is increased, this increases the osmotic pressure and the particles are forced apart.

The entropic effect is a consequence of the compression of the polymer layer decreasing the number of configurations the polymer can adopt.

Both effects have been quantified but the theory is not as good as that of double layers or van der Waals interactions.

Question 31

How can polymers be used to flocculate particles?

Answer 31

Under some circumstances polymers can flocculate particles.
Under very dilute polymer concentrations (so that the particles are not fully covered by polymer) aggregation occurs by one polymer molecule bridging between two particles.

Also if the polymer is non adsorbing, when the gap between two particles becomes less than the dimension of the polymer, there will be less polymer between the two particle surfaces than in the bulk.
Thus solvent will flow from the dilute to the concentrated region. This is brought about by the particles moving towards each other.

This is known as depletion flocculation.

Question 32

In ideal polymer solutions, ideal mixing may occur so dH = 0.
What other types of mixing may take place?

Answer 32

In practice few systems obey Raoult’s Law and deviations may occur.

1. Athermal Solutions where DH=0 but DS no longer given by equation 3.6 in lecture notes
2. Regular Solutions where DS has the ideal value, but DH is finite
3. Irregular Solutions where both DH and DS are non-ideal

Question 33

What does the interaction of polymers depend on?

Answer 33

We can see from graphs that the type of interaction (attraction or repulsion) is dependent on the value of the Chi parameter.

The Chi parameter itself is dependent on temperature, thus flocculation (aggregation)
of particles is also dependant on temperature.

Since Chi decreases as T increases for most solvent soluble polymers, the same is true regarding flocculation.

For water soluble polymers the opposite is true.

Question 34

How does surface tension arise?

Answer 34

As a result of imbalance of intermolecular interactions.

At the liquid/vapour interface, cohesive forces towards the liquid bulk are not balanced and the molecules are attracted inwards.

Whilst in the bulk liquid, cohesive forces are balanced in all directions and molecules experience no net force.

Question 35

What is the work of cohesion?

Answer 35

Work required to pull apart a volume of unit cross sectional area (A=1)

W.cohesion = 2*γ(a)

Where γ is surface tension

Question 36

What is the work of adhesion?

Answer 36

Work required to separate a unit area of interface between two phases (a and b) to form a new surface of each phase.

W.adhesion (a,b) = γ(a) + γ(b) - γ(ab)

Where γ(ab) is the interfacial energy per unit area (or interfacial tension)

Question 37

How is the spreading coefficient, S, found?

Answer 37

S = γ(ab) - (γ(a) + γ(b))
= W.adhesion - W.cohesion

If S > 0, spreading is spontaneous.
E.g. with oil and water, spontaneous spreading means that oil adheres to the water more strongly than it coheres to itself.
S = γ(wv) - (γ(ov) + γ(ow))

Question 38

What is the Fowkes equation?

Answer 38

The Fowkes method is used to calculate the surface free energy of a solid from the contact angle with several liquids. In doing so, the surface free energy is divided into a disperse part and a non-disperse part.

γ(ab) = γ(a) + γ(b) - 2sqrt(γ(ad)γ(bd))

It is based on assumption that cross-interaction term across interface (work of adhesion) is due exclusively to dispersion forces.

Question 39

What are the different liquid surface tension measurement techniques?

Answer 39

Du Noüy ring method

Wilhelmy plate method

Pendant drop method

Spinning drop method

Question 40

Describe the Pendant drop method for liquid surface tension measurements:

Answer 40

The shape of the drop hanging from a needle is determined from the balance of forces which include the surface tension.

The surface or interfacial tension can be related to the drop shape by an equation,
Bo = ΔρgR^2/γ

If Bo is too large, the drop can detach from the needle (indicating very low surface tension).

Question 41

How does fluid contact angle indicate hydrophobic/philic nature?

Answer 41

Hydrophilic (or oleophilic if oil): 0 < θ < 90

Hydrophobic (or oleophobic if oil): 90 < θ < 180

Superhydrophobic: θ = 180

Question 42

What are the different surfactant types?

