Lecture #1 begins here Introduction

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Lecture #1 begins here Introduction What are Colloids? The word “colloid” is not widely used in ordinary English, although you can find it listed in most dictionaries. This word was coined in the 1860’s by Thomas Graham from the Greek word kolla meaning glue. Today the word is used in the technical literature to describe something related to glue but also quite different. For the purpose of this course, I will define colloid as follows: (lyophobic♣) colloid – a state of mixtures of two immiscible components in which one component is finely divided as bubbles, drops or particles in a second (usually continuous) phase By “finely divided” we mean that at least one dimension is in the range 10–9 m to 10–6 m. The size is perhaps the most important aspect of the definition – as we will see in a moment – because it is the size of the particles which distinguishes this state of mixtures from others. For comparison, let me give a similar definition for a more familiar state of mixtures: thermodynamic solution - a state of mixtures of two (miscible) components in which the molecules of one component (the solute) are each completely surrounded by molecules of the second component (the solvent). The important difference is that in a solution the solute molecules are completely separated from each other, whereas in a colloid, clusters of molecules of one component are surrounded by a sea of the other component. The following figures shows three different states for mixtures of two components indicated by the black and white colors:

♣ lyophobic literally means solvent hating or solvent fearing. Wiktionary defines it as “having no affinity for the dispersion medium and thus easily precipitated.”

Copyright© 2016 by Dennis C. Prieve

separated phases

colloid

thermodynamic solution

Consider two components like benzene and water. If we vigorous stir the mixture with a mechanical blender we can produce an emulsion of oil droplets in water. This is a simple and familiar example of a colloid. emulsion – droplets of one liquid finely divided in a second immiscible liquid (e.g. benzene in water) The reason that these two components are immiscible in each other is that molecules of benzene would prefer to be in the vicinity of other benzene molecules rather than near molecules of water. While molecules in the interior of colloidal particles are "happy" because they are in contact with "friends," those on the surface of the particle (droplet) are "unhappy" because they can see "foes." The collective happiness of the entire group can be increased if all the oil droplets combine to form one large droplet. Then fewer molecules lie on the boundary between the two phases. In this case, the colloidal state is not thermodynamically stable because it tends to revert to the original state of the mixture before mixing in which the oil and water for two separated layers of liquid. This represents the third state of mixtures of two components: totally separated phases. Other household examples of colloids include: • • • • •

salad dressing, mayonnaise (L/L) paint, ink, xerographic fluids (S/L) fog (L/G) smoke (S/G) magnetic recording tape (S/S)

Most of these familar household products are useful only if they maintain their colloidal state. For example, if the polymer particles in latex paint were to combine to form a continuous layer of polymer in the bottom of the can (with a layer of water on top), you would be unable to spread the polymer on the wall. PDF generated April 27, 2016

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So maintaining the colloidal state (which is often thermodynamically unstable) is an important problem in colloid science. Another important problem is the inverse of this problem: getting a colloidal dispersion to revert to separated phases. Once you spread the paint over the wall, you want the polymer particles to combine together to form a continuous film which will protect the substrate from the environment. In this course, we will try to understand why colloids are generally unstable, how they can be partially stabilized, and once stabilized how they can be destabilized. We will also learn how to characterize colloids and we will learn about other kinds of behavior that they display. Some Model Lyophobic Colloids

Some Lyophilic Colloids Besides mixtures of immiscible phases, there are also soluble macromolecules which have colloidal lengths or which self-assemble into structures having colloidal lengths. These are called lyophilic♦ colloid – a thermodynamic solution of a macromolecule having colloidal dimensions or a small molecule which self-assembles to form structures having colloidal dimensions Examples of macromolecules which are quite soluble in water are DNA, proteins, polysaccharides (starch) and poly(acrylic acid).

The above are 2 µm spheres of ZnS also known as sphalerite. Below are 1 µm cubes of CdCO3.

Small molecules which can self-assemble into colloidal structures are called surfactants♥ – small molecules which display an affinity for the interfaces between two immiscible phases By affinity, I mean tending to adsorb at these interfaces. A typical surfactant is sodium dodecylsulfate or SDS:

The above two micrographs were taken from Hunter Vol. 1, Fig. 1.5.2. Of course, most colloids of commercial interest are of irregular shape and not uniform in size. Below is a micrograph of silver particles downloaded from http://silver-colloids.com/Reports/SEM-Images.html.

The zigzag portion of this molecule contains 12 methyl (CH2) groups which form an alkane chain. If we eliminate the sulfate group, we have dodecane, which is an oil (immiscible in water but miscible in other oils). But owing to the polarity of the sulfate group (SO4–), SDS is quite soluble in water.

♦ lyophilic means solvent loving. Wiktionary defines it as “having an affinity for the dispersion medium and thus not easily precipitated.” ♥ surfactant is a contraction of surface active agent


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Not only do surfactants adsorb at interfaces, but they also tend to self-assemble to form larger structures in solution, as illustrated below

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surface (or interface) – the boundary between any two immiscible phases While we tend to think of surfaces or interfaces as being sharp, at the molecular level they are diffuse regions having a thickness on the order of 1 nm.

interface

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phase #2

»1nm

Adsorption of surfactants, tends to reduce the interfacial energy (surface tension) which can slow the rate of aggregation of thermodynamically unstable lyophobic colloids. Later in the course, we will find that addition to surfactants to lyophobic colloids is essential to formulating many commerical products. The circular aggregate of SDS molecules is called a micelle – an aggregate of surfactant molecules While SDS tends to form spherical or cylindrical micelles (depending on conditions), other surfactants (e.g. phospholipids) form other shapes such as bilayers:

Regardless of whether we are talking about lyophilic or lyophobic colloids, the most important part of the definition is size of the structure: size matters a lot. There are several reasons which will be revealed as the semester progresses. Let me give you one reason now. As particle size decreases, the fraction of molecules of the particle which reside at the surface increases. To prove this, consider two concentric spheres whose radii differ by t = 1 nm, the “thickness” of the interface. Let’s compare the volume of the interfacial shell Vshell =

4 3

π(R + t) − 3

4 3

πR3

with the volume of the interior sphere Vinner=

4 3

πR3

Taking their ratio 3 π ( R + t ) − 43 πR3  t  1 = = +   −1 4 πR3 Vinner  R 3

Vshell

4 3

3

Taking t = 1 nm, we obtain the following values for this ratio:

A spherical bilayer is called a vescicle. When the structure is formed using phospholipids, it’s called a liposome. All cell membranes are bilayer sheets of phospholipids. Importance of the Surface The course title concerns “colloids” and “surfaces.” The word “surface” means about the same in physical chemistry as it does in ordinary English:


R (m)

Vshell Vinner

10–3 (1 mm) 10–4 10–5 10–6 (1 µm) 10–7 10–8 10–9 (1 nm)

3×10–6 3×10–5 3×10–4 0.003 0.03 0.33 7

For particles of 1 mm (e.g. sand grains), the fraction of molecules in the interfacial region is negligible.


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For particles of 1 µm, the fraction is starting to become significant. For nanometer particles, there are actually more molecules in the interfacial region than in the interior. Think of this interfacial region as a third phase whose thermodynamic properties lie somewhere between those of the interior phase and the exterior phase. As particle size decreases, the thermodynamic properties of the mixture changes, with the interfacial region playing an increasingly important role. Surface Tension The interface contributes to the Gibbs free energy of a mixture. That contribution is proportional to the interfacial area; the proportionality constant is called the interfacial tension – Gibbs free energy per unit area of interface between any two immiscible phases surface tension – Gibbs free energy per unit area of surface between a liquid and its vapor (or air) In equation form, these definitions correspond to

 ∂G  γ = mix   ∂A T , P, N j where γ is surface or interfacial tension, Gmix is the Gibbs free energy of the mixture, A is interfacial area, T is temperature, P is pressure and Nj is the number of moles of each component j. Surface tension as an energy Recall from thermodynamics that systems tend to spontaneous transform themselves into that state having the minimum energy. The simplest experiment which demonstrate the existence of surface tension is the following.♠ Consider a soap film supported by a rectangular wire frame in which one edge of the frame is moveable:

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moveable slide wire soap film F

l

x rigid wire frame Top View gas liquid gas Side View The soap film satisfies the definition of a colloid because at least one dimension (the thickness) is in the range of 1 nm to 1 micron. The soap film has two faces, so the total interfacial area is A = 2xl and the energy associated with the interface is this area times the interfacial tension: E = Aγ = 2xlγ

(1)

Thus surface tension can be thought of as the energy of the interface per unit area: E = γ = 72.7 mJ/m2 for water at 20ºC A

Surface tension as a force Owing to the thermodynamic tendency to minimize the energy, the soap film wants to retract in order to reduce the interfacial area A. To overcome this tendency, we need to apply a force F to the slide wire as shown in the figure above.

soap film dx

♠ This also demonstrates the importance of surfaces to colloids.


To calculate the magnitude of this force, consider a differential change in the x-dimension of the soap


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film denoted by dx. If the associated change in interfacial energy is dE, the force is given by

F=

dE = 2l γ dx

where the second equation is obtained by substituting the derivative of (1). Rearranging

γ=

F = 72.7 mN/m for water at 20ºC 2l

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Then the smaller the particle, the higher the density of information storage: large size → low info density But there is a limit to how small the particles can be. For small particles, the magnetic dipole undergoes significant thermal fluctuations, which means that any stored information would be lost: small size → (super)paramagnetic Thus there is an optimum size.

Recall that the “2” in this equation arises between for a soap film, there are two air-water interfaces which contribute to the force, each of length l. Notice also that the length l is measured along a direction which is perpendicular to the direction of the force. Thus surface tension is also a force per unit length. This is a real force whose affect can be seen in this slide wire experiment. If the slide wire is not pinned the force will spontaneously pull the slide to the left in order to reduce the area of the film. Surface tension causes a number of fascinating phenomena, which are demonstrated in the film by Lloyd Trefethen (MechE Dept, Tufts U, 1963) shown in class. Particle Characterization

Another example of a colloidal product is latex paint which consists of 200 nm spheres typically of some acrylic polymer. Here the shape isn't so important. The emulsion polymerization process happens to produce spheres. Thin films of these polymers would tend to be clear. TiO2 particles are sometimes added as a pigment to produce an opaque white film.

Shape Reading: Berg §5.D.4 on page 382f. The performance of a colloid as a product is often determined by physical properties -- size, shape, density -- as well as chemical composition. One example is the iron oxide particles commonly found in audio and video recording tape as well as floppy disks, hard disks etc. These particles are composed of γ-Fe2O3 (maghemite), which is a particular phase (denoted by the γ) of this ferric oxide which has good magnetic properties. Size and shape are also important. The best recording properties are obtained using needleshaped (acicular) particles typically 20×120 nm in size.

Another common shape is the platelet. Many clays used for ceramics naturally occur as hexagonal plates typically on the order of 100x1000 nm in size. Barium ferrite is a magnetic material which also exists as hexagonal platelets, but much smaller in size. This material maintains its ferromagnetic properties at much smaller size and so it is of considerable interest for magnetic recording. One difficulty is getting it dispersed.

Generally, small particles are preferred since, ideally, each particle could store a single bit of information. Copyright© 2016 by Dennis C. Prieve


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The magnetic dipole is usually aligned with the shorter geometric axis with the ferrites whereas the dipole is aligned with the long axis in ferric oxides.

y

side view

x

z

Strong magnetic attractions cause these particles to aggregate with the dipoles aligned north-to-south. This produces long necklace-like structures with the ferric oxides and poker-chip stacks with the ferrites. The stack of poker chips is much harder to disperse.

top view

x Rotating this ellipse around the x-axis results in a cigar-shaped object called a prolate ellipsoid. A 3-D version of this is:

Ellipsoids or Spheroids A (triaxial) ellipsoid is a generalized version of a sphere. The mathematical equation of a sphere in rectangular coordinates (x,y,z) is

x2 + y 2 + z 2 = R2 where R is the radius of the sphere. The equation of an ellipsoid is 2

2 2  x   y   z   +  = 1   +  Rx   R y   Rz 

which has three different radii. The volume of a ellipsoid is given by

If instead, we rotate around the y-axis we obtain a pill-shaped objected called an oblate ellipsoid. y

V= 43 πRx R y Rz

A formula also exists for the surface area but it involves special functions (http://mathworld.wolfram.com/Ellipsoid.html). A sphere has all three radii equal. If only two of the three radii are equal, the object is called an ellipsoid of revolution, or a spheroid. There are two different types of spheroids: prolate and oblate. Both are obtained by rotating an ellipse around one of the principal axes.

side view

x

z

top view

x

The equation of an ellipse is the 2-D analog of that for an ellipsoid: 2  x   y   +   Rx   R y

2

  = 1  

An ellsipse having Rx = 3Ry is shown below:



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measure the vertical height of each particle (four horizontal bars mark the top and bottom of each of two particles near the right edge of Figure 1). This height (the vertical distance between the two marks) is one measure of particle size called the Feret diameter. Of course, another possibility is to choose the horizontal width of each particle (or the dimension measured along some other angle). While the horizontal and vertical dimensions might not be the same for a given particle, the distribution of widths or heights should be nearly the same if 1) the sample is large enough (so that each size is well represented) and 2) the orientation of particles in the image is random (flow can orient particles so orientation is not always random). Particle Size Distribution Reading: Berg §§5.D.1-3 beginning on page 371.

A third alternative is to measure the projected area of each particle and calculate the diameter of a circle which has the same area. In what follows, we choose the vertical Feret diameter as our measure of size. Next we have to decide how to present this data. We could just list the raw diameter of each particle sorted from smallest to largest value. However, there are more than 580 particles in the image of Figure 1 (not all are visible at the resolution of this bitmap), so this is a long list and it would be hard to make sense of it (a good example of “not being able to see the forest because of the trees”). A common solution is to subdivide the total range of sizes into a number of intervals and then count the number of particles in each interval. This process is called classification of the particles: we are assigning each particle to a class and then counting the members of each class. Below is the classification of vertical Feret diameters determined from the image of Figure 1:

Figure 1: Optical micrograph of particles from a Min-USil® 15 dispersion, taken from page 372 of Berg.

Most sols are not composed of identical particles. Nearly always there will be a range of particle sizes. Today, we use image-analysis software to assess the size distribution from images like the one in Figure 1; e.g. Image J, a public domain Java program available from NIH.♦ Since each particle has a different shape and is 3D — although the image shown in Figure 1 is a 2-D projection — how do we assign a unique size to each particle? There are several possibilities. One is to ♦ http://rsb.info.nih.gov/ij/docs/



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The cumulative fraction is also useful. It is defined as ni

qi =

i

∑ fk

k =1

This gives the fraction of particles which has size d ≤ di or smaller. The cumulative histogram for Figure 1 is shown below:

median

Figure 2: Classification of particles sizes from the image in Figure 1.

Notice that the “mark” di assigned to each class is the center of the interval of diameters associated with that class. For example, the class corresponding to 0.5-1.0 µm is assigned a “mark” of di = 0.75 µm and there are ni = 91 members of this class. The number fraction (also sometimes called the number frequency) of particles in each class is calculated as

fi ≡

ni

∑ nk

(2)

Figure 4: Cumulative histogram.

