Colloids: Concentrated Dispersions and Rheologyadahlin.com/onewebmedia/TIF015/colloids_2.pdf2017-09-18 2 A bucket of paint typically contains 15 000 m2 of interfacial area between

2017-09-18

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1Soft Matter Physics

Colloids: Concentrated Dispersions and Rheology

2017-09-19

Andreas B. DahlinLecture 2/6

Jones: 4.3-4.5 Hamley: 3.4-3.7, 3.15

[email protected]

http://www.adahlin.com/

Many colloidal dispersions are highly concentrated.

• What are the interactions forces between colloids?

• When do colloids aggregate and can we control it?

• What are the rheological properties of concentrated dispersions?

2017-09-19 Soft Matter Physics 2

Colloidal Interactions

mailto:[email protected]

http://www.adahlin.com/research/

2017-09-18

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A bucket of paint typically contains 15 000 m2 of interfacial area between the colloids

and their solvent.

One can suspect that suspensions generally aggregate to reduce interfacial energy. On

the other hand the contact area is not high…

Can the repulsive forces create a sufficiently high kinetic barrier to prevent aggregation

on ordinary timescales?

What other attractive forces come into play?


Aggregation

×

Attractive:

• Depletion interactions (if another species is present).

• Van der Waals (always).

Repulsive:

• Electrostatic (depends on ionic strength).

• Entropic (excluded volume).

In addition, chemically modified colloids (e.g. with polymers as we will see) will have

more attractive or repulsive forces.


Interaction Forces

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Repetition: Entropy (Statistical Mechanics)

Entropy is about probabilities and the number of microstates associated with a certain

macrostate. The microstates are not observable! Entropy is lack of information.

Two dice have 36 microstates with equal probability. Entropy can be observed in the

sum macrostate.

Example: Probability of getting macrostate 7 with two dice is 1/6 (6 out of 36

microstates). The probabilities for getting 2 or 12 are only 1/36 each.

1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12


Repetition: Boltzmann’s Entropy Formula

Most general entropy formula:

The probability of microstate i is pi. Boltzmann’s

constant kB = 1.3806×10-23 JK-1 relates entropy to

free energy via temperature.

If all W microstates are equally probable p = 1/W for

all i and n = W. We can get the famous formula:

n

i

ii ppkS1

B log

WkWW

kWW

kSW

i

W

i

loglog11

log1

B

1

B

1

B

Wikipedia: Ludvig Boltzmann

The logarithmic dependence essentially comes from combinatorics: If there are WA states

in system A and WB states in system B the total number of states is WAWB, but entropy

becomes additive: SA + SB = kBlog(WA) + kBlog(WB) = kBlog(WAWB)

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Alice and Bob have two kids…

I: One is a boy.

p(the other is also a boy)?

I: The older is a boy.

p(the younger is also a boy)?

I: (nothing more)

p(both are boys)?


Test: Entropy of Gender


Repetition: Volumes and Entropy

Consider volume expansion again from the viewpoint of statistical mechanics.

We can discretize the space available into a certain number of positions, each with

volume dV, where a gas molecule can be located. The entropy change is then:

dV

Vi

Vf

i

fB

i

fB

i

fBifBiBfB

log

d/

d/loglogloglogloglog

V

Vk

VV

VVk

W

WkWWkWkWkS

dV

Typically used when dealing

with classical problems in gas

expansion etc.

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Depletion Interactions

Imagine that a flocculant, another type of solvated objects, is added to the dispersion.

The new objects are intermediate in size between the colloids and the solvent molecules.

There will be a depletion zone with thickness d just around the colloids with zero

concentration and d must be at least equal to the radius of the flocculant.

Aggregation is entropically favorable because it increases the volume available to the

flocculant with Vdep which is the overlap of the depletion zones.

Vdep d

R0


Osmotic Pressure

The solvent and the flocculant will strive for homogenous concentrations.

Osmotic pressure pushes colloids together.

When Vdep > 0 there will be a net force pushing the colloids further together because

there is no depletant in between, so the solvent concentration is higher.