Answer 42

Non-ionic (stable in varying pHs) [used in detergents, food products, pharmaceuticals, cosmetics, pesticides, etc.]

Anionic (-ve head) [used in pesticide formulations and toothpastes]

Cationic (+ve head) [used in hair and fabric conditioners]

Zwitterionic/Amphoteric (+ve head connected to another -ve head) [ found in cell membranes and used in drug delivery and cosmetics]

Question 43

Where do surface tension forces originate from?

Answer 43

Intermolecular forces

Cohesive forces come from:
1. Van der Waals interactions
- between polar molecules: permanent dipole-permanent dipole (Keesom or orientation interaction)
- between a polar and a non-polar molecule: permanent dipole-induced dipole (Debye or induction interaction)
- between ALL molecules: fluctuating dipole-induced dipole (London or dispersion interaction)

2. Polar, metallic, hydrogen bonding

Each interaction contributes to total surface tension.

Question 44

What is an advancing and receding contact angle?

Answer 44

Advancing Contact Angle: The maximum angle formed between the tangent to the liquid drop’s interface and the solid surface when the drop spreads on the surface.

Receding Contact Angle: The minimum angle formed between the tangent to the liquid drop’s interface and the solid surface when the drop contracts or recedes on the surface.

Question 45

What are surfactants?

Answer 45

Surfactants (surface-active agents) are chemical compounds that decrease the surface tension or interfacial tension between two substances.

The hydrophobic ‘tail’ can form van der Waals
bonds with non-polar grease molecules, whilst the polar ‘head’ can form hydrogen bonds with water.

Question 46

What value of Hamaker constant (A) and Spreading coefficient (S) indicate that a substance will spread spontaneously / favourably?

Answer 46

S > 0

A < 0

Question 47

It is observed that 10-2 M sodium sulfate will aggregate charged stabilised particles, whilst 0.6M sodium sulfate is needed to aggregate particles stabilsed by a copolymer containing polyethyleneoxide, why?

Answer 47

In 10-2 M Sodium sulfate the electrical double layer is very short, around 2nm, and so the particles are not sufficiently repulsive to prevent the particles from aggregating.

The addition of polymer stops this aggregation, via a steric mechanism.

However as the electrolyte concentration increases, the polyethyleneoxide becomes less soluble as there is a competition between the hydrated ions of the electrolyte and the polymer for water, which will hydrogen bond.

At sufficiently high electrolyte concentrations this competition is such that at room temperature the polymer comes out of solution.

Thus also the adsorbed polymer layer also begins to precipitate, but as it does so it collapses onto the particle surface, making the polymer layer very thin (also the interactions between the polymer layers become attractive) so that the PEO stabilised particles no longer remain stable and aggregate.

Question 48

If you had to choose to study a colloidal suspension either by light or electron
microscopy, which would you prefer and why?

Comment on the possibility to get a sharp picture of a colloidal particle with radius 30 nm which is not immobilised.

Answer 48

Most often, electron microscopy is preferable.

Particles will move due to Brownian motion an average distance of about 3 µm, i.e. about 60 times its diameter, within 1 second, which is a typical time to take a picture. The picture will thus be blurred.

Further more it turns out that an electron microscope needs a very high vacuum to work, such that only dry samples can be investigated, thus it is not actually possible to take pictures of very small particles in liquids.
What is normally done is that samples are frozen, very quickly to form a dispersion in a frozen liquid (ice).
The sample is then micro-tomed, the ice allowed to sublime to some extent and then the surface replicated with a very thin carbon film.

This carbon film is then put in the AFM. This process is called Freeze Fracture. It is fraught with artefacts, so you have to be very careful.

Note the AFM would not work as the particles are in a liquid, the AFM can only image surfaces, although it would work if you could put particles on the surface in some way, although of course now its immobilised.

For these reasons you cannot currently get pictures of particles smaller than 100 nm or so in liquids.

Question 49

Describe wetting vs wettability:

Answer 49

Wetting - process of displacing a fluid with another one

Wettability is the result i.e. the equilibrium position that the liquid/fluid takes on the surface, and the measure of the contact angle

Question 50

How does surface energy affect hydrophobicity?