Different Averages Such a distribution can be characterized by its mean d n and its standard deviation σ, which are calculated as:

∑ ni di

fi  n d = dn ≡ i = ∑ in di n ∑i i ∑i

k

A plot of the fraction fi of each class i versus the size di of that class is called a histogram (of number frequencies). The following plot is histogram of diameters for Figure 1:

i

=

fi di ∑=

(3)

i

2.12 µm

i

and

= σ

∑ fi ( di − d )

2

i

The above mean is called the number average diameter. The mean d n is one measure of the center of the distribution while the standard deviation σ is a measure of the width of the distribution.

Figure 3. Histogram if particle sizes.


In addition to the mean, two other averages are defined. The mode of the distribution is the class mark having the greatest number of members. For the distribution in Figure 3, this would be 1.25 µm. The median of the distribution is the size at which 50% of the particles are smaller and 50% are larger or qmedian = 0.5. From Figure 4, we see that the di corresponding to the median is between 1.25 µm and


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1.75 µm, let’s say 1.5 µm. Thus the mean, mode and median are three different measures of the center of the distribution.

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= F (s)

f ( x) =

 ( x − x )2   exp  − 2σ2x  2πσ2x  1

(4)

which is called a normal distribution. For a normal distribution, mean, mode and median are equal. Eq. (4) yields the familiar bell shape of a Gaussian when you plot f vs x on linear coordinates (see red curve in Figure 5).

1 + erf ( s

where

erf ( s ) ≡

1

2)

2

−∞

Shape of Distribution If the width of the histogram is narrow compared to the mean, the distribution of particle sizes is often fit reasonably well by a Gaussian shape:

s

= ∫ f ( t ) dt s

 t2 

0



∫ exp  − 2 dt 2π 

is called the error function. Notice that erf is an odd function: erf(−s) = −erf(+s). To see that the mean, mode and median are equal for a Gaussian, notice that at s = 0 (i.e. at the mean) we have the maximum value of the red curve (which represents the mode) and we also have F(0) = 0.5 (which represents the median). All three definitions coincide. When the particle size distribution is quite broad, a better fit is sometimes obtained using a log-normal distribution:

f ( x) =

 ( )2  exp  − ln x − ln x  2 2 2σln 2πσln x   x 1

(5)

Eq. (5) gives the bell shape of a Gaussian when you plot f vs x on semi-log coordinates, with the x-axis being the logarithmic one. More Averages

Figure 5. Shape of a Gaussian (normal) distribution.

In Figure 5, we have transformed the size x into a dimensionless variable: s is the deviation of x from the mean x , normalized by the standard deviation σx:

s≡

x−x σx

Then (4) becomes

= f (s)

 s2  exp  −   2  2π

1

The blue curve in Figure 5 is the cumulative distribution calculated as the partial integral of the histogram:

1) In Figure 2 and the subsequent analyses based on it, we classified size using the vertical Feret diameter. Another possibility in the analysis of the same image in Figure 1 is to use the number of pixels inside the outline of any particle as the measure of “size.” Of course the number of pixels would be proportional to the area Ai of the projection of the particle onto the image plane. Assuming we have some calibration to convert the number of pixels into an actual area Ai, then our classification would lead to a table like Figure 2 except the number ni is the number of particles having a projected area Ai in the some narrow interval of areas. Using this new table we could compute the average area of the particle distribution:

∑ ni Ai A= i

∑ ni i

If we treat each particle as a sphere, the projected area Ai is related to diameter di by Ai =


(6)

π 2 di 4

(7)


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Similarly we might interpret the mean area A in terms of a mean diameter d s using the same formula A=

2 π ds 4

( )

diameter is not the same as the number average d n computed from (3). Substituting (7) and (8) into (6) after dividing both sides by π/4 and taking the square root, yields:

∑ ni di2 i

∑ ni i

Let’s assume that the set of ni’s in this second classification by area is the same as the ni’s in the first classification by vertical Feret diameter.♥ Then we can apply (2) to the above equation to obtain 12

d s =  ∑ fi di2     i 

= 2.51 µm

3 π i = dv ∑ ni 6

( )

The analog of (9) is 13

∑ ni di3 ∑ ni Ai di = d sv

= 2.91 µm

(10)

♥ Since interval widths of diameter in Figure 2 are of equal width, this would require that the interval widths of area would not be of equal width. Instead classes corresponding to larger areas would have larger widths in area.


i = ∑ ni di2

= ∑ ni Ai

i

∑ fi s di

(11)

i

i

i

Ai = πdi2

is the surface area of one sphere having diameter di and where fi s ≡

ni Ai ∑ n j Aj j

is the fraction of the total surface area born by particles of class i. The superscript “s” is added because this fraction is not the same the number fraction defined by (2). The average defined by (11) is sometimes called the Sauter mean (after the German scientist Josef Sauter who published on the sizing of fuel droplets in internal combustion engines in the late 1920’s). Eq. (11) represents a different kind of average in

i

dv =  ∑ fi di3     i 

3) One of the experiments in the CPS Lab for assessing surface area of fine powders is the BET method.♠ The amount of nitrogen gas adsorbed onto the sample is carefully measured for a known mass of powder. To the extent that nitrogen adsorbs to form a monolayer and the adsorption area per molecule is known, this method yields the surface area of the powder per gram of sample. If the density of the solid is known, then this can be converted into surface area per unit volume of solid. The reciprocal of this can be interpreted as an average diameter of the powder particles (assuming spherical shape):

where

2) Another experiment for sizing particles is the Coulter counter (see section beginning on page 101). In this device the electric current through a tiny pore is monitored while individual particles pass through the pore. The presence of an insulating particle in the pore increases the resistance and decreases the current through the pore by an amount which is proportional to the volume of the particle. By passing a large number of particles through the pore (one at a time), we can determine the distribution of particle volumes. The average volume is given by

= V

dn < ds < dv

(9)

where the specific numerical value is obtained using the fi’s and di’s in Figure 2.

∑ niVi

These three measures of the average diameter are not equal. Generally the larger the exponent on d, the more heavily larger diameters are weighted in the average; thus

(8)

We added the subscript “s” because this mean

ds =

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that the weighting factor is the area fraction fis rather than the number fraction fi. In (3), (9) and (10), the weighting factor is the number fraction fi. These three definitions differ in what measure of particle size is being weighted: diameter, area or volume. In (11), the size is diameter and the weighting factor is area fraction.

♠ BET stands for Brunauer, Emmett and Teller, who are the authors of “Adsorption of Gases in Multimolecular Layers,” J. Amer. Chem. Soc. 60, 309-319 (1938).


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4) Another important weighting factor is volume

f iv ≡

v

fraction fi or mass fraction. As it turns out, turbidity measurements on polydisperse polymer samples yields the weight-average molecular weight of the polymer (see section beginning on page 42). The volume average diameter is = dvv

Vi =

π 3 di 6

j

Dividing both numerator and denominator by

niVi ∑ n jV j

(12)

(13)

j

ni d3 ∑ nk i = f iv

5) Static light scattering by very small particles (i.e. Rayleigh scatterers) is proportional to the 6th power of the diameter or the volume squared (see section beginning on page 38). Using volume squared as the weighting factor for diameter yields what is called the Z-average♣ diameter: ni di7 ∑ niVi2 di ∑ = ∑ ni di6 ∑ niVi2

As before, each definition yields a different estimate of the “average” and definitions having a larger exponent on d weight larger diameters more heavily and therefore yield a larger “average.” In the homework problems you will need to convert from one weighting factor to another. The following example illustrates how this can be done.

fi di3 = ∑ f j d 3j

= nj ∑ n d 3j j ∑ k k

fi di3 C

j

k

Now the value denominator of the second equation is the same for all classes i so we just define this as the constant C. Solving the result for the unknown number fraction:

is the fraction of the total volume born by particles of class i. If the density of all the particles is the same, then this is also the mass fraction or weight fraction.

dz =

∑ nk k

is the volume of one sphere having diameter di and where f iv ≡

j

and applying (2) yields a relationship between the volume fraction and the number fraction:

ni di4 ∑ niVi di ∑ = = ∑ f iv d i 3 n d ∑ i i ∑ niVi

where

niVi ni di3 = ∑ n jV j ∑ n j d 3j

f iv

fi = C

di3

(14)

The value of C can be deduced by requiring the sum of the number fractions to equal unity: f jv f C = = ∑ j ∑ d3 1 j j j

Thus

1

C=

f jv

∑ d3 j

j

Substituting this result into (14) yields the solution:

f iv fi =

∑ j

di3

f jv d 3j

Example: Suppose we are given volume fractions

fiv for each diameter mark di (whose values are also available) and are asked to find number fractions fi. Solution: substituting (12) into (13):

♣ “Z” is for zentrifuge, which is German for centrifuge. The z-average also arises in analyzing ultracentrifuge experiments.



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Lecture #2 begins here Brownian Motion Reading: Berg Sect. 5.F.1 and 2 Colloidal dispersions exhibit Brownian motion. Brownian motion was predicted by Einstein in his PhD thesis. This was also a subject of several of his The scientific first papers, starting in 1905.♥ significance of this work is that it provided a method for measuring the size of individual molecules (i.e. Avogadro’s number). Later observations of Brownian motion by Perrin (when combined with Enstein’s theory) yielded a value for Avogradro’s number which was consistent with earlier values. The success of this totally independent method laid to rest any remaining doubts about the molecular nature of matter. Brownian motion - random movement of a particle as a consequence of thermal agitation Suppose we were able to track, for example, a single gas molecule as it undergoes collisions with other gas molecules. The trajectory might look something like that shown below (except in 3-D). The trajectory of a Brownian particle in a liquid would look similarly random except that the meanfree path (length of straight paths between collisions) is much shorter.

1-D Random Walk To simplify the mathematics, let's confine our attention to 1-D motion -- for example, motion along a capillary tube. Assumptions:

The figure below summarizes the results of three Brownian dynamic simulations (the green, red and blue curves) performed according to the above rules:

x =0 x

x 2 = 2 Dt x2

N t The x-axis is the number N of coin tosses while the yaxis gives the position x (after that many coin tosses) in unit steps from the origin. The black curve is the average of 10 simulations (7 are not shown). Clearly after any number N of coin tosses, the final position x varies from one simulation to the next. What is the distribution of x values if an infinite number of simulations were performed? In the following paragraphs, we answer this question using elementary statistics. We will see that the average x for an infinite number of simulations is zero for any number of coin tosses (half the x’s are >0 and the other half are 0 we clearly need to take more heads than tails. Indeed, the final position depends only on the number of heads relative to the number of tails:

but

probability

0

m=H–T

N = 1: possible locations are m=+1 or -1, which are equally probable, so:

N=H+T

Solving simultaneously for H and T:

N +m 2   N −m  T= 2  H=

if m = 1 if m ≠ 1 (including 0)

1

N=1

(16)

Substituting (16) into (15):

( )

0.5

0 − 10

−5

Arbitrary N: suppose I toss a coin N times. The probability of a H heads and T tails can be determined from elementary statistics as:

m

 1 P ( m,1) =  2 0

0.5

1   probability of   N! 2 heads, tails H T =   H !T !  in any order   

N=0

−5

N=2

( )

if m = 0 if m ≠ 0

1

0 − 10

1

0 − 10

Now let's try to determine P(m,N) subject to the assumptions stated above.

1 P ( m, 0 ) =  0

probability

–N ≤ m ≤ +N

−5

0

5

10

N = 2: there are four possible sequences for these 2 events 1st 2nd step step m P(m,2) L L -2 1/4 L R 0 1/4 R L 0 1/4 R R +2 1/4 Each of these outcomes has equal probability, leaving

1  2 P ( m, 2 ) =  1 4 0

if m = 0 if m = 2 otherwise

♠ A vertical bar on the right of a paragraph denotes that this material was skipped during the lecture.


N  N ! 12  if N + m = even   N + m   N − m  and m ≤ N    ! ! P ( m, N ) =   2   2   if N + m = odd 0  and m ≤ N  0 if m > N 

The zero result for N+m = odd can be inferred from (16). Since the number of heads H and the number of tails T are integers, then (16) implies that 2H = N+m and 2T = N–m must both be even numbers. In particular N+m cannot be an odd integer, so we assign zero probability to this outcome. We also assign zero probability to m>N. If all the tosses result in heads, then the final location is m=N. Similarly if all of the tosses result in tails, then the final location is m=–N. There is no way for m to ever be more positive than +N or more negative than –N.


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14

N→∞: When N becomes very large we can use Sterling's approximation for calculating the factorial of some large numbers:

ln ( n !)=

Spring, 2016

Then the discontinuity [in (17)] between odd and even values of N+m will never be resolved. So let’s smooth out (17) by replacing it with

( n + 12 ) ln ( n ) − n + 12 ln ( 2π) + O ( n−1 )

and a second approximation (valid when m  N ), which is a Taylor series expansion of the logarithm function around unit value for its argument: 2  m 3  m 1 m  m ln 1 ±  = ± −   + O    N 2 N   N    N

2

 m2  exp  −   2N 

(20)

is the final position of the particle after N steps. Also let τ denote the time interval for elementary steps; then t = Nτ

1

 m2   2 exp even −   if N + m =  (17)  2N   0 if N + m = odd

(19)

where the factor of ½ in front arises from averaging odd and even values. Let l denote the length of an elementary step; then x = ml

With these two approximations, our expression for probability becomes a little easier to calculate:

  2 P ( m, N ) ≈  πN  

1

1 2  P ( m, N ) ≈  2  πN 

(21)

Probability Density

is the time required for N steps. Finally let ∆x denote the spatial resolution of position measurements. Based on the arguments above, we expect ∆x ≫ l. The probability of finding the particle between x and x+∆x is the sum of

Our goal is to derive Einstein's famous equation: in random walk the mean-square displacement grows in direct proportion to the time of the walk:

{

x

2

= 2 Dt

(18)

Now let's try to relate (17) to Brownian motion in gases and liquids. What is the length of a step? How big is N? To answer these questions, it is convenient to keep in mind the kinetic theory of gases. The distance a gas molecule travels between collisions is the mean free path: −7  10 m for gases near STP mean-free path =  −10  10 m for liquids

Thus the step sizes can be of atomic dimensions, which would be virtually impossible to follow. The rate of stepping can be equated with the collision frequency: 1010 sec−1 for gases near STP dN =collision freq ≈  13 −1 dt for liquids 10 sec

In the experiments I have in mind, we would be tracking a single colloidal particle through a light microscope using video microscopy, which might be capable of 30 frames/sec. Thus there is no hope of seeing individual steps; we can only sample at intervals of many steps.