We have Vdep = 0 when the center to center distance r > 2[R0 + d].

Note that r < 2R0 is impossible (hard wall).

osmotic pressure

Limitation: Does not explain why

colloids would get close enough to

start with…

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Depletion Interaction Energy

For two spheres with radius R and center to center

distance r, Vdep is given by geometry:

In our case we have R = R0 + d. Upon aggregation r = 2R0

so we get:

The osmotic pressure is approximated with an ideal gas:

The energy can then be approximated as:

2dep 2412

π, rRrRRrV

Wolfram MathWorld: Sphere-Sphere Intersection

R R

r

dRd

RdRRdRV 463

π2224

12

π0

22

0000dep

deposmVPG

TCkP Bosm concentration of flocculant

Simplified physical explanation is charge fluctuations:

Temporary interacting dipoles.

From quantum mechanics: The interaction energy U

of two uncharged atoms at separation distance r is:

For the planar surfaces of two semi-infinite blocks

separated by a distance d the interaction energy per

unit area is:


Van der Waals

6

constant

rrU

2

flat

π12 d

Z

A

dU

Here Z is the Hamaker constant, which depends on material etc. Z ≈ 10-19 J

Van der Waals interactions are responsible for several important sticking phenomenon

like plastic kitchen films and the Gecko lizard!

Wikipedia: van der Waals force

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For other geometries the energy-distance relation

changes! If d is smaller than the objects involved the

geometric Derjaguin approximation can be used.

Sphere and planar surface:

Two spheres:


Sphere Interactions

dR

d

ZRFdU

d6

dR1

R2

d << R1

d << R

2

flat

6π2

d

ZR

A

URRdF

2

21

21flat

21

2121

6π2

d

Z

RR

RR

A

U

RR

RRRRdF

d

Z

RR

RRFdU

d621

21

We now turn to repulsive forces: Many colloids are stabilized by being charged.

Recall that surface potentials are screened by ions. The repulsion is thus not simply

electrostatic. It is mainly an osmotic pressure effect (repulsive unlike depletion).

For two plates with d comparable to κ-1 the diffuse layer theory (last lecture) can be used

to show that the pressure (force per area) is:

No derivation here, sorry…

Actually P is just proportional to exp(-κd)!

d

ψ0 ψ0


Repulsion Between Charged Interfaces

d

Tk

e

Tk

e

Tk

e

Tk

e

TCkdP

exp

expexp

expexp

64

2

B

0

B

0

B

0

B

0

0B

J.N. Israelachvili

Intramolecular and Surface Forces Academic Press 2011

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Interaction Energy for Spheres

The interaction energy per area for moving the plates together is then:

We can do the same approximation as for the van der Waals force. For sphere-surface:

For two spheres:

dPP

A

dU

d

1flat

dPRA

URdF 1flat π2π2

dPRdU 2π2

dPRR

RR

A

U

RR

RRdF 1

21

21flat

21

21 π2π2

dPRR

RRdU 2

21

21π2

now representing

repulsion


Simplifications and Assumptions

Expression for P(d) is strictly only valid for 1:1 electrolytes, but not so bad for other

electrolytes as long as |ψ0| < 80 mV. (But you must still calculate a new Debye length for

each electrolyte!)

If |ψ0| < 25 mV the formula for osmotic pressure simplifies:

The simplified Grahame equation (previous lecture) can then be used:

This simplifies the expressions for U(d) and works for any electrolyte!

Remember: Grahame only relates ψ0 to surface charge density σ.

ddP exp22

0

2

0

ddP

exp2

0

2

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The DLVO theory (Derjaguin and Landau, Verwey and Overbeek) combines van der

Waals attraction and double layer repulsion:

Van der Waals energy always gives d-1 dependence.

Double layer energy always approximately proportional to exp(-κd).

Some limitations:

• No hydrophobic effect (solvent entropy)!

• The eventual repulsion at very small d is not included.

• Adsorbed ions! At what d do we have ψ0?