Answer 50

Low surface energy encourages hydrophobicity.

Question 51

What is a Wenzel wetting state?

Answer 51

A hydrophobic wetting state on a rough surface where the liquid integrates into the microstructure.

Wenzel wetting state occurs when a liquid fully wets a rough surface, spreading across it uniformly.
Surface roughness amplifies the apparent contact angle, leading to enhanced adhesion and spreading of the liquid over the surface.

Question 52

What is a Cassie Baxter wetting state?

Answer 52

A hydrophobic wetting state on a rough surface where the liquid adsorbs to the material but does not integrate into the microstructure.

Question 53

List examples of good hydrophilic and hydrophobic groups (used in surfactants):

Answer 53

Hydrophilic:
Sulphonate
Sulphate
Carboxylate
Phosphate
Hydroxyl
Ether
POE Polyoxyethylene
Polyols
Ammonium

Hydrophobic: mostly made up of hydrocarbons
Natural fatty acids
Olefins
Fluorocarbons
Silicones
Alkylaromatics

Question 54

What’s a Gibbs monolayer and a Langmuir monolayer?

Answer 54

Gibbs: A monolayer formed by a compound that is soluble in one of the phases separated by the interface on which the monolayer is formed.

Langmuir: aka an insoluble monolayer is a one-molecule thick layer of an insoluble organic material spread onto an aqueous subphase

Question 55

How does the critical micelle concentration (CMC) vary with alkyl chain length?

Answer 55

CMC decreases exponentially with alkyl chain length

Some amphiphiles have very low CMC and always form micelles in water (example: phospholipids)
▪ They are called insoluble surfactants
▪ Can form monolayers at water/air or water/oil interface
▪ Can be deposited on interface from a volatile spreading solvent (e.g. chloroform)
▪ Spread monolayers are called Langmuir monolayers
▪ (in contrast with monolayers formed by spontaneous adsorption, called Gibbs monolayers)

Question 56

What’s the Marangoni effect?

Answer 56

The mass transfer along an interface between two phases due to a gradient of the surface tension.

Since a liquid with a high surface tension pulls more strongly on the surrounding liquid than one with a low surface tension, the presence of a gradient in surface tension will naturally cause the liquid to flow away from regions of low surface tension.
In simple cases, the speed of the flow is given by u = Δγ/μ
Where Δγ is the difference in surface tension and μ is the viscosity of the liquid.

Question 57

What is surface tension vs surface pressure?

Answer 57

Surface Tension:
Force acting per unit length along the surface of the liquid.
Arises due to the cohesive forces between molecules in the liquid phase.

Surface Pressure:
It’s the pressure exerted by the weight of the atmosphere above that location.
In the context of surface tension, surface pressure can also refer to the change in surface tension as a function of the area of the liquid surface available to each molecule in a solution.

Question 58

What’s an emulsion?

Answer 58

An emulsion is a mixture of two or more liquids that are usually immiscible but, under specific transforming processes, will adopt a macroscopic homogeneous aspect and a microscopic heterogeneous one.

In an emulsion, one liquid is dispersed in the other.

Question 59

What is the continuous and dispersed phase of an emulsion?

Answer 59

Continuous: outer phase

Dispersed: inner phase (droplets in continuous phase)

Emulsions can also be classified by:
- Temperature
- Type/conc of emulsifier (surfactant)
- Viscosity ratio of the two phases
- Volume ratio of inner: outer phases

Question 60

How are ultra-low surface tensions achieved at oil-water interfaces?

Answer 60

By adding:
- co-surfactants
- certain ionic surfactants
- non ionic surfactants

Question 61

Describe microemulsions:

Answer 61

Thermodynamically stable emulsions formed with 10-50 nm droplets.

The drops are too small to scatter visible light, so tend to appear optically clear.

Question 62

What is the Bancroft rule?

Answer 62

The continuous phase will be the phase in which the emulsifier becomes more soluble.

Question 63

What is the hydrophilic-lipophilic balance (HLB)?