}

m=

x +∆x l

∑

probability of finding it = between x and x + ∆x

m=

P ( m, N )

x l

(22) 1

2   ∆x 1  2  2 ≈ exp  − m    l 2  πN   2N 

where the second equation substitutes (19) and is approximate because it treats the exponential as a constant over this narrow (differential) interval of m. Dividing this sum of probabilities by the spatial interval ∆x yields the probability density p(x,t). After substituting (20) and (21), we have

∑P ≈ p ( x, t ) ≡ ∆x

1

 τx 2   2 exp  −  (23)    2πl 2 t   2l 2 t  τ

The net effect of the approximations is to take us from a discrete description of the random walk process to a continuous description (i.e. m was constrained to take on integer values whereas x is not integer). Unlike P(m,N) (which is discontinuous), p(x,t) is a smooth continuous function. In effect, we have averaged out the singularities in P by accepting a coarser description of the random walk process.♣

♣ Actually the singularities in (17) are an artifact of assuming uniform step size. In reality, each elementary step has a different length; then these discontinuities disappear.


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15

Some properties of p(x,t): it has units of m–1, it has Gaussian shape (see below) and m =+∞

∑

P (= m, N )

m =−∞

Similarly we can calculate the mean square displacement which does not vanish: ∞

∞

= ∫ p dx 1

= x2

t

l2 t τ

(24)

Notice that the mean-square displacement is proportional to t and the proportionality coefficient has units of m2 sec–1 – just as in (18).

t1

1-D Diffusion from a Point Source

2

Now we will try to show that the coefficient of t in the above expression is just the usual diffusion coefficient (times two). First let me carefully define what I mean by diffusion coefficient (D). According to Fick's 1st law of diffusion:

t2 > t1 1

t3 > t2 0 −2

−1

2

1

0

x of finding it = p ( x, t ) dx {probability between x and x + dx }

In effect, we have replaced ∆x by a differential quantity dx, which just acknowledges that ∆x is sufficiently small that the exponential function of x is essentially constant over the interval between x and x+dx. To calculate the average value of x(t) for a large number of random walks, we take the probability of finding a particular x, multiply by x, and add up the result for each x:

= x (t )

N x ( x, t ) = − D

= ∫ x p ( x, t ) dx 0

Brownian motion - random walk of a single particle diffusion – average motion of an ensemble of particles undergoing Brownian motion independently of each other The “ensemble” is a set of particles (or small molecules) which is statistically large enough so that the average behavior is reproduced if the diffusion experiment is repeated.

−∞

Substituting p(x,t) from (23) and integrating, the result is zero mean x, independent of time t. This integral must vanish because p(x,t) is an even function while xp is odd: 0

0

t

0

t1

♦ δ(x) is called the “delta function” and it’s defined such that for any continuous function F(x), the following relationship holds:

− 0.1 − 0.2 − 0.3 −2

−1

1

0

x


Substituting (25) into the above equation yields Fick's 2nd law of diffusion:

We are going to look at the particular solution of this equation which satisfies the following initial condition♦

t2 > t1

0.1

∂C ∂N x + = 0 ∂t ∂x

∂C ∂2C =D ∂t ∂x 2

t3 > t2

0.2

(25)

Conservation of particles (a mass balance) for diffusion in the x-direction only requires

∞

0.3

∂C ( x, t ) ∂x

where Nx is the flux (rate of transport per unit area) of particles in the x-direction. What is the difference between Brownian motion and diffusion?

We can use (23) to re-write (22):

x⋅ p ( x, t)

2 = ∫ x p ( x, t ) dx

−∞

−∞

3

p ( x, t)

Spring, 2016

2

+∞

∫

F ( x ) δ ( x − x0 ) dx ≡ F ( x0 )

−∞


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16

C ( x, 0= )

t = 0:

Spring, 2016

n δ ( x) A

and boundary conditions: C ( ±∞, t ) → 0

x → ±∞:

This mathematical problem describes the following experiment. n particles (the ensemble) are initially distributed uniformly over the x=0 plane inside a tube of cross sectional area A. At time zero they are released and diffuse away in the +x and –x directions.

area A

x The particular solution to this mathematical problem is

 x2  exp  −   4 Dt  4πDt

n A

= C ( x, t )

Comparing this result to (23), we see that the two results have the same dependence on x and t. Indeed the two equations are identical if we make the following substitutions:

D=

and

= p ( x, t )

l2 2τ

(26)

C ( x, t ) =[ ] m −1 n A

Moreover, substituting (26) into (24) yields (18):

x 2 = 2 Dt

(18)

for Brownian motion in 1-D. The same result applies to the other two dimensions:

= x2

= y2

= z2 2 Dt

The corollaries for diffusion in 2-D or for diffusion in 3-D are

r 2 = x 2 + y 2 = 4 Dt r2 =

x 2 + y 2 + z 2 = 6 Dt

whose proof will be left for a homework.



06-607

17

Lecture #3 begins here

v∞ =

Sedimentation We have just seen how tracking the Brownian motion of single colloidal particles with the aid of a microscope can allow us to measure the diffusion coefficient D. Einstein went on to use D to calculate Avogadro’s number NA from Rgas T Df

(27)

where Rgas = 8.31 J mol-1 ºK-1 is the universal gas constant, T is absolute temperature and f is the friction coefficient of the same particle having diffusion coefficient D. So what’s this friction coefficient? and what’s the basis for the equation above? Why does the gas constant appear when we are not dealing with a gas? Eq. (27) results from comparing two independent approaches to describing sedimentation equilibrium. In the following sections we will describe these two approaches. But first we need to characterize sedimentation rate. The Friction Coefficient Suppose we apply a force Fapp to a colloidal particle initially at rest. The force causes the particle to accelerate in the direction of the applied force. Once it begins to move through the fluid, the particle also experiences a drag force, which can be predicted from Stokes law: −Fdrag = 6 πµR v

(28)

f

f

(30)

While the nature (gravity, electrical etc.) of Fapp is not important, the most common force is gravity. For a sphere of density ρs immersed in a fluid of density ρf, the gravitation force acting on the sphere is the difference between its weight and the weight of the fluid displaced by it (Archimedes law):

Fapp = Fgrav =

4 3

πR3 ( ρs − ρ f ) g

(31)

where g is the acceleration of gravity, a vector pointing to the center of the earth and having magnitude of 9.81 m/sec2. Sedimentation Equilibrium: Analysis #1 The first approach to describing sedimentation equilibrium is via mass transfer theory. Suppose we disperse n identical colloidal particles in a test tube of cross-sectional area A and allow them to sediment for a long time. As soon as their concentration x begins to build up near the bottom of the test tube, diffusion will occur which opposes further sedimentation. Eventually an equilibrium concentration profile is established such that the downward sedimentation rate is exactly balanced by upward diffusion at every point in the test tube:

Fgrav dC flux = − C −D = 0 f dx       diffusion vx  

(32)

sedimentation

for spheres of radius R in a fluid having viscosity µ. Notice that this force acts in the direction opposing the velocity v. A similar relationship applies for nonspherical particles, where the proportionality constant multiplying v is call the friction coefficient f. The drag force offsets some of the applied force so that the sphere then accelerates more slowly until it reaches its final or terminal velocity v∞ at which the drag force is equal to the applied force but acts in the opposite direction: −Fdrag = 6 πµR v ∞ = Fapp

1 Fapp f

where the new proportionality constant (1/f ) is called the hydrodynamic mobility.

Reading: Berg Sect. 5.F.3

NA =

Spring, 2016

(29)

The minus sign appears in front of the first term because we are calculating the flux in the +x direction (upward) and the gravity force is acting downward. The shape of concentration profile can be obtained by integrating this equation

 Fgrav  = C ( x ) B exp  − x  Df 

(33)

where B is some integration constant chosen so that the integral of the particle concentration times the cross-sectional area A of the container equals total number n of particles added initially

Solving for the terminal velocity:



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18

xtop

A

∫

container, resulting in a nonuniform concentration at equilibrium.

C ( x ) dx = n

xbot

Although easily calculated, the exact value of B is not important to the current problem.

The second approach to describing sedimentation equilibrium is via thermodynamics. Recall that a given chemical component i of a mixture distributes itself between phases such that at equilibrium: (34)

where µi is called the chemical potential of component i. Within a given phase, each component will distribute itself such that its chemical potential is the same at each point, or

∇µi = 0

or

µi = const

= Rgas T d ln Ci

where Φi is the potential energy of component i arising from that force. Within a given phase, each component will now distribute itself such that the sum of its chemical potential and the potential energy is the same at each point; (35) becomes

∇ ( µi + Φi ) =0

(36)

So within a given phase, a solute will distribute itself such that its local concentration is uniform throughout the phase. Example: if we place crystals of highly soluble potassium permanganate (purple in color) in a container of water, we see intense color variations as the crystals dissolve in the water: see image at left below.

or

µi + Φi = const

For colloidal particles in our test tube, which experience a gravitational force given by (31), their potential energy can be calculated from

(35)

For an ideal solution, the chemical potential is calculated using

( d µi )T , P

For components which experience an interaction force from a source outside the system, we need to modify the criteria for phase equilibrium: (34) becomes µi + Φi in phase #1 = µi + Φi in phase #2

Sedimentation Equilibrium: Analysis #2

µi in phase #1 = µi in phase #2

Spring, 2016

Φi ( x ) = N A Fgrav x

(37)

where we have multiplied by Avogadro’s number to get the same units of energy per mole of particles as in (36) (recall that RgasT is energy per mole). At equilibrium in one dimension (x), thermodynamics requires

d µi d Φ i d + = 0 ( µi + Φ= i) dx dx dx Substituting (36) and (37):

RgasT

d ln C 0 + N A Fgrav = dx

solving for dlnC, then integrating from x=0 to x=x: ln C ( x )

∫

ln C ( 0 )

d ln C = −

N A Fgrav x Rgas T

∫ dx 0

Integrating:

 N A Fgrav  = C ( x ) C ( 0 ) exp  − x   Rgas T   But if you wait long enough (for equilibrium), the color becomes uniform (see image at right above), indicating uniform concentration of the dye throughout the water. This uniform concentration is consistent with the prediction of (35) and (36). Unlike small ions or molecules, colloidal particles can have significant mass which (in a gravitational field) causes them to settle near the bottom of the


(38)

Thus thermodynamics predicts the final concentration must decay exponentially in concentration with elevation x — just as mass transfer predicted [recall (33)]. Now (33) and (38) are describing the same final state of the system. If no mistake has been made, the exponents in (33) and (38) must be the same, so:


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19

Fgrav

=

Df

N A Fgrav Rgas T

(39)

Solving this for NA yields (27): NA =

Rgas T Df

The Brownian motion experiments suggested by Einstein were performed by Jean Baptiste Perrin in 1910, which yield a value for D. Perrin further measured the exponential decay of concentration predicted above for sedimentation equilibrium (as well as the sedimentation rate) using the same colloidal particles whose Brownian motion had been observed. The sedimentation experiments yield a value for f (see “Perrin” on page 20). These experiments lead to the first accurate value of Avogadro’s number. But more importantly, Einstein’s theory and Perrin’s experiments laid to rest any lingering doubts about the molecular theory of matter. Perrin was awarded the 1926 Nobel Prize in physics. Sedimentation Length Dividing Avogadro’s number into the gas constant yields Rgas J = 1.3805 × 10−23 NA °K

Of course, today the value of Avogadro’s number and Boltzmann’s constant are well known fundamental constants whose values can be looked up in any scientific handbook. One important application of (41) is to combine it with Stokes law f = 6πµR

(27)

In 1905 (the year of Einstein’s paper on Brownian motion), the values of the gas constant and absolute temperature were known.♦ Also the friction coefficient f of a colloidal sphere could be calculated from Stokes law (28) (published in 1851) if the size R of the particle were measured. Thus a measurement of the diffusion coefficient D of the colloidal particle is all that is needed to evaluate Avogadro’s number from (27). And Einstein suggested that D could be inferred by observing the Brownian motion of single colloidal particles.

k =

Spring, 2016

(40)

(42)

for a sphere of radius R in a fluid having viscosity µ, to obtain a formula for calculating the diffusion coefficient of spherical colloidal particles:

D=

kT 6πµR

(43)

which is called the Stokes-Einstein equation. This equation also gives a reasonable estimate of diffusion coefficients of molecular solutes, although it is based on a continuum theory (in other words, R was assumed to be large compared to molecular size in Stokes continuum theory of drag). Another use of (41) is to simplify the sedimentation equation described by (38) which can be re-written as:  Fgrav x  −x l C ( x ) =A exp  −  =Ae kT  

We define the sedimentation length l such that

Fgrav l ≡ kT Thus l represents the vertical distance over which the particle concentration changes by a factor of e = 2.71. It also gives an idea of how thick the sediment layer becomes if the particle concentration remains dilute: say n/A → 0. Because the weight of a particle scales with the cube of its radius, this sedimentation length is quite sensitive to particle size. Consider particles of a hypothetical material having a specific gravity of 2. The sedimentation length in water for particles of different size are shown in the following table. 2R

l

which is called Boltzmann’s constant. kT = 4.11×10–21 J (at 25ºC) is the characteristic thermal energy of single molecules or single colloidal particles. (27) and (39) can then be re-written as

1 mm (sand grains)

10–6 nm (sub-atomic)

0.1 µm (colloid)

1 mm

Df = kT

1 nm (gas molecule)

1 km (thickness of earth’s atmosphere)

(41)

♦ Boyle’s law (V is proportional to 1/P) was published in 1662 while Charles’ law (V is proportional to T) dates to the 1780’s. Kelvin’s paper defining absolute temperature was published in 1848.


(44)

The calculated lengths l vary from sub-atomic to a kilometer.


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20

While (38) and (44) were obtained for a gravitational potential, the more general result is  Φ ( x)  = C ( x ) C ( 0 ) exp  −  kT  

(45)

which is called Boltzmann’s equation, where Φ(x) is some arbitrary potential energy per particle or per molecule. Determining Particle Size from Sedimentation Equilibrium Today we use measures of the concentration profile C(x) in sedimentation equilibrium to infer the mass of the sedimenting particles:

Spring, 2016

Since kT is known, Fgrav can be determined which (if the particle is a sphere) can then be converted into a particle size R using (31) Fgrav =

4 πR 3 ρ − ρ ( s f )g 3  m

(31)

ρs

For nonspherical particles, the volume of the sphere can be replaced by its mass m divided by its density ρs. Then the regressed slope of the equilibrium concentration profile can be interpreted as  ρf  g slope = −m 1 −  ρs  kT 

(46)

With the two densities, g and kT all known, the slope can used to infer the particles mass, m, independent of shape. Determining Particle Size from Sedimentation Rate In addition to observing the equilibrium concen– tration profile, we might also infer particle size by measuring the rate of sedimentation. For a single particle, the sedimentation velocity is related to particle size by

x

= v

C(x)

Fgrav kT

  g 

(47)

where f is the drag coefficient, which is also a function of particle shape.

Taking the log of both sides of (44):

ln [C ( = x )] ln A −

Fgrav m  ρ f = 1 − f f  ρs

x

so plotting the concentration on semi-log coordinates should yield a straight line:

Regardless of shape, sedimentation equilibrium can be used infer the particle mass m from (46). Once the mass is known, the friction coefficient f can be inferred from sedimentation rate experiments using (47). Perrin thus used sedimentation to infer a value of f. Centrifugal Sedimentation Rate Reading: Berg Sect. 5.E.4

According to (44), the slope of this line should be negative and its magnitude equals the ratio of the gravitational force to kT: slope = −

For small particles or large macromolecules, the gravitational sedimentation length is too large to be measured; also the sedimentation speed is too small to be measured. One solution is to perform these measurements using a centrifuge, which can increase g by as much as 106-fold.