Combining Van der Waals and Double Layer

dUdUdU dlvdWtot


DLVO Curves

Curve types depending on ionic strength:

• High energy barrier.

• Secondary minimum.

• Monotonic decrease.

Secondary minimum indicates reversible

semi-aggregation!

Example for:

ψ0 = 20 mV

T = 300 K

ε = 80

R = 50 nm

Z = 10-19 J

0 100 200 300 400 500-15

-10

-5

0

5

10

15

κ-1 = 100 nm

κ-1 = 50 nm

κ-1 = 20 nm

κ-1 = 1 nm

d (nm)

U/[

k BT

]

minimum

dR R

just proof of principle,

only accurate when d << R

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Aggregation

When the attractive forces overcome the repulsion the colloids aggregate:

Weak attraction gives crystals (hexagonal close packed or face centered cubic).

Strong attraction gives fractal forms. The ideal packing is never reached!

crystalline fractal


Sedimentation of Aggregates

Aggregates have a much higher tendency

to sediment due to their weight.

This explain the formation of a river delta!

The salty seawater screens the electrostatic

repulsion and causes aggregation of

colloids in the river freshwater. The

aggregates sediment and gather up on the

bottom. Water is forced to spread out.

Also an important separation technique:

Induce aggregation and collect the

sediment!

Selenga River

http://www.geology.com/

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Hard Spheres

The basis for understanding concentrated suspensions which do not aggregate is the hard

sphere model.

Extremely simplified yet reasonably accurate energy-distance relation:

U(r > R) = 0

U(r < R) → ∞

center to center distance (r)

inte

ract

ion

en

ergy

0 R0

Similar but not identical to an

ideal gas…

Charged colloids can still be

modelled by adding the Debye

length to the radius:

R = R0 + κ-1

κ-1

stays at

infinity ↑

stays at

zero ↓


Temperature Independence

When dealing with colloids as hard spheres the only energy levels are zero or infinity, so

temperature does not come into play.

Conclusion: Changing T does not influence a dispersion. Instead the critical parameter is

the volume fraction Φ of hard spheres (not of colloids).

Can hard sphere colloids form crystals in concentrated (Φ approaching 0.5) dispersions?

• There are no attractive forces and no enthalpic effects.

• There is no phase transition temperature.

• There can only be an entropy change.

T Φ

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Video: Colloidal Self Assembly


Entropic Repulsion

Clearly the hard spheres will collide with each other all the time as they diffuse around.

Random packing has a maximum of Φ = 0.64.

It turns out that an ordered state of Φ = 0.545 is preferable for hard spheres at Φ > 0.494.

Each colloid has more space locally despite the long range order (∆S > 0).

This is a true phase transition: Suspensions with 0.494 < Φ < 0.545 are unstable and

separate. But the process is (almost) independent of temperature.

60°

Φmax = 0.64

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Close Packed Crystals

Also Φ > 0.545 the equilibrium state for the colloids remains crystalline. As Φ increases

the distance between colloids simply decreases. Φmax ≈ 0.74 for crystalline spheres.

However, a glass can be formed if Φ ≈ 0.58 due to the high viscosity of the dispersion.

(Not the same mechanism as for fractal aggregates!)

crystal glass

A summary of dispersion phases:


Hard Sphere Phase Diagram

Φ0.49 0.54 0.58 0.64 0.74

← random

“liquid”

impossible →

(packing limit)

impossible

(phase separation)

glass may

form

crystal

“solid”

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Estimate at which volume fraction colloids of radius 100 nm form crystals in water

containing monovalent salts at ionic strength of 10 μmolL-1 (very pure fresh water) and

100 mmolL-1 (comparable to salt water).


Exercise 2.1

The Debye length is calculated with:

C0 = 10 μmolL-1 = 10-2NA m-3 or C0 = 100 mmolL-1 = 102NA m-3

e = 1.602×10-19 C, kB = 1.381×10-23 JK-1, ε0 = 8.854×10-12 Fm-1, ε = 80 and assume T =

300 K. This gives κ-1 ≈ 97 nm and κ-1 ≈ 1 nm.

The volume fraction of colloids (Φc) is related to the volume fraction of hard spheres

(ΦHS) as:

The transition to the ordered phase is at ΦHS = 0.494 and R = 100 nm. This gives Φc =

0.06 at the low and Φc = 0.48 at the high ionic strength (strong effect at long κ-1).