Answer 63

A number assigned to a surfactant describing its hydrophilic-lipophilic balance at 20C.

HLB = xH - yL + 7

Where x and y are the number of hydrophilic and lipophilic groups respectively

H and L are the associated values related to a particular species (from a table)

Question 64

What’s the effect of the hydrophilic-lipophilic balance (HLB) number on emulsification?

Answer 64

Low HLB numbers (< 4) show no dispersibility in water, and would be good water-in-oil (W/O) emulsifiers.

Higher HLB numbers (12 and above) are translucent to clear, and make good detergents and solubilisers in O/W systems.

Question 65

What makes an emulsion appear transparent?

Answer 65

1. If the droplet size is much smaller than the wavelength of light (microemulsion)
2. If the refractive index of the liquids are matched

Question 66

Describe the rheology of concentrated emulsions:

Answer 66

They exhibit non-Newtonian rheology e.g. shear thinning and yield stress.

Its viscosity will depend on: volume fraction, inner/outer phase viscosity, droplet size, droplet deformability, and the nature of droplet-droplet interactions.

Question 67

List key issues with emulsion stability:

Answer 67

Coalescence
Flocculation
Creaming
Breaking

Question 68

How can an emulsion be stabilised against coalescence?

Answer 68

Lower interfacial tension

Marangoni effect - local fluctuations dampened by local gradients of interfacial tension

Mechanically strong interfacial films with large values of interfacial elasticity E

High viscosity of continuous phase retards drainage of thin films

Stable colloids with repulsive interactions between droplets, preventing aggregation

Question 69

What is a Pickering emulsion?

Answer 69

Systems composed of two immiscible fluids stabilized by solid particles.

Instead of relying on traditional emulsifiers, Pickering emulsions utilize solid particles to stabilize the interface between the two immiscible liquids.
These particles adsorb onto the oil-water or air-water interface, creating a protective barrier that prevents coalescence.

If oil and water are mixed and small oil droplets are formed and dispersed throughout the water (oil-in-water emulsion), eventually the droplets will coalesce to decrease the amount of energy in the system.
However, if solid particles are added to the mixture, they will bind to the surface of the interface and prevent the droplets from coalescing, making the emulsion more stable.

Question 70

What does the Weber number, We, show?

Answer 70

It describes the balance between flow-induced droplet break-up and surface tension forces resisting drop deformation

We = Drag / Cohesive Forces

If the We number exceeds a critical value (at a certain viscosity ratio), the drops break into smaller droplets, We>(We)c

Question 71

What is the Capillary number?

Answer 71

The capillary number, Ca, is defined as the ratio of viscous to interfacial forces and is used to study the microscopic displacement of the polymer.

• The larger the Ca number, the greater the deformation
• Adding surfactants favours emulsification as surface tension decreases and viscosity increases

Question 72

Define foam

Answer 72

Air dispersed in a liquid as a stable mass of bubbles

Question 73

What are the key parts of foam structure:

Answer 73

Lamellae (flat faces)
Vertex
Plateau borders (straight edges between borders)

Question 74

What 3 things are needed to form a foam?

Answer 74

Gas
Liquid
Surfactant

Foam triangle

Question 75

Why are surfactants important in foam formation?

Answer 75

Foaming requires a stable gas-liquid interface, provided by surfactant molecules

At the gas-liquid interface, surfactants reduce the liquid surface tension keeping lamellae surfaces apart.

Micelles can form at high concentrations.

Question 76

How are bubbles added to a foam?

Answer 76

Sparging (blowing in gas)

From solution (evolution, boiling, cavitation)

By entrainment (plunging jets, turbulence, vortex formation)

Question 77

What is a Plateau and Vortex?

Answer 77

Plateau: 3 foam films meeting at lines (always meet at 120o)

Vortex: 4 plateau borders meeting at a junction (always meet at tetrahedral angle around 109o)

More than 3 films at a plateau border or 4 plateau borders at a vertex are unstable

Question 78

What occurs in froth flotation?

Answer 78

Hydrophobic particles stick to the bubbles and float to the surface.

Make the valuable mineral hydrophobic using surface chemistry, and add air bubbles. Via the above, an overflowing froth is formed that holds the valuable particles.
Waste stays in the liquid and can be disposed.

Chemically and physically complex.

Question 79

Why are emulsions, Pickering or surfactant based, thermodynamically unstable?
Given that they are thermodynamically unstable, why can emulsions be formed (both Pickering and Surfactant types) which are stable for more than 12 months?

Answer 79

• The lowest energy state will minimise the surface area of any drops as this lowers the total surface energy.
• The lowest energy state corresponds to all the droplets coalescing, so that eventually the two phases separate.
• However, the interactions between two drops in a “stable emulsion” are going to be repulsive.
• If this energy barrier is high, then the drops will not be able to get close enough to coalesce.

Question 80

You are to make a Pickering emulsion (particle stabilised emulsion) using silica nanoparticles, which are hydrophilic, and negatively charged, water and diesel.
Unfortunately, it does not work.

Why is this so and how could it be overcome?

Answer 80

• It is unstable because the particles are too hydrophilic
• They do not go to the interface and so no stable emulsion is made
• The particles need to be more hydrophobic
• Ideally this would give a contact angle of 90o

Note: the continuous phase is the one in which the surfactant/emulsifying particles are more stable, so in this case the continuous phase is the water

Question 81

Cetyltrimethyl ammonium bromide, CTAB, is a common cationic surfactant.
In water it produces approximately spherical micelles, but in the presence of 2 molar sodium chloride forms long rod-like or wormlike micelles, why is this so?

Answer 81

The change of micelle shape from a water system to a 2M NaCl system is due to the packing fraction (the relative size of the head group to the tail)

P=Vs/(as L) Where V is the tail volume, a is the optimal surface area occupied by the surfactant monomer at the micelle-water interface, and L is the maximum extended chain length

In water, there is a repulsion between the head groups on the surface of the micelle, so the head group is larger in area than the tail group, so packing as a sphere is favoured.

Adding salt screens the range of repulsion between the head groups, making the effective surface area less encouraging and packing in rods

Question 82

A 30% oil in water mixture is to be emulsified. It is found that 1% of Sodium dodecyl sulfate added to the mixture works well, whilst a 1% dodecan1-ol added to the mixture does not, even though in both cases the interfacial tension between the oil and water drops from 30mN/m to 1 mN/m.

Why is this so?

Answer 82

• Simply lowering the interfacial tension will not give rise to a stable emulsion
• It is necessary that there is repulsion between the drops.
• SDS has sulfate groups which are charged and repel each other through the electrical double layer.

Question 83

In a system with two glass spheres immersed in various sodium chloride solutions at pH 5.5, in 10 mM and 100 mM NaCl, they are entirely repulsive.

Why is this so?

Answer 83

• The electrical double layer interaction dominates over Van der Waals attraction
• If the spheres were pressed very hard together, the attraction would become larger than EDL repulsion
• They have not been pressed that hard

Question 84

In a system with two glass spheres immersed in various sodium chloride solutions at pH 5.5, why is the repulsion interaction longer ranged at 10 mM compared to 100 mM?

Answer 84

Due to screening of the electrical double layer EDL, which is larger at higher salt concentrations.

Question 85

What is the EDL?

Answer 85

Electrical double layer

The EDL is essentially a wall of ions that surrounds the surface in a way that balances its charge and screens the potential from the surface.

Question 86

What does screening refer to with regards to colloids and particle interactions?

Answer 86

The phenomenon where the presence of ions in the solution reduce the strength/magnitude of the electric field surrounding charged particles.

This reduction in the electric field occurs because the ions in the solution partially neutralize the charges on the surface of the particles, effectively shielding them from interacting with other charged particles or surfaces over long distances.

When ions from the solution are present in high concentrations, they can penetrate the diffuse layer of the EDL and interact with the charged particles.
These ions effectively shield the charged particles from each other by neutralizing their charges to some extent.
As a result, the strength of the electric field surrounding the charged particles is reduced, and the interaction between them becomes weaker over longer distances.

The screening effect reduces the range over which charged particles can interact with each other. In solutions with higher ion concentrations, the screening effect is stronger.

Question 87

In a system with two glass spheres immersed in various sodium chloride solutions at pH 5.5, why are there only attractive forces at 500 mM?

Answer 87

The EDL forces have been reduced so much that the interaction is totally dominated by Van der Waals forces, so there is no repulsion.

Question 88

A dilute xanthan solution (xanthan is a polysaccharide) meets this rheological
criterion, but on its own is generally not suitable as a drilling fluid, why is this so?

Answer 88

Xanthan does not form a filter cake in permeable regions, so drilling fluid could penetrate permeable areas.

Question 89

What’s depletion flocculation?

Answer 89

A depletion force is an effective attractive force that arises between large colloidal particles that are suspended in a dilute solution of depletants, which are smaller solutes that are preferentially excluded from the vicinity of the large particles.

These attractive interactions bring the dispersed particles together resulting in flocculation.

The depletion force is described as an entropic force because it is fundamentally a manifestation of the second law of thermodynamics, which states that a system tends to increase its entropy.

Question 90

Why is the frequency of oscillation (repulsion and attraction) greater at higher polymer concentration i.e. at 48.5 mM compared to 4.8 mM?

Answer 90

Polystyrene sulfonate is a polyelectrolyte, and so the negative charges on the polymer expand the polymer (It’s as if the negative charges repel each other, pushing the polymer chains apart).
This means there are 10 times as many counter ions at the higher polymer concentration. This interaction will be screened more at high concentrations, so the polymer will be smaller and the frequency of oscillations will be higher.

The negative charges on the NaPSS polymer attract positively charged ions (counter ions) from the solution. At higher concentrations of NaPSS, there are more negative charges, which means there will be more counter ions attracted to the polymer. The presence of these counter ions screens or reduces the repulsion between the negative charges on the NaPSS polymer chains. Think of it like adding a layer between two magnets – it reduces the repulsion between them.
The oscillations in the system are related to how close or far apart the colloidal particles are from each other. At higher NaPSS concentrations, where the polymer molecules are smaller due to screening by counter ions, the particles can get closer to each other before the repulsion between the polymers becomes significant. This means that they oscillate more frequently between attraction and repulsion as they move closer together and farther apart.

Question 91

Why is the magnitude of the oscillations (of particle attraction and repulsion) larger at higher polymer (polyelectrolyte) concentrations, i.e.
at 48.5 mM compared to 4.8 mM?

Answer 91

At higher concentrations, the depletion interaction (which is osmotic in nature) is stronger.

Question 92

How does electrical double layer (EDL) thickness vary with concentration?

Answer 92

As concentration increases, thickness decreases.

As the concentration of ions in the solution increases, more ions are available to neutralize the surface charge of the particle. This results in a more compact electrical double layer with a smaller thickness.

At very high concentrations, the electrical double layer thickness may reach a limiting value, beyond which further increases in concentration have minimal impact on the double layer thickness. This is often referred to as the “Debye length”

Debye length (thickness of electrical double layer) = 1/κ

As concentration increases, κ increases, so 1/κ decreases, so the EDL gets smaller/thinner

Question 93

Quantify the dispersion contributions to the surface tensions of pentane and hexadecane. Justify your answer:

Answer 93

Hydrocarbons only have dispersion interactions, so the total hydrocarbon surface tension is equal to the dispersion interaction.

This is because alkanes are nonpolar, and their surface tension is primarily due to dispersion forces since they lack the ability to engage in hydrogen bonding or significant dipole-dipole interactions.
So, their dispersion contribution to surface tension is their total surface tension value.

Question 94

What key factors determine foam/bubble stability?

Answer 94

Surface viscosity (higher is better)

Film elasticity (higher is better)

Zeta potential (higher repulsion between ionic surfactants in the film walls prevents the collapse of bubble walls)

Question 95

How does chain length impact the conc’ and surface tension in foams?

Answer 95

As chain length increases, the concentration of surfactant needed to reduce surface tension decreases.