Fgrav kT



06-607

21

Spring, 2016

front becomes more diffuse with time. This is why the concentration profile initially displays a sharp rise from zero concentration to the plateau value, but images at later times show a more general rise in concentration.

A centrifuge is a device in which a sample is rotated in a plane around some axis. If the distance of the sample from the axis of rotation is r and the rotation rate is ω (radians per time), then the centrifugal acceleration is

gcentrifuge = r ω2

(48)

Substituting this quantity for g in (31) yields the centrifugal force on particles:

Fcentrifugal = mgcentrifuge 4 πR3 3

( ρs − ρ f ) rω2

(49)

Today’s ultracentrifuges (the size of a washing machine) are capable of accelerations up to 106 g which greatly extends the range of particle sizes which can be measured. As with gravitational sedimentation experiments we can either measure 1) the concentration profile C(r) at sedimentation equilibrium or 2) the terminal velocity generated by the centrifugal force. Let’s examine the determination of sedimation velocity first. Sedimentation rate of molecules in an ultracentrifuge is often analyzed by imaging the evolving concentration profile though a Schlieren lens, which reveals gradients in the refractive index of the solution. When the concentration of solute varies with position, so does the index of refraction, thus creating a gradient in refractive index. In Figure 6a below, we see the evolution of the concentration profile with time (the four curves represent the profile at different times). As sedimentation occurs (from left to right) far from equilibrium, the solution at the top of the sample (small r) becomes free of solute; the abrupt increase in concentration marks a “front” between a solutefree solution and the solution bearing solute moves downward (toward larger r) at the rate of sedimentation of the molecules.

Figure 6:

For dilute solutions of polymers or other solutes, the refractive index usually increases linearly with solute concentration. See the plot of refractive index versus molar concentration of NaCl just below. 1.39

Refractive Index of Solution

=

Refractive Index of NaCl Solutions

1.38 1.37

1.36 y = 0.00873x + 1.33399 1.35 1.34 1.33 0

1

2

3

4

5

CNaCl (mol/L)

While the dependence appears to be linear over the entire range of concentration (5 M is a saturated salt solution at room temperature), a closer look will reveal some downward curvature in the data at moderate to high concentration. The Schlieren lens is able to image the gradient in refractive index n. Indeed the intensity I of light at some point in a Schlieren image is proportional to dn/dr which in turn is proportional to dc/dr. An

Owing to the presence of molecules of different size, which then settle at slightly different speeds, the Copyright© 2016 by Dennis C. Prieve


6

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22

intensity scan I(r) of an actual Schlieren image of a sample in an ultracentrifuge is shown below.♥

Spring, 2016

Centrifugal Sedimentation Equilibrium Multiplying both sides of (47) by f and then replacing g by gcentrifuge given by (48) yields:

 ρf Fcentrifuge =m 1 −  ρs

I r Such scans can also be considered plots of dc/dr vs r. Four such plots of dc/dr vs r (taken at 4 different times) are shown in Figure 6b above. In particular, the r-coordinate of the peak of this intensity (or the peak in dc/dr vs r) gives us the instantaneous location of the front.

Lecture #4 begins here By tracking the speed of the front, we can infer the average sedimentation velocity dr/dt. Replacing v in (47) by dr/dt and replacing g by rω2, we obtain

dr Fcentrifuge m  ρ f = = 1 − dt f f  ρs

 2  r ω 

(50)

The resulting speed is usually reported as the sedimentation coefficient defined by s≡

dr dt 1 d ln r 1 ln ( r2 r1 ) = = 2 2 dt ω r ω ω2 t2 − t1

(51)

The units of s are sec. In particular 10–13 sec = 1 Svedberg = 1 S which is named after Theo Svedberg who won the 1926 Nobel in chemistry the same year as Perrin won the physics prize. While (51) shows how the experimental data can be translated into s, the interpretation of s in terms of physical properties is obtained by substituting (50) into (51)

d Φ centrifuge  2  r ω =− dr 

This force can be written as the negative derivative of a corresponding potential energy. Integrating from some reference position rref = 0 to some final r, we obtain the centrifugal potential energy of some particle at position r:

m  ρf Φ centrifuge ( r ) = − 1 − 2  ρs

 2 2  r ω 

and Boltzmann’s equation (45) concentration profile at equilibrium:

 mω2 = C ( r ) C ( 0 ) exp   2kT

 ρf 1 − ρs 

yields

 2  r   

the

(53)

Example: (Prob. 2 from H&R, Chapt. 2) A preparation of virus particles in water reach sedimentation equilibrium at 25°C after 40 hr in a centrifuge rotating at 12,590 rpm. C(r) is reported in arbitrary units (and plotted below). 2a) Evaluate the mass of the virus particles (ρs = 1.370 g/cm3) and estimate R0 and f0 of the particles. Does the sample appear to be monodisperse? 2b) The sedimentation coefficient for this preparation is 2.7 S. Evaluate f and f /f 0. Solution: Centrifugal sedimentation differs from gravitation sedimentation in that the force depends on position. According to (53), a plot of lnC vs r2 should be linear. The data given yields 2.2

or

dr dt m  ρ f = s = 1 − f  ρs r ω2

  

(52)

Notice that, even if r varies significantly across the cell, the value of s should be independent of r and can be calculated from the displacement of the Schlieren peaks using (51) and translated into m/f using the equation above.

2.0 1.8

ln(C) 1.6 1.4 1.2 1.0 0.8 43

♥ taken from H. K. Schachman, Ultracentrifugation in Biochemistry, Academic Press (1959)


44

45

46

47

r 2 (cm2)


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The line is a best-fit linear regression having the following slope:

ln C2 − ln C1 = 0.349 cm −2 r22 − r12

= slope

mω2 2kT

 ρf 1 −  ρs

 −2 =  0.349 cm 

negative potential energy there. In summary, we would expect the average m to increase with r for polydisperse samples. The first three points of the graph above appear to have a different slope from the remaining points and Hiemenz expects you to follow up on this. Below are shown separate regressions for the two sets of points:

According to (53), the slope can be interpreted as

slope=

Spring, 2016

(54)

Some of these quantities can be calculated from the info given in the problem statement:

12, 590 rev radians = ω 2π = 1.318 × 103 sec−1 min rev

2.2 2.0 1.8

ln(C) 1.6 1.4 1.2

J   −21 kT = 1.3807 × 10−23 J  ( 298° K ) = 4.11× 10 °K  

1.0 0.8

We are told the protein has a density of 1.370 gm/cm3. So

1−

ρf ρs

1− =

1 0.270 = 1.370

Solving (54) for the apparent mass m of one molecule:

= m 6.11× 10−20 gm Avogadro’s number of these molecules equals the apparent molecular weight of the protein:

M= 6.02 × 1023 m = 3.68 × 104

gm mol

Knowing the mass and density of one sphere allows us to calculate its apparent radius: 13

 3m  = R0 = 2.20 nm   4πρs  and apparent friction coefficient from (42):

f0 = 6πηR0 = 3.69 × 10−8

gm sec

where we have substituted a viscosity of η = 8.9×10-4 Pa-sec (0.89 cp) at 25ºC. Question 2a) also asks whether the sample appears to be monodisperse (i.e. one size). If multiple sizes were present, heavier particles would end up at larger r where their potential energy is more negative while smaller particles would be able to have some presence at smaller r despite the less


43

44

45

46

47

r 2 (cm2)

While you can make some case for two sizes from the graph above, that case is not compelling. The fact that the first “blue” point lies below the red line detracts significantly from the case. If the sample were polydisperse, you would expect the slope to increase monotonically with r. Instead, the slope (of a line drawn) between the 3rd and 4th point actually decreases. Moreover the single regression line on the earlier graph is a reasonable fit even for the first three points: positive and negative residuals (deviations from the line) are randomly distributed along the regression line. This suggests that the less than perfect fit is more likely due to random uncertainty in individual values rather than to two sizes. Thus I think Hiemenz is trying to read more into this data than can be justified. My experience suggests that a simpler interpretation (i.e. one size) is often more reliable and defensible than the more complicated interpretation (i.e. two sizes). Question 2b) gives some additional information: the sedimentation coefficient s of this sample is known to be 2.7 S (= 2.7×10–13 sec). We are asked for the (true) value of the friction coefficient f. Once the mass m is known from sedimentation equilibrium, the friction coefficient can be deduced by measuring the sedimentation speed. s is defined by (52); solving for f and substituting s = 2.7×10–13 sec:

m  ρf f = 1 − s  ρs

 −8 gm  =6.12 × 10 sec 


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f = 1.66 f0

This is called hydration (in the case of water) or solvation (in the more general case).

So why is f not equal to f 0? There are two common reasons: nonspherical shape and solvation. We will now discuss each.

R

and

∆R

Nonspherical Shape Reference: Berg (5.86) and (5.87) One reason for the difference is that f 0 was calculated from the mass assuming the particle is a sphere. No such assumption was made in calculating f from the sedimentation coefficient. Nonspherical shape usually increases the friction coefficient for the same volume (or mass) of particle. The geometry of ellipsoids was described in Ellipsoids or Spheroids on page 6. For nonspherical particles, the formulas for volume and friction coefficient f used above are not valid. For example, for prolate (β < 1) or oblate (β > 1) ellipsoids, the friction coefficient is calculated from  1 − β2   2  β2 3 ln 1 + 1 − β f  = β f0  β2 − 1   23 −1 β2 − 1   β tan

Adding a rigid layer of water on the outside of the sphere would also increase the friction coefficient (by increasing the hydrodynamic radius of the sphere) without changing the net weight of the sphere. The net weight is the true weight less weight of displaced water:

mnet =

Fgrav g

=

4 πR3ρ + 4 π  R + ∆R 3 − R3  ρ ) s 3 (  f 3       core

(

fluid shell

)

3 − 43 π R + ∆R ρ f    displaced fluid

for β < 1 (prolate)

=

for β > 1 (oblate)

where f0 = 6πµR0 is the friction coefficient for a sphere of the same volume as the ellipsoid and β = b/a, where a is the radius along the axis about which the ellipsoid is rotated and b is the other radius.

4 πR3 3

( ρs − ρ f )=

 ρf ms 1 − ρs 

  

Attaching a layer of water increases both the true weight and the weight of the displaced water by the same amount, leading to no change in the net weight. On the other hand, increasing the diameter of a spherical particle with a rigid solvation layer will increase the friction coefficient; for a sphere f = 6πµ ( R + ∆R )

1.6 1.5

Prolate

1.4

Oblate

f/f0 1.3 1.2 1.1 1

0.1

1

10

Axis Ratio, b/a

Solvation A second reason is that water molecules can attach themselves to particles or large molecules.



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1.2

Of course, both explanations might apply. Fig. 2.9 in H&R (above) shows different combinations of these two explanations which can together yield a given value for f /f0. For example, our ratio of 1.66 can be obtained if the particles are ellipsoids having an axis ratio of 12:1 (prolate) or 0.07:1 (oblate) and zero grams water per gram of protein; or spheres with a/b = 1 could yield this f /f0 if 2.7 grams of water is bound to every gram of anhydrous solid.



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Light Scattering In the last chapter, we saw how sedimentation experiments can be used to size colloidal particles. Another technique for determining particle size is light scattering. In this chapter, we will develop some of the principles of light scattering. Nature of Light Light consists of electric and magnetic fields which oscillate with time at a frequency on the order of 1015 sec–1; these fields also oscillate with position. Visible light has a period in the oscillation with position on the order of 0.5 µm which is in the middle of the colloidal size range. These electric and magnetic fields are governed by Maxwell’s four equations (1873):♣ Coulomb’s law: Ampere’s law:

Faraday’s law:

∇.D = ρe

∇ ×= H

(55)

∂D +j ∂t

(56)

∂B ∂t

(57)

∇×E = −

∇.B = 0

(58)

Maxwell’s equations which are to electrodynamics what the First and Second Laws are to thermodynamics, or what the Navier-Stokes equations are to incompressible, Newtonian fluid mechanics: they govern all behavior of these quantities. The symbols appearing in these equations have the following names and units: Units

Spring, 2016

distribution of charge is described by ρe which represents the local concentration of charge per unit volume. André Marie Ampère observed (1820-1825) that magnetic fields can be induced either by a steady electric current or by an time-varying electric field. (56) is the continuum manifestation of Ampere’s law. Michael Faraday observed the converse: that timevarying magnetic fields can also induce electric fields. (57) is a continuum manifestation of Faraday’s law. Finally, (58) is the analog of Coulomb’s law (55) for magnetic fields, except magnetic monopoles do not exist in nature so the analog of the monopole charge density ρe is zero. Maxwell’s equations involve five vectors (D, H, E, B and j) and a scalar (ρe), for a total of 5×3 + 1 = 16 scalar unknowns. The four vector equations constitute 4×3 = 12 scalar equations. Clearly more information or equations is needed. We need to supplement Maxwell's equations with constitutive equations: a statement of how matter responds to oscillating electric and magnetic fields. Some examples of constitutive equations in mechanics include Hooke’s law (for how an elastic solid responds to stress) or Newton’s law of viscosity (for how a viscous fluid responds to stress). The simplest electromagnetic constitutive equations for an isotropic material are: D = εE, ε(ω) = electric permittivity B = µH, µ(ω) = magnetic permeability

(59)

Just as Newton's law of viscosity is a linear relation between stress and rate of deformation, these constitutive relations also describe a linear response. ε and µ represent physical properties of the material, which happen to depend on the frequency (ω) of the oscillations in time.

Symbol ρe j

Name space charge density electric current density

coul m-3 coul s-1 m-2

E

electric field intensity

volt m-1

Special Case: No Free Charges (e.g. in a Vacuum)

H

magnetic field intensity

D

electric displacement

amp turn m-1 coul m-2

B

magnetic induction

Light propagates from the sun to the earth through a vacuum. Suppose, as a limiting case, we consider the medium to be a vacuum or else a perfect insulator (a.k.a. dielectric) with no free charges present and no current densities:

weber m-2 = volt s m-2

(55) is the manifestation of generalized Coulomb’s law for a continuum. Instead of a set of discrete point charges, charges are smeared out and the ♣ Students who have had an undergraduate transport course should already be at least a little familiar with vector operations involving the “del” operator (∇). A thorough treatment of the subject is contained in Vector Fields.pdf which is available in the same section of BlackBoard as our lecture notes.


j = 0 and ρe = 0

(60)

Substituting (60) and the constitutive equations (59), Maxwell's equations (55) through (58) simplify to

∇.E = 0

∇×H = ε

∂E ∂t

(61) (62)


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∂H ∂t ∇.H = 0

∇ × E = −µ

(63)

∂ (∇ × H ) ∂t

In place of ∇×H, we substitute the right-hand-side of (62):

∇ × ∇ × E = −µε

∂2E

1

∂2 E ∂ t 2

is the position vector and k is called the wave vector. The wave vector k points in the direction of propagation of the light and its magnitude is related to the wavelength of the light, as we will soon see. Suppose we align the x-axis of our Cartesian coordinate system with k so that k = kex

∂ 2t

but since ∇.E=0 [according to (61)], the curlsquared equals minus the laplacian:♣ ∇ 2 E = µε 

which is the kind of light produced by a laser pointer. In this equation, ω is the angular frequency (radians/sec) of oscillation,

r = xe x + ye y + ze z

Taking the curl of both sides of (63):

∇ × ∇ × E = −µ

Spring, 2016

(64)

c2

which is the familiar wave equation.

(67)

where k is the magnitude (length) of the wave vector k. Then the argument of the cosine in (66) becomes ωt − k.r =ωt − kx

To deduce the direction and speed of propagation of the wave, let’s try to track the location of particular crest of the wave. Since cosine takes on its maximum value at zero value for its argument, that must be one particular crest of the wave:

ωt − kx =0


Solving for the x-coordinate of the crest yields: A general property of wave equations is that the coefficient of the time-derivative is related to the speed of propagation of the wave, which we will denote as c. In particular, the coefficient is related to c by

µε = or

c=

1 µε

1 c2

= speed of wave propagation

(65)

Linearly Polarized Plane Wave One particular solution of the wave equation, which is relevant for light, is a linearly polarized plane wave E= ( r, t ) E0 cos ( ωt − k.r )

(66)

♣ For any vector field u, the following mathematical identity can be applied: ∇×(∇×u) = ∇(∇.u) – ∇2u When ∇.u = 0, the first term vanishes. This and many other mathematical identities can be found in Appendix I of Vector Fields.pdf available on BlackBoard.


x=

ω t k

Thus we see the crest is moving in the +x direction [which is also the direction of the wave vector k according to (67)] at speed

dx = dt

ω = c k

(68)

which relates the speed c to the wave vector’s magnitude. Now cosine has a second maximum when its argument is 2π. So two adjacent crests are given by

ωt − kx1 ( t ) =0

ωt − kx2 ( t ) = 2π The distance between these two adjacent crests is the wavelength of the light, which we denote as λ. Subtracting the two equations above results in ωt cancelling out, leaving

2π λ x1 ( t ) − x2 ( t ) == k

(69)

One final property of the wave vector k: in order for (66) to satisfy (61), k must be perpendicular to E0 or


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k.E0 = 0♠ Since k points in the x-direction [recall (67)], let’s align the y-axis with E0 because this will make the above dot-product zero. Then (66) can be simplified as

E= ( r, t ) E0 cos ( ωt − kx ) e y

(70)

In order to satisfy (63), the magnetic field must be k × E ( r, t ) kE0 = cos ( ωt − kx ) e z ωµ µω

H = ( r, t )

k=

=

ω c

Spring, 2016

at a given position r undergoes a periodic linear trajectory with time. To see this, consider the timedependence of E at a particular point in space, say at r = 0.

E = ( 0, t ) E0 cos ( ωt ) This is a vector with a fixed direction whose magnitude and sign vary with time. The head of such a vector would trace out a linear trajectory with time, which is why such a wave is said to be “linearly polarized.”

(71) E0 cos ( ωt − kx ) e z µc

where the first equation applies the 2nd and 3rd of the relationships in the footnote♠ to (63); the second equation is obtained by substituting (67) and (70); and the third equation is obtained by substituting (68). Comparing (67), (70) and (71), we see that E, Η and k are mutually orthogonal vectors. H

E

k

There are also circularly polarized and elliptically polarized light waves, although we will not encounter them in this course. (66) is called a plane wave because, at a particular t, the surfaces of constant E are planes; in particular, surfaces having E = E0 are wave fronts. For plane waves, these fronts are planar with planes orthogonal to the wave vector k and spaced one wavelength apart. To see this, note that

= k.r kr cos θ Consider a plane which is orthogonal to k.

Because the fluctuations in E and H are perpendicular to the direction of propagation, light is a transverse wave. By comparison, sound is a longitudinal wave: sound is the oscillation in position of individual molecules in a direction parallel to the direction of propagation of the sound. The electric field in (66) is said to be linearly polarized because the head of the electric field vector ♠ Using complex variables, the following three relationships can be derived: ∇.E = –k.E, ∇×E = –k×E and ∂E/∂t = iωE. To derive these, replace (66) by

{

}

= E ( r, t ) E0 Re exp i ( ωt − k.r ) 

Recall that the real part of the complex exponential is just the cosine.


All points r on this plane have the same value for rcosθ (and that value is the length of the red line in the figure above): as θ increases, r increases but cosθ decreases in such a way that their product remains constant. Since k.r = const on this plane, so is E. There are also spherical waves which are produced by a point source of light emanating equally in all directions: see Radiation from an Oscillating Point Dipole on page 34. One mathematical form for a spherical wave is Er = Eφ = 0 and Eθ ( r , θ= , φ, t )

A cos ( ωt − kr ) r


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Figure 7: Electromagnetic spectrum.

which satisfies (66) in spherical coordinates (r,θ,φ). Surfaces of constant Eθ are spheres. Energy Flux and Intensity of Light An important property of light waves is that they transport energy without any associated transport of mass. In the heat & mass transfer course usually offered to chemical engineers, this transport mechanism is called radiation. In terms of the above description, the energy flux of a linearly polarized plane wave can be calculated as

Polarization of Matter by Light Reference: Born & Wolf, Chapt. II.♥ How does matter respond to an electric field? Example #1: Hydrogen Atoms Perhaps the simplest problem to consider is a single hydrogen atom in a vacuum:

Π ( r, t ) ≡ E × H =

E02 k cos 2 ( ωt − kx ) e x µω

=

E02 cos 2 ( ωt − kx ) e x µc

(72)

where we have substituted (70), (71) and (68). Notice that the direction of propagation of energy is the direction in which the wave vector k points. Averaging over either one cycle of time or over one wavelength yields 2π

= Π

1 2π

ω

ω

∫

Π = ( t ) dt

0

E02 W = ex [ ] 2µc m2

We can think of a hydrogen atom as a proton (the black dot) surrounded by a spherical electron cloud (the circle) of negative charge have a magnitude to that of the proton. In the absence of any applied electric field, the center of negative charge coincides with the center of positive charge. When a static electric field is applied, the positive nucleus is pulled in direction of the electric field, while the oppositely charged cloud is pulled in the opposite direction. A displacement between the centers of positive and negative charge occurs:

(73)

Electromagnetic Spectrum Figure 7 lists the variety of electromagnetic radiation from gamma rays to radio waves. Visible light occupies a narrow portion of this broad spectrum.


♥ M. Born and E. Wolf, Principles of Optics, Cambridge Univ. Press, 7th ed, 1999.


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The result is that a dipole moment has been induced in the atom. If x is the vectorial displacement between the centers of positive and negative charge, and e is the quantity of charge (in this case, it’s the elemental charge e on one proton), then the induced dipole is a vector given by p ≡ ex The magnitude of this dipole is zero when E is zero and increases linearly as E increases (at least for small E). The direction of p is parallel to the direction of E: p = αE

(74)

where α is a material property called the atomic or molecular polarizability. Thus, even molecules which do not possess a permanent dipole moment acquire a dipole moment through the shift in their electron cloud by the applied electric field. The oscillation of this dipole with time (since E oscillates with time) is what leads to “scattering” much like the signal from a radio transmitter which uses a dipole antenna. In the case of a single spherical electron cloud, the polarizability is given by (see Hwk3)

α= 4πε0 R3 = 3V ε0

(75)

where R is the radius of the cloud and 8.854 × 10−12 ε= 0

Coul2

is the electric permittivity of vacuum, which is the proportionality constant in Coulomb’s law:

1 q1 q2 4πε0 R 2

For charges embedded in media other than vacuum, for that medium. ε is also related to the speed of light: see (65). The ratio of the speed of light in vacuum c0 to the speed of light in some other medium c is defined as the refractive index of the material

µε c0 == c µ0 ε0

Example #2: a Dielectric Sphere Suppose we have a dielectric particle of one phase dispersed in a second phase. The analog of (75) can be deduced by calculating the apparent dipole moment induced in a spherical particle of radius R and permittivity εs by a uniform electric field externally applied in the surrounding phase (e.g. water) having permittivity εf. Assuming both phases are dielectrics the result turns out to be (see Hwk3)

εs − ε f α particle = 4πε f R3    ε s + 2ε f

ε ≥1 ε0

(76)

Most nonmagnetic media have the same value of magnetic permeability µ so this property cancels out. Eq. (76) applies at frequencies for which absorption


(77)

3V ε f

The important result is that the polarizability of small particles is proportional to their volume. While this was also true of (75), the proportionality constant is different. This difference arises because the electron cloud of an H atom was treated as a good conductor whereas the particles (and the surrounding fluid) are treated as insulators. In particular, different electrical properties show up in the derivations (see Hwk3). Now ε is a property of a continuum whereas α in (75) is a property of single molecules or atoms. The two properties are related by the Clausius-Mossotti equation (also called the Lorentz-Lorenz formula): = α

ε0 in Coulomb’s law is replaced by the permittivity ε

n≡

of light is negligible. When absorption occurs, ε and n are complex numbers and their relationship also involves the absorption coefficient.

Clausius-Mossotti Equation

2 N-m   farad m

F=

Spring, 2016

3ε0 ε − ε0 3ε0 n 2 − 1 = N ε + 2ε 0 N n2 + 2

(78)

where N is the number of molecules per unit volume, each of which has polarizability α; the second relation results from substituting (76). Comparing (78) with (77), we find the expressions are very similar: εf has been replaced by ε0 and the particle volume V has been replaced by 1/N. While the two substitutions suggested in the previous paragraph help to rationalize the mathematical form of (78), these substitutions are not selfevident. Indeed this is not how (78) is derived. For example, V in (77) is the volume of the particle being polarized (a condensed phase consisting of many molecules); whereas 1/N in (78) is the total volume (interior plus exterior) occupied by an ensemble of gas molecules (not just the volume of their interiors).


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For a more rigorous derivation, see §4.5 on p159 of Jackson♦ or §2.3 on p89 of Born & Wolf.♣ Eq. (78) works well not only for gases but also condensed phases. To see this, let’s look at some experimental measurements taken with liquids and gases. These experimental measurements are tabulated in terms of the molar refractivity A, defined as

N N A n2 − 1 cm3 = =[ ] A≡ A α N n2 + 2 3ε0 mole

(79)

where NA is Avogadro’s number. All the tables below taken are from Born & Wolf.♣

♦ J.D. Jackson, Classical Elecctrodynamics, 3rd ed., Wiley, (1999) ♣ Born & Wolf, Principles of Optics, 7th ed., Cambridge U Press (1999).



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In Table 2.1 (just below) shows the molar refractivity of air at various pressures. Recalling the ideal gas law — ˆ ♥ — the number of gas molecules nˆ per unit volume V is given by the equation next to the table below. pV = nkT Note that N is proportional to p. Increasing the pressure increases the number concentration N of gas molecules as well as the refractive index n.

N=

nˆ p = ∝p V kT

Although the concentration N of gas molecules varies by two orders of magnitude — which causes the refractive index to rise — the molar refractivity is virtually constant. This suggests that the refractive index is indeed changing with N in the manner given by (79). The second table compares A for various liquids with A for the vapor of the liquid:

Notice that A measured for liquid or vapor of the same compound are virtually equal. This means that the polarizability of molecules does not depend upon the state (vapor or liquid).

♥ Since we are using the number of molecules instead of the number of moles, we replace the universal gas constant R by Boltzmann’s constant k.



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The third table looks at mixtures of sulfuric acid and water in various proportions:

The right-most column contains the A for the mixture inferred from measurements of refractive index. The column just to the left represents A calculated as the weighted average of the A’s for the pure components; the weighting factor is the molecular number fraction. The agreement between these two columns is remarkable in all proportions. This suggests that each molecule contributes independently of the others to the refractive index of the solution. Finally we compare atomic refractivity with molecular refractivity:

The right-most column in Table 2.5 is obtained by adding the reflectivity for each atom (from Table 2.4) in the molecule. For example, the value of 6.74 for HCl is obtained by adding 1.02 for H to 5.72 for Cl. These calculated values are compared to the A’s


inferred from measurements of refractive index (next to last column). Once again, the agreement is remarkable (the largest error is 11% for CS2). This suggests that each atom of a molecule contributes independently of the others.


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Comment: Values of the molar refractivity will depend on wavelength (color) of the light used for refractive index measurements. By using visible light (light source = sodium D lamp), the refracttivities reported in the tables above correspond to electronic polarization of the atoms and molecules as described in the section on page 29 above. At lower (sub-microwave) frequencies, polar molecules like H2O or CS2 exhibit much higher α’s owing to orientational polarization arising from orientation of permanent dipole with any applied electric field. Many of the comparisons made above would fail for sub-microwave frequencies.

Lecture #6 begins here Radiation from an Oscillating Point Dipole Reference: Lipson, Lipson & Lipson, §5.3 and §13.2♠

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mutually orthogonal unit vectors in spherical coordinates.

(80) is called a spherical wave. The wave vector k for this scattered radiation is not unique, but depends on the position of the observer and is given by = k ker ( θ, φ )

where the scalar magnitude k still satisfies (68) and (69): = k

Reading: Berg Sect. 5.H We have just seen that static electric fields can induce a dipole moment in all material — even if the atoms or molecules do not possess a permanent dipole of their own. Next, we will see that oscillation in this dipole generates electromagnetic radiation in all directions. Let the induced point dipole oscillate with a frequency ω at the origin of spherical coordinates at r = 0:

= p p0 cos ( ωt )

This oscillating dipole itself radiates energy in all directions, much like a radio or TV transmitter or even your cellphone. All of these devices generate radiation at a particular frequency by sending an oscillating current through an antenna, sometimes called a dipole antenna. Instead of a macroscopic dipole antenna, we are concerned with an atomic size oscillating dipole.

ω 2π = λ c

(81)

In other words, a spherical wave is radiating outward (from the oscillating dipole) in all directions simultaneously. The magnetic field associated with (80) is again calculated from (63) as 3 k × E k p0 cos ( ωt − kr ) sin θ eφ = r µω 4πµε0 ω

H rad = ( r, t ) =

k 2 cp0 cos ( ωt − kr ) sin θ 4π

r

eφ

where in the last step we have substituted (65) µε0 = c–2 and (68) ω = kc. Notice that Erad (acts in θdirection), Hrad (acts in the φ-direction) and k (acts in the r-direction) are still mutually orthogonal vectors. The local energy flux associated with this spherical wave is given by [compare with (72)]:

The particular solution to the wave equation (64) Π= E × H which describes the far-field (kr ≫ 1) solution to the k 4 p02 c cos 2 ( ωt − kr ) sin 2 θ (82) watt radiation from an oscillating “point”= dipole is = er [ ] 2 2 2 16π ε0 m r k 2 p0 cos ( ωt − kr ) sin θ volt rad [ ] (80) = E ( r, t ) = eθ The total power emanating from this point can be r 4πε0 m calculated by integrating the flux over any closed where r is the magnitude of the position vector r surface A containing the oscillating dipole (whose origin coincides with the location of the = P ∫= n.Π da [ ] watt oscillating dipole) and where er, eθ and eφ are the A

♠ A. Lipson, S.G. Lipson and H. Lipson, Optical Physics, Cambridge Univ. Press, 4th ed., 2011.


where da is a differential element of area on the closed surface A and n is a unit vector normal to da


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which points out of the region enclosed by A. Averaging the result over one period of the oscillations and substituting (82): 2π

P =

1 2π

ω

k=

k 4 p02 c = P yields= 12πε0

2π λ

ω

∫ P (t ) dt 0

4π3 p02 c =[ ] watt 3ε0 λ 4

Spring, 2016

The amplitude of the oscillating dipole is proportional to the strength of the electric field of the incident light according to (74) and the proportionality constant is given by (75). This links the scattered intensity given by (82) to the incident intensity given by (73). Time averaging Πr given by (82) over one cycle (cos2 is replaced by 1/2) and replacing p0 by αE0 yields k 4 α 2 E 2 c sin 2 θ Π rsca = 0 r2 32π2 ε0

(83)

In the last step, we eliminated k = 2π/λ by substituting (81). Note that the result is inversely proportional to the 4th power of the wavelength λ of the scattered light, which is the same as the wavelength of the incident light.♠ Visible light varies in wavelength from about (see Figure 7 on page 29).

Dividing the result by Πx given by (73) and substituting k = 2π/λ from (81) and c2 = (µε0)–1 from (65): Π rsca π2 α 2 i ≡ = sin 2 θ 2 4 2 inc I0 ε0 λ r Πx

violet or blue:

λ = 400 nm

which is dimensionless.

to red:

λ = 700 nm

Scattering Cross-Section

P red  700 −4 0.11 = =  P blue  400  Clearly red light has a longer wavelength and scatters much less than blue light: noon sun

blue blue red

To get a better idea of the how much energy is being scattered, let’s replace p0 in (83) by αE0 and divide the result by the time-average energy flux in the incident light given by (73). This results in cancellation of E0:

P Π inc x

white white

setting sun

earth This is why the daytime sky appears blue and sunsets (and sunrises) appear red. During daytime, we don’t look directly at the sun, but at white light which is scattered off of molecules of air and water in the atmosphere. Since the blue portion of the spectrum is scattered more intensely, the sky appears blue. On the other hand, at sunset we are looking almost directly at the sun. So we see the light which has not been scattered away. Thus the setting sun appears red. ♠ This equality of the wavelength of the incident and scattered light is general for Rayleigh scattering, which is also called elastic scattering. An example of inelastic scattering is Raman scattering in which the wavelength of the scattered light differs from that of the incident light (the scattered wavelength is usually longer).


(84)

=

(

4π3 α E0

)

2

c 2µc

3ε0 λ 4 α= 4 πε0 R3

=

E02

64π5 R 6 λ4

c2 =

1 µε0

=

4π3α 2 3ε02 λ 4

≡ Acs

In the third equation above, we have substituted (75) for the atomic polarizability. The resulting ratio has units of area and can be thought of as the crosssectional area Asc of the incident light whose energy is being scattered by the atom. Assuming this area is a circular, how does its radius compare to the radius R of the scattering atom? 4

R = 64π4   ≈ 10−12 2 λ πR Acs

(85)

The Bohr radius of the H atom is 5.3×10–11 m and a typical wavelength for visible light is 0.5×10-6 m. Substituting these quantities into the above equation yields 10–12 — an extremely small fraction. By contrast, if we considered the incident light to be a stream of photons whose radius is negligible


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compared to the H atom, and their scattering is what would be expected for the ballistic motion of hard spheres, then the above area ratio is unity. In other words, any photon inside of a circular tube of radius R whose axis passes through the H atom would be expected to collide with the H atom and be scattered. Allowing photons to be larger in size would only increase that area ratio. That the area ratio in (85) is much smaller than unity for real scattering suggests that real scattering is very different from ballistic scattering of hard spheres. Scattering from Two Atoms: Interference So far we have only considered scattering from a single atom. Suppose now we have two atoms exposed to the same incident light wave. Is the energy flux of scattering twice as large as for a single atom? The answer turns out to be between 0 and 4 times larger, depending on the relative location of the two atoms and the location of the observer (because of constructive or destructive interference). To calculate the total scattering flux from two atoms, we must add the electric field E (or magnetic field H) emanating from each of the two oscillating dipoles at the location of the observer — the energy fluxes Π themselves are not generally additive. In order to calculate this sum, we need to consider the geometry of a typical light scattering experiment shown below.

θsca

Spring, 2016

The angle between these two vectors is the scattering angle θsca. The scattering angle θsca is not the same as the polar angle θ in spherical coordinates which is the angle measured from the axis aligned with the electric field for the incident light (see sketch on page 34). The figure below shows the two atoms as the two red dots with the vector r providing their relative position. ninc and nsca are unit vectors which are aligned with kinc and ksca respectively.

Figure 8: Geometry needed to calculate phase angle δ.

The dashed line labelled “reference plane (incident wave)” is the x=0 plane for the incident light [Einc(x,t) given by (70)]. The dashed line labelled “detector plane” is the face of the detector (usually a PMT) and it is tilted an angle θsca relative to reference plane. Atom #2 is further from the incident light source (the reference plane for incident wave) than atom #1 by a distance ninc.r. This causes a phase lag (in radians) of δ1 = kninc.r

The light source is typically a laser. The sample is held in a 1 cm diameter glass cylinder a few cm long. The detector is usually a photomultiplier tube (PMT) which is held by a goniometer, a device which allows the detector to be rotated in a horizontal plane around a vertical axis through the center of the sample to detect light scattered at various angles θsca relative to the direction of the incident light, which is stopped by a light trap. Usually the goniometer rotates in a horizontal plane. This plane contains both the wave vector kinc for the incident beam as well as the wave vector ksca for the scattered rays. The plane containing both of these wave vectors is called the scattering plane.


where k = 2π/λ for the incident light. But the same atom is actually closer to the detector plane by a distance nsca.r. This causes a negative phase lag of δ2 = –knsca.r

The total phase difference is obtained by adding δ(r) ≡ δ1 + δ2 = k(ninc–nsca).r

(86)

We will assume that the detector responds to the instantaneous sum of the two oscillating electric fields evaluated at on the detector plane:

( t ) E1 ( t ) + E2 ( t ) Edet = where E1 and E2 are two scattered electric fields like (80) emanating from the two atoms and evaluated at


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Spring, 2016

the detector plane at the right end of the red lines. Let’s assume the two atoms have the same polarizability. While both electric fields have the general form of (80) the main difference is a phase angle δ:♥

= Edet ( t ) Esca cos ( ωt ) eθ + Esca cos ( ωt + δ ) eθ

= Esca [ cos ( ωt ) + cos ( ωt + δ )] eθ

(87)

Now let’s use this electric field to calculate the associated magnetic field: Figure 9: Total scattering by two atoms normalized by the scattering of one, given by (90). δ is the phase angle in (86).

k × Edet H det ( t ) = sca µω = H sca [ cos ( ωt ) + cos ( ωt + δ )] eφ

Notice that the flux ratio varies between zero at δ=π (destructive interference) to four at δ=0 or δ=2π (constructive interference). If we averaged the ratio over all values of δ, the average would be two.

and energy flux as in the analysis leading to (82)

Π= det Edet × H det

Scattering from Small Molecules: Atomic Polarizabilities are Additive

= Esca H sca [ cos ( ωt ) + cos ( ωt + δ )] er 2

In the analysis of the previous section, we assumed the α’s for the two atoms are equal. Suppose the polarizabilities of the two atoms differ; let them be α1 and α2. The analog of (90) is

Integrating this over one period of oscillation: 2π

Π det = r 2

1 2π

ω

ω

∫Π

det r

( t ) dt

(88)

0

= Esca H sca (1 + cos δ ) The comparable result for scattering from only one atom can be obtained by dropping the second cosine from (87) (the cosine containing δ) and repeating the above two calculations, which leads to Π det r

1 = Esca H sca 1 2

(89)

The factor of ½ which appeared above is just the average of one cosine squared. The ratio of energy flux from two atoms compared to the flux from one is given by (88) divided by (89):

Π det r Π det r

2

= 2 (1 + cos δ )

Π det r

2 1

α 2 + 2α1α 2 cos δ + α 22 = 1 α12

(90)

which is plotted below:

♥ As drawn in the figure, the two red lines denoting the path from the two atoms to the detector have quite different lengths yet we have taken r to be the same in (80) for both electric fields. In actual experiments, the detector is much further from the atoms than shown in the figure; then the percent difference between the two r’s is negligible.

(91)

which is the ratio of energy flux from two atoms having α1 and α2 relative to the energy flux from a single atom having α1. Notice that if α1 = α2, then (91) reduces to (90). Inside a small molecule (e.g. HCl ), the two atoms are very close to one another (perhaps 10–10 m) compared to the wavelength of visible light (perhaps 10–6 m). Then δ ≈ 10–4 and (91) can be approximated by

Π det r

1


Π det r

Π det r

2 1

δ=0

≈

( α1 + α 2 )2 α12

Which suggests the scattering by one HCl molecule is the same as one H atom (atom #1) plus one Cl atom (atom #2); in other words, molecular polarizability is just the sum of the atomic polarizability. Generalizing this to molecules having more than two atoms: α molecule =∑ αi atom i


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This simple rule is obeyed by the molecules tested in Table 2.5 on page 33. As noted in the comment following Table 2.5, this result applies only for electronic polarizability arising from the distortion of electron shells. It does not apply to orientational polarizability arising from any permanent dipole possessed by the molecule. Scattering from Small Particles: Rayleigh Scattering Reading: Berg §5.H.3 Instead of a small molecule consisting of n nearby atoms, we now consider a nanoparticle consisting of n nearby molecules. If all the molecules are close enough together to see the same electric field, their induced dipoles oscillate in unison. Then the particle polarizability should be n times the molecular polarizability; this result is proportional to the particle volume:

α particle = nα ∝ V This conclusion is consistent with (77) in which αparticle is proportional to the volume of the scatterer.

Such particles are called Rayleigh scatterers. So how small does the particle have to be so that all the oscillating dipoles are in unison? Essentially the largest phase angle δmax (for any pair of molecules in the particle) must be small enough to that cosδmax ≥ 0.9

or

δmax≤ 0.45

(92)

This value for the cosine is somewhat arbitrary. Recall from (86) that the phase angle is given by δ(r) ≡ δ1 + δ2 = k(ninc–nsca).r

(86)

where the quantities are defined in the figure on page 36 and k = 2π/λ. We are interested in the largest value of δ, which would arise when –nsca = ninc and both unit vectors are parallel to r. Then the dot product becomes 2r where r is the magnitude of r. The largest distance between two molecules in the particle (assumed spherical) would be equal to the diameter of the sphere (say 2R). Thus the largest value of δ is

2π 2R δmax = 4 R = 4π λ λ To satisfy (92), we must require

4π

2R ≤ 0.45 λ

or

2R ≤ 0.036 λ


(93)

Spring, 2016

the particle diameter must be very small compared to λ. Using λ = 0.5 µm, the particle diameter 2R must be less than 18 nm to be a Rayleigh scatterer. Scattering from Multiple Brownian Particles: Independent Scatterers Up to now we have considered scattering by multiple atoms or small molecules which are close enough together to have their induced dipoles oscillate in unison. Suppose we have a large ensemble of these Rayleigh scatterers spread uniformly at a concentration of N particles per unit volume throughout some region which is large compared to the wavelength. First consider the total scattering from two arbitrary particles of this ensemble. Although these two particles might be very far apart (i.e. r ≫ λ), their scattering at any instant of time might interfere either constructively (i.e. δ = 0 or 2π) or destructively (i.e. δ = π) or anywhere in between, where δ is still calculated from (86). Over time, Brownian motion will cause r and δ to change and the total scattering from the two particles will fluctuate with time. This is the basis of dynamic light scattering which will discussed later (see page 49). Averaged over time or δ, the scattering of two Brownian Rayleigh particles is given by (90) as

Π det r Π det r

2

=

2 (1 + cos δ )

cos δ =0

=

2

(94)

1

The scattering from two particles is just twice that from one. In other words, they are acting independent scatterers — displaying neither constructive nor destructive interference. If the ensemble is large and uniformly distributed, then the probability of any two particles having a particular δ at a particular time is independent of δ. Averaging over δ again results in (94) for the average scattering by two particles. Thus on average Brownian particles behave like independent scatterers. If there is a total of n particles in the ensemble, the average scattering is simply n times that of one particle. Substituting the particle polarizability of (77) into the energy flux ratio given by (84) (replacing ε0 by εf) and multiplying the result by the number of independent scatterers per unit volume N:


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is i = N I0 I0 2

9π2 v 2p  ε s − ε f  sin 2 θ = N [ ] m −3   λ 4  ε s + 2ε f  r 2 where vp is the volume of one particle.

(95)

2

As in the previous section, each gas molecule acts as an independent scatterer, so the scattering by an ensemble is the scattering from one molecule times the number of molecules in the sample. For dilute gases, the refractive index n is only slightly larger than unity (e.g. air at STP has n = 1.0003). Then we can approximate (78) by 3ε0 n 2 − 1 3 ε0 n 2 − 1 ≈ N n2 + 2 N 3 ε0 2ε 0 ( n + 1) ( n − 1) ≈ ( n − 1) = N  N

= α

(96)

2

Since n−1 is proportional to N, the concentration of gas N will cancel out as it should if α is the property of a single atom or molecule. This is why the molar refractivity A in Table 2.1 (on page 32) is independent of pressure. Now (84) gives the scattering from a single atom or molecule. Multiplying (84) by N molecules per unit volume and substituting α from (96) we obtain

is I0

gases: =

i 4π2 ( n − 1) = N θ [ ] m −3 sin 2= 4 2 I0 λ r N 2

(97)

gives the dimensionless scattering intensity (i.e. the ratio of the scattered flux to the incidence flux) per unit sample volume. Scattering from Liquid Solutions For condensed phases like liquids, the molecules are too close together to behave like independent scatters. Instead, for every molecule whose dipole oscillates with a particular phase there is another molecule which oscillates 180º out of phase causing destructive interference in the total scattering. A perfectly uniform pure liquid then scatters no light. Nonetheless experiments do reveal weak scattering from pure liquids. Scattering in pure liquids is caused by inhomogeneities in the local density of the liquid caused by random thermal agitation. Solutions having a solute dissolved in the liquid also scatter some light. Some of this scattering comes from density fluctuations in the pure liquid Copyright© 2016 by Dennis C. Prieve

and some comes from fluctuations in the concentration of solute. The theory is quite involved♣ and goes beyond the scope of this course. We will just give the final result:

2 is 4π  n ( dn dC )  kTC 2 = = sin θ [ ] m −3 4 2 I0 λ r ( ∂πosm ∂C )

Scattering from Dilute Gases

gases:

Spring, 2016

where πosm and n is the osmotic pressure and refractive index of the solution having mass concentration C of solute. Once again the above ratio represents the dimensionless scattering intensity (i.e. the ratio of the scattered flux to the incidence flux) per unit sample volume. Comment: θ in these notes is measured relative to the direction of the electric field amplitude E0 [see figure on page 34 right below (80)]. By contrast θ in (5.109) or (5.123) of Berg is the scattering angle (assuming unpolarized incident light). The two angles are related below on page 43 when angle dependence is discussed. Refractive index of solutions are often linear functions of the mass concentration of solute. The following figure is for sucrose solutions.♠ 1.5 1.48 1.46 1.44

n 1.42 (Refractive 1.4 Index) y = 0.1440x + 1.3330

1.38 1.36 1.34 1.32 0

0.2

0.4

0.6

0.8

1

1.2

C (g of sugar per cm^3 of solution)

The largest concentration in the above graph corresponds to a very concentrated solution (80% sugar by weight, 20% water). Yet dn/dC is nearly constant over the entire range. A similar linear variation of refractive index with concentration was shown for NaCl solutions on page 21.

♣ See §5.3b on p204 of H&R. ♠ data from CRC Handbook of Chemistry & Physics, 94th ed., Table 5-29-92: “Concentrative Properties of Aqueous Solutions.”


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At infinite dilution, all solutions behave ideally and the osmotic pressure is given by van’t Hoff’s law: RT πosm = C M

where M is the molecular weight of the solute. At higher concentrations, the osmotic pressure becomes a Taylor series expansion in the concentration, which is called a virial expansion. At low concentrations, we can truncate after the second term, leaving

Spring, 2016

(98) represents the scattering in a particular direction. In the turbidity experiments below, we are interested in the total scattering, regardless of direction. Integrating (98) over any closed surface A (completely enclosing the sample volume) gives: is =  ∫ I0 2da A

r dΩ

is 2

∫ I0 r 2π

=

d Ω

sin θ d θ d φ π

is 2 ∫ d φ ∫ I0 r sin θ d θ 0 0  2π

C  πosm = RT  + BC 2  M 

where θ and φ are the two angles in spherical coordinates.♠ Substituting (98):

where B is called the 2nd virial coefficient which contains information of the intermolecular interaction between solute molecules. Using this second expression for the osmotic pressure, the scattering becomes

is KC 3 ∫ I0 da = 2π 1 M + 2 BC ∫ sin θ d θ A 0   

is KC sin 2 θ = I 0 1 M + 2 BC r 2

where♥

4π2  n ( dn dC )  K≡ N Aλ 4

4

= (98)

2

In the limit of zero concentration of solute C, the scattering goes to zero as expected. Comment: For an actual scattering experiment in the limit C→0, you would measure some residual scattering arising from density fluctuations in the solvent. Thus the turbidity τ defined below should be interpreted as the increase in the turbidity (over that of the solvent) arising from the solute. While we have introduced the equations in this section as applying to soluble species (or lyophilic colloids), the same equations also apply to small particles, droplets or bubbles of an immiscible phase (lyophobic colloids), provided the particle size is small enough that Rayleigh theory is applicable. Such mixtures have an apparent refractive index which is generally higher than that of the pure solvent and its value n is proportional to the mass concentration of the immiscible phase C.

♥ This definition for K differs slightly from (5.123) in Berg in that the angular dependence is not included in K. Also note that θ in these Notes is defined differently than θ in Berg: the two angles are reconciled in Static Light Scattering Experiments: Angle Dependence on page 43.


π

3

(99)

8π KC = ≡ τ [ ] m −1 3 1 M + 2 BC

This represents the total scattered power (per unit sample volume) divided by I0 (the incident power per unit area). So the net units are m-1. We can think if this as the fraction of the incident power which is scattered per unit path length of the incident light passing through the sample (assuming the path length is very short so that the drop in incident power is negligible). This result turns out to be the turbidity τ defined experimentally below. Static Light Scattering Experiments: Turbidity Reading: Berg §5.H.3 One type of experiment is to measure the intensity of light transmitted through the sample Iout relative to the intensity of the entering light Iin.

♠ dΩ in (99) is a differential element of solid angle. Solid angles are expressed in steradians. Note that, if we integrate over a sphere of radius r :

∫ da = ∫ r

2

d Ω = 4πr 2 so

we say there are 4π steradians in a sphere which is the 3-D analog of having 2π radians in a circle.


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Spring, 2016 Solving for τ:



= τ

Iin

Iout

Source

Sample Generally, the amount of light transmitted will be less than that incident for two reasons: 1) some of the light might be absorbed by molecules in the sample or 2) light might be scattered by the sample. We will assume that no light is absorbed. Then Iout = Iin – Is where Is is the total scattering intensity from the sample volume. We expect that the light intensity will decay monotonically from Iin to Iout as light propagates along the x axis through the sample. Denote the local intensity of the unscattered light as I(x). Consider the change dI in intensity across a slice of sample dx which is thin enough so that the incident intensity changes by a negligibly small amount dx

I(x)

I(x) + dI

The fraction of the incident light intensity I(x) which is scattered can be obtained by multiplying (99) by the path length dx. Keep in mind that scattering will decrease the remaining intensity so dI is negative; thus dI = − I ( x ) τ dx

or

dI = −τI ( x ) dx

The particular solution after imposing the boundary condition I(0) = Iin is I ( x ) = I in e−τx

In particular

I out ≡ I (  ) = I in e−τ


I Iin 1 1 [ ] m −1 = ln in ln =  I out  I in − I s

where Is = Iin – Iout is the total intensity of scattering. The equation above for τ is the analog of Beer’s law for the specific absorption coefficient (except that the decrease in intensity here is due to turbidity – not absorption). In most scattering experiments, the fractional change in intensity is very small, so the logarithm function can be replaced by a one-term Taylor series approximation:

I  I 1  τ = − ln 1 − s  ≈ s   Iin  Iin

for

Is 1 Iin

where τ is given by (99), the turbidity becomes

= τ

8πKC HC = 1  1  + 2 BC 3  + 2 BC  M M  

(100)

3 8πK 32π  n ( dn dC )  ♣ H≡ = 3 3N Aλ 4 2

where

Two types of experiments can be performed. Expt. #1) Measure τ vs. λ.♦ If a spectrophotometer is used, we can vary the wavelength of the incident light. Our theory is based on turbidity being caused by scattering. Then τ ∝ λ–4

A plot of log τ versus log λ should be linear with a slope of –4. Obtaining this result is strong confirmation that turbidity is due to scattering. The other major source of turbidity is absorbance of light by the sample. But absorbance (the analog of turbidity when light is being absorbed rather than scattered) typically displays non-monotonic behavior with wavelength: peaks of absorbance occur at specific wavelengths at which absorbance takes place.

♣ This is the same H defined by (5.121), although its relationship to K is not the same: see footnote next to definition of K on page 40. ♦ τ in these notes is actually the contribution from the solute. In other words, τ = τsolution – τsolvent. This subtraction usually is performed by the spectrophotometer which measures both the turbidity of the solution and of pure solvent.


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Spring, 2016

∑ τi

Expt. #2) Measure τ vs. C. If our solute is a polymer of unknown molecule weight M, this data will allow us to determine M. The equations above express concentration C in mass per volume, which can be determined without knowing the molecular weight. (100) can be rearranged to yield a straight line in its dependence on C:

The average molecular weight is defined by the intercept of (101):

HC 1 = + 2 BC τ M

The contribution to turbidity of each fraction can also be calculated from (101):

(101)

Below is plotted turbidity data for polystyrene solutions in methylethyl ketone obtained at 25ºC using light having λ = 436 nm.♦ Results for three different fractions (having different molecular weights) are shown:

τtot =

(103)

i

HCtot 1 = τtot M

M =

and

HCi 1 = τi Mi

τtot HCtot

τi =HCi M i

or

(104)

(105)

Substituting (102), (103) and (105) into (104):

∑ τi τtot i = M = = HCtot H ∑ Ci

∑ H Ci M i i

H ∑ Ci

i

≡ M w (106)

i

Thus the appropriate average is the weight-average molecular weight of the mixture (since the contribution from each fraction is weighted by the mass concentration of that fraction). For contrast, omotic pressure meaurements yield the numberaverage molecular weight of the mixture, which is defined as

∑ ni M i Mn ≡ i The linearity of the points is consistent with expectations based on (101). This equation suggests that the intercept of the straight lines is the molecular weight of each fraction (116, 180 and 270 kDa) and the slope could also be used to infer B.

i

To see that this is different from (106), we substitute Ci = niMi/V in (106):

∑ ni M i2 Mw ≡ i

Polydisperse Samples Suppose that each fraction is polydisperse. In other words, suppose that in volume V, we have ni moles of polystyrene having molecular weight Mi where there are many classes i. What type of average molecular weight is obtained with this turbidity experiment? To answer this question, we assume that each fraction contributes independently of other fractions to the total turbidity and it is only the total concentration that is measured: = Ctot

∑ ni

= Ci ∑ ∑ i

i

ni M i V

♦ from H&R, Fig. 5.7, p211.


(102)

∑ ni M i

(107)

i

For polydisperse samples, Mw > Mn. Their ratio is defined as the polydispersity index:

Mw ≥1 Mn which equals 1 for a monodisperse mixture and is >1 for a polydisperse mixture. Larger values mean greater polydispersity. Average molecular weight is of interest for solutions of soluble polymers. For (lyophobic) colloidal particles, we are usually interested in their average diameter (if they are spherical) or their average volume if they are nonspherical. The M’s in (107) are molecular weights (g/mol). Dividing both PDF generated April 27, 2016

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sides of (107) by Avogadro’s number — and then dividing both numerator and denominator on the right-hand side by Avogadro’s number — we convert the molecular weights into the mass per particle m:

∑ ni mi2

Spring, 2016

When we say that Rayleigh light scattering yields the Z-average, we have in mind an experiment which monitors changes in scattering intensity, such as the flocculation experiments (see Application to Flocculation Studies) or Dynamic Light Scattering Experiments below on page 49.

mw ≡ i

∑ ni mi


i

Dividing next by the particle’s density, we convert the mass per particle into the volume per particle v:

∑ ni vi2 vw ≡ i

∑ ni vi

(108)

i

This is (5.36)♥ in p378 Berg’s book, although he doesn’t give its derivation. Z-Average vs Mass Average We have just shown that, with polydisperse samples, turbidity yields the mass-average molecular weight. Earlier we noted [see (95) for example] that light scattering is proportional to the volume-squared (or mass-squared) of Rayleigh scatterers, which suggests that light scattering should yield the Zaverage [see item 5) on page 11]:

Static Light Scattering Experiments: Angle Dependence Light scattering experiments today typically use a laser as a light source. The sample is held in a 1 cm diameter glass cylinder a few cm long (the small black square labelled “sample chamber” in the schematic below). The detector is usually a photomultiplier tube (PMT) which is held by a goniometer, a device which allows the detector to be rotated around a vertical axis through the center of the sample to detect light scattered in various angles relative to the direction of the incident light.

θsca

∑ ni vi3 vz ≡ i

∑ ni vi2 i

which is quite different from the mass average (or volume average) vw given by (108). To reconcile this apparent contradiction, recall that our turbidity experiment obtains the molecular weight from the intercept of a plot of HC/τ versus C. This was needed because we cannot measure the turbidity at zero concentration (it is zero) so instead we extrapolate the ratio C/τ to C=0 (to eliminate the effect of the second virial coefficient B in the equation for osmotic pressure). While the contribution to τ itself from each scatterer is proportional to its volume-squared (or mass-squared), the contribution of a single scatterer to the ratio τ/C (i.e. the turbidity per unit mass) is proportional to its mass to the first power. Thus the intercept yields the mass average rather than the mass-squared average.

Usually the goniometer rotates in a horizontal plane. This plane contains both the wave vector kinc for the incident beam as well as the wave vector ksca for the scattered rays. The plane containing both of these wave vectors is called the scattering plane. The angle between these two vectors is the scattering angle θsca. The scattering angle θsca is not the same as the polar angle in spherical coordinates θ which is the angle measured from the axis aligned with the electric field for the incident light (see sketch on page 34). The angular dependence of scattering depends on the polarization of the incident light.

♥ used in Hwk #1.



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Spring, 2016

Case 1) If the incident beam is a LPPW with the electric field oriented horizontally (in the scattering plane), then (see sketch above)

90 120

60 0.8

θsca = π/2 – θ

0.6

150

If instead, Einc in the sketch above points upward rather than downward (but Einc still lies in the plane of the page), the relationship between the two angles is

30

0.4 0.2 180

0

θsca = θ – π/2

In either case, owing to the squaring of the sine, (84) predicts that the scattering intensity behaves like

(

)

2 Π r ∝ sin= θ sin 2 π2 − θsca = cos 2 θsca (109)

( )

2 π Π r ∝= sin 2 θ sin = 1 2

(110)

which is a constant for different scattering angles (shown as the black circle in the polar plot below). Case 3) Light from the sun (and most light sources other than lasers) is unpolarized, which means that the electric field of the light is not oriented in a particular direction (except that E is always perpendicular to k but this still leaves a degree of freedom). Then the angular dependence of scattering turns out to be the arithmetic average of horizontal and vertical polarization [i.e. (109) and (110)]:

Πr ∝

1 + cos θsca 2 2

This is the same angle dependence appearing in (5.109) or (5.123) of Berg. The scattering profiles for all three cases are plotted in the polar diagram below:

300

240 270

which is plotted as the red curve in the polar plot below (resembles a figure eight on its side). Case 2) On the other hand, if the incident LPPW is polarized such that Einc is vertical, then the scattering plane coincides with the equatorial plane of the sphere which has θ = π/2 for all points. Varying the scattering angle varies the second angle of spherical coordinates (i.e. the azimuthal angle φ). If θ = π/2 for all scattering angles, then (84) predicts that the scattering intensity behaves like a constant:

330

210

E horizontal unpolarized E vertical

Scattering from Larger Particles: Rayleigh-Debye Scattering Reference: van de Hulst, p85f and Kerker, p414. Reading: Berg §5.H.4 Rayleigh scattering requires that all the dipoles of each particle must be oscillating in unison. This will be true only if each atom in the particle sees the same electric field. Generally, this requires that the largest dimension of a particle must be very small compared to the wavelength of the incident light. Recall (93): 2R < 0.036 λ

(111)

Atoms in larger particles are not oscillating in unison. As a result, the scattering from various dipoles do not interfere constructively as Rayleigh scattering theory assumes. Rayleigh recognized these effects and took them into account in a series of papers published in 1910, 1914 and 1918. Further contributions were made by Debye in 1915. The basic problem is to compute the total phase difference between radiated light reaching the detector from an arbitrary pair of atoms in the particle. This total phase difference will determine if interference is constructive (as assumed in Rayleigh scattering theory) or destructive, or somewhere in between. The total phase difference was calculated in (86) as δ(r) ≡ δ1 + δ2 = k(ninc–nsca).r

(112)

where the vectors are defined in Figure 8 on page 36. Copyright© 2016 by Dennis C. Prieve


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ninc-nsca θ 2

Spring, 2016

θ 2

The figure above shows the vector addition (subtraction). From the geometry of the right triangle formed by bisecting the isosceles triangle, we can deduce that the length of the difference vector is

{

0.8

0.6 P 0.4

}

0.2

0

To make it dimensionless, we divide by volume squared:

P ( θ)

iδ

2

corrected

(114)

uncorrected Rayleigh

where P(θsca) is called the form factor. Form Factor for Spheres Rayleigh (1914) worked out the result for a sphere of radius a:


6

8

10

The function P(u) is plotted above. It seems to decay almost monotonically to zero.

(113)

This form factor is used to correct the angular dependence of simple Rayleigh theory for the effect of interference arising because not all the atoms in a particle are oscillating in unison. For example, scattering from solutions is described by (98) in terms of is/I0. Correcting this by multiplying by P(θsca) gives is ( θ ) P ( θsca ) I0 

4

1

The result depends on the scattering angle for a particular shape and size of particle representing the region V.

ic ( θ, θsca ) = I0  

2

u

dV

V2

Q

1

2 iδ ( r − r ) iδ ∫ dV ∫ dV e 1 2 = ∫ e dV

∫e =

(116)

where Q is magnitude of the scattering vector k(ninc– nsca). Here P is a dimensionless correction factor (not the total power radiating from the scatterer).

θ = 2 sin 2

and so on. Once we have δ, the phase factor is calculated as eiδ. This is then summed over all pairs of atoms in the two bodies:

r2

(115)

λ   2  

Q

nsca

ninc − n sca

2

 θsca  4π  θsca  sin  where u ≡ 2k sin  a = a  2  

ninc

r1

 3  P ( θ= sca )  3 ( sin u − u cos u )  u 

sphere

0.1 0.01 P

1 .10

3

1 .10

4

1 .10

5

1 .10

6 0

5

10

15

20

25

30

u

But if we plot the same data on semi-log coordinates (above), we see quite different behavior. Note the appearance of many maxima separated by deep cusps (values of u at which P vanishes). This predicts some complicated scattering patterns when the particle size exceeds a certain value. If we are only interested in the scattering relative to that scattered forward in the direction of the incident light, we could substitute is/I0 = cos2(θsca), which was used for the plots below, in which the effect of particle size on the scattering patterns is illustrated.


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2 . a = 0.02 µm

90 120

2 ⋅a = 0.6 µm 120

60 0.8 0.6

150

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60 0.8

30

0.6 0.4 0.2

150

0.4 0.2 180

90

0

30

180

0

330

210

210 270

240

corrected scattering intensity uncorrected scattering intensity


Increasing the particle radius a bit more to 0.6 µm (above), which is just a little larger than the wavelength in water (0.6 µm/λ1 = 1.3), causes the backscattering pattern to disappear from the figure above left while the forward scattering pattern has become significantly distorted: with scattering falling off more sharply with angle than Rayleigh’s simple theory predicts. 2 ⋅a = 0.6 µm

90 60

120

120

0.8 0.6

150

30

60 30

−4

1 × 10

0.2 0

180

210

330

300 270

corrected scattering intensity uncorrected scattering intensity Raising the particle size an order of magnitude to 200 nm (above) causes the backscattering to become significantly weaker than the forward scattering, but even forward scattering is beginning to show some deviation from cos2(θsca).


90 1 0.01

150

0.4

240

300 270

The above polar plot shows the scattering pattern for 20 nm polystyrene (PS) latex spheres (n = 1.67) in water (n = 1.33) when illuminated by a HeNe laser (λ0 = 623.8 nm in vacuum or λ1 = λ0/n1 = 469 nm in water). These particles are very small compared to the wavelength: 20nm/λ1 = 0.043. The pattern in red is virtually indistinguishable from cos2(θsca) (given in blue), i.e. the simple Rayleigh theory. Recall from (111) that the simpler Rayleigh theory was valid for particles whose diameter was less than 1/20 = 0.05 of the wavelength.

2 . a = 0.2 µm

330

300

240

1 × 10

180

−6

0

210

330 240

300 270


Actually, backscattering has not completely vanished as can be seen in the figure above right in which the radial axis has been made logarithmic.


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Inferring Size from Angle Dependence For Rayleigh scatterers, the angular dependence of scattering contains no information on the size of the scatterer. But for particles which are large enough to have a form-factor different from unity, the angular dependence does contain information about the size. For example, the plot below shows the angular dependence of scattering intensity calculated for polystyrene (PS) latex spheres of different size (2a = 0, 0.2, 0.4, 0.6, 0.8 and 1.0 µm) in water. The incident light from a HeNe laser has λ0 = 515.5 nm in vacuum (or λ = 387.6 nm in water) and is horizontally polarized. Then the uncorrected angular dependence from Rayleigh scattering is cos2(θsca)

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1 ln P ( θsca ) ≈ − a 2 Q 2 5

(118)

Then a plot of lnP vs Q2 in the limit Qa→0 should be linear and its slope is –a2/5. A plot of lnI vs Q2 is called a Guinier plot. For vertical polarization (θ = π/2 for any θsca) the angle dependence of lnI is the same as for lnP. Here is the data above plotted on those coordinates: a increasing

Relative Scattering Intensity

1

a increasing and we see on these coordinates, the plots are linear (at least for small Q).

0 .5

Form Factor for Random Coils, Rods, Disks

0

20

40

60

80

S c a tte r in g A n g le

The form factor for other shapes has also been worked out and the results are shown in the figure below (p423 of Berg):

The strong intensity dependence on size has been removed by normalizing the scattering results by the scattering at zero degrees. Nonetheless, you can clearly see that size affects the rate of decay with scattering angle at small angles. In the limit of small scattering angles, we can approximate the form factor (115) using a Taylor series P (u ) = 1 − 15 u 2 + O ( u 4 )

Truncating after the u2 term and substituting u from (116) P ≈ 1−

where

16π2 a 2 5λ 2

Q=

1 θ  sin 2  sca  = 1 − a 2Q 2 (117) 5  2 

where the x-axis is u defined by (116) except that the sphere radius a is replaced by the disk radius, half the rod length or the radius of gyration of random coils. Note that the functional form of P(u) differs for different shapes.

4π θ  sin  sca  λ  2 

was introduced in (116) and is the magnitude of kinc – ksca. Taking the ln of both sides:



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The analog of (118) (i.e. the Guinier plot) for random-coil polymers♠ and rods is given by polymers:

1 ln P ( θsca ) ≈ − Rg2 Q 2 3

where Rg is the radius of gyration of the molecule; and

ln P ( θsca ) ≈ −

rods:

1 2 2 LQ 36

where L is the length of the rod. Scattering from Solutions: Zimm Plots To obtain the molecular weight M of an unknown polymer solute, we can vary either the scattering angle θsca or the mass concentration C of the polymer. The Rayleigh scattering result is given by (98); multiplying by the appropriate form factor for a random coil polymer: ic = I0

is KC sin θ P ( θsca )= P ( θsca ) I0 1 M + 2 BC r 2 2

(98)

Recall (from page 44) the relationships between θ and θsca for the three primary forms of polarization: sin2θ = cos2θsca for horizontal polarization of the incident light, sin2θ = 1 for vertical polarization, and sin2θ = (1+cos2θsca)/2 for unpolarized light. Experimental data is usually plotted in terms of the Rayleigh ratio which is defined as Rθ ≡

r2

ic ? KC P ( θsca ) = 2 I M + 2 BC 1 sin θ 0

(119)

In the same way that turbidity was transformed to create a quantity that was linear in C [see (101)], we can also transform the Rayleigh ratio

KC ? 1 2 BC = + Rθ MP ( θsca ) P ( θsca )

(120)

The equations above marked with “?” are not quite correct. Recall the derivation P(θsca) from (113): the weighting factor for the phase factor eiδ was taken to be unity (independent of position) when ♠ The shape of a “random coil polymer” resembles a Brownian random walk in 3-D. Just like the mean-square displacement is proportional to time, the mean-square radius of gyration is proportional to the number of monomer segments (like the number of steps). See §2.7 in H&R for more information.


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integrating over the volume. This is fine if the oscillating dipoles are uniformly distributed through the volume. However at higher concentration where ideal solution can no longer be assumed (the conditions in which the second term of the virial expansion is needed to calculate πosm), intermolecular forces give rise to a nonuniform radial distribution function. Then (113) is not appropriate. The phase factor eiδ in (113) should be weighted by the probability of finding another scatterer at that location. In 1948, Bruno Zimm♣ considered the radial distribution function in the calculation of the Rayleigh ratio. He showed that the correct form of (119) is actually 2 Rθ =K ( MPC ) − 2 B ( MPC ) + 

Expanding KC/Rθ in a Taylor series in C and truncating after two terms: KC 1 = + 2 BC +  Rθ MP ( θsca )

which is subtly different from (120) in that the form factor P does not appear in the second term. Substituting the counterpart to (117) which is appropriate for random coils: ♥

  2 2 KC 1  16π Rg θ  1+ sin 2  sca   + 2 BC (121) = Rθ M    2   3λ 2    1 2 2   Rg Q 3   where Rg2 is the mean-square radius of gyration of a random polymer coil (another measure of its size). The minus sign in (117) becomes a plus sign because the term in square brackets represents the inverse. Below is a graph of scattering data (black points) in which both scattering angle and polymer concentration are varied. The data is for cellulose nitrate dissolved in acetone and is Fig. 5-56 in Berg.

♣ Bruno H. Zimm, “The Scattering of Light and the Radial Distribution Function of High Polymer Solutions,” J. Chem. Phys. 16, 1093 (1948). ♥ P.C. Hiemenz & T.P. Lodge (Polymer Chemistry, 2nd ed., Marcel Dekker, 1984) only apply this correction factor (in square brackets) to the first term — not the second. Also see http://en.wikipedia.org/wiki/Static_light_scattering


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∑ Ci M i = M

i

∑ Ci

≡ Mw

i

Application to Flocculation Studies Later in this course, we will discuss aggregation (or flocculation) of colloidal dispersion. Light scattering can also be used to study the rate of this process. If I use (77) instead of (75) in (74) and (84), then multiply by the number of particles per unit volume N, the Rayleigh ratio behaves like

Rθ ∝ N α 2 ∝ NV 2

The nearly vertical lines through the black points represent regressions of data for the same concentration but different scattering angles. The triangles at the bottom of these lines represent the extrapolation to zero scattering angle. The nearly horizontal lines represent regressions of data for the same scattering angle but different concentrations. The triangles at the left end of each of these lines represents the extrapolation to zero concentration. The regression of either set of triangles should yield the same y-intercept. The value of this intercept is the limit of (121) for both C → 0 and θsca → 0:

lim

C →0 θsca →0

KC 1 = Rθ M

(122)

The slopes of the two lines through the two sets of triangles also have meaning which can be inferred from (121): 2 2 dy ( θ, C = 0 ) 16π Rg = d sin 2 θ2 3λ 2 M

dy ( θ =0, C ) = 2B dC Polydisperse Samples The arguments leading to (106) can also be applied to (122) with the Rayleigh ratio Rθ replacing the turbidity τ. Thus in the case of polydisperse sample, the molecular weight inferred from the intercept is the mass-average:

where V is the volume of one scatterer. During flocculation two independent scatterers combine to form a single scatterer having twice the volume

If all singlets combined in this way we would be obtain half as many particles with each one having twice the volume. But according to (123), this would change the scattering by a factor of (1/2)22 = 2. As aggregation proceeds, the scattering increases at a rate that related to the rate of aggregation.

Lecture #8 begins here Dynamic Light Scattering Experiments Reading: Berg §5.H.9 In Scattering from Multiple Brownian Particles: Independent Scatterers on page 38, we pointed out that the total scattering intensity from an ensemble of particles would fluctuate with time owing to Brownian motion of the individual particles, which causes random changes in the many phase angles δ. In the analysis of static light scattering we monitor the average scattering intensity as a function of scattering angle or solute concentration. However the fluctuations themselves contain information regarding the size of the scatterer. The analysis of the fluctuations is called dynamic light scattering. For example, the figure below (taken from Berg, p439) shows the raw scattering signal for three different sizes of polystyrene latex particles

0.085 µm


(123)

0.220 µm

1.011 µm


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While each signal is some random fluctuation around a mean value, larger particles clearly have slower fluctuations than smaller particles. These fluctuations arise because the PS particles undergo Brownian motion so the relative distance between any two particles in the sample volume changes with time. This changes the nature of the interference in light scattered from any two particles from complete constructive interference to complete destructive interference. The rate of change is related to the diffusion coefficient of the Brownian particles. The extraction of the diffusion coefficient from such noisy data involves computing the autocorrelation of the scattering intensity  t0 +T  1  C ( τ ) ≡ lim  i t i t + τ dt ( ) ( )  ∫ T →∞  T  t 0 

(124)

where i(t) is the instantaneous measured scattering intensity. This function describes how strongly the intensity measured at some later time t+τ is correlated with the intensity at current time t. For example, if i(t+τ) is strongly correlated to i(t) (for example if τ=0), then their product will be i 2(t) and we obtain C (0) = i2

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C ( Q, τ )

2

(

)

i2 − i = 1+ exp −2Q 2 Dτ ♦ 2 2 i  i     ξ g 2 ( Q,τ ) where ξ is an instrument constant and

Q=

θ 4π sin sca λ 2

(126)

Experimental data are most often plotted as

G ( Q, τ ) ≡

C ( Q, τ ) − i i2 − i

2

2

(

= exp −2Q 2 Dτ

)

(127)

which should behave as a single exponential decay if the scatterers are monodisperse. This is the G appearing in (5.161) and (5.162) of Berg, but not that in (5.158). Example #1: The figure below shows the autocorrelation function measured at 25ºC with Malvern’s ZetaSizer Nano (in the CPS lab) for a sample of nearly monodisperse polystyrene latex particles purchased from Duke Scientific with a nominal diameter of 60 nm. The instrument uses back scattering (θsca = 178º) from a HeNe laser (λ0 = 632.8 nm). Deduce the average particle size from this data.

On the other hand, at very large delay times, we expect that i(t+τ) will be completely uncorrelated to i(t). For half of the t's in the integral, i(t+τ) will be greater than the mean i and for the other half i(t+τ) will be smaller than the mean. If there is no correlation, we just replace i(t+τ) by i . Then (124) yields C ( ∞ ) =i

2

(125)

For virtually all functions i(t), i 2 > i 2 and C(τ) decays monotonically from one limit to the other as the delay time τ increases. For monodisperse particles (all have same size), the normalized autocorrelation function decays exponentially with the delay time τ:

♦ The exponent in this equation and (127) differs by a factor of “2” from (5.157) and (5.158) in Berg. I believe Berg’s exponent describes the correlation function for the electric field not the intensity. Intensity correlation function defined by (124) is what is usually measured Squaring of the electric field gives rise to the extra factor of 2.



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1.0

0.9

0.8

Correlation Coefficient

0.7

0.6

0.5

0.4

0.3

Figure 10

0.2

The resulting data are well fit by a single straight line having a slope of [suggested by (127)]:

0.1

0.0 0.1

−2Q 2 D = −1.049 × 104 sec−1 10

1000 Time (µs)

Solution: Notice the log scale on the x-axis. This is helpful because the delay time τ can be experimentally varied by orders of magnitude, which can reveal more than one decay owing to scatterers of very different size. Here we are seeing only one decay which is consistent with a monodisperse sample. Also notice that as τ→0, G goes to 0.924 instead of unity. Apparently something other than Brownian