Exercise 2.1

2/1

B0

2

02

Tk

eC

3

1HS

HS

cHSc

R

RΦ

V

VΦΦ R

R + κ-1

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Repetition: Non-Newtonian Liquids

Newtonian

shear thinning

shear thickening

strain rate (∂e/∂t)

shea

r st

ress

(σ)

0

0

shear thickening

strain rate (∂e/∂t)

vis

cosi

ty (η)

0

0

shear thinning

Newtonian

Bingham plastic

t

e

t

e

Viscosity may depend on strain rate!


Viscosity of Dispersions

We now turn to rheology!

Using the hard sphere model the viscosity of suspensions can be predicted.

If Φ is not too high (single colloids) Einstein said:

This also means the suspension remains Newtonian!

Only based on hydrodynamics. The ”hard sphere” border is in at the zeta potential, not at

the Debye length extension!

Φ5.210

Φ < 0.1

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The characteristic diffusion time is the average time it takes for a colloid to diffuse a

distance equal to its own size. From Brownian motion this time is:

Using Einstein-Stokes we get:

The characteristic flow time is given by the inverse shear rate. Taking the ratio we get

the Peclet number:

For Pe < 1 the colloids have time to rearrange themselves by diffusion. The flow drags

the whole suspension.

For Pe > 1 the flow breaks the colloids free into an inhomogeneous dispersion.


The Peclet Number

D

R2

diff

Tk

R

B

3

0diff

π6

Tk

R

B

3

0π6Pe

velocity

gradient new

structure


Shear Thinning in Dispersions

W.B. Russel, D.A. Saville, W.R. Schowalter

Colloidal Dispersions Cambridge University Press 1989

Data from different colloids

in different solvent appear

on the same graph!

At very high shear rates,

aggregates form by

hydrodynamic coupling and

a small shear-thickening

effect is usually observed.

Note that Φ is not specified.

low shear

plateau

high shear

plateau

Pe = 1

eventually

increased η

changed by

shear stress

ηlow/η0

ηhigh/η0

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High and Low Pe Limits

W.B. Russel, D.A. Saville, W.R. Schowalter

Colloidal Dispersions Cambridge University Press 1989

Φ influences η differently

depending on whether we

are in the low or high Pe

limit.

Empirical formula for

dependence on Φ: Pe < 10-3

Einstein

2

0

high

71.01

Φ

2

0

low

63.01

Φ

glass

transition

Pe > 1

A water-based varnish is composed of a dispersion of polymer spheres with diameter

200 nm. It is brushed onto a surface as a film of thickness 200 μm. How fast must the

brush be moved to achieve a high degree of shear thinning?


Exercise 2.2

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To get substantial shear thinning, the Peclet number should be unity:

We have R = 100 nm, η0 = 10-3 Pas. Assume T = 300 K. This gives:

The shear rate is the velocity gradient, which we assume is linear. If d = 200 μm:

v = ∂e/∂t×d = 0.043… ms-1

So the brush should move about 4 cms-1 (reasonable).


Exercise 2.2

1π6

B

3

0

t

e

Tk

R

1-

393

23

s ...63.2191010010π6

3001038.1

t

e

v

d


Reflections and Questions

?

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Exercise 2.3

Derive the “low potential” and “any electrolyte” version of DLVO theory for two

spheres of radius R. (For the double layer repulsion, start with the simplified expression

for P(d) and use the Derjaguin approximation in the same manner.)

→

d

ZdRdU

12expπ2

2

00

Documents

Colloids: Concentrated Dispersions and Rheologyadahlin.com/onewebmedia/TIF015/colloids_2.pdf2017-09-18 2 A bucket of paint typically contains 15 000 m2 of interfacial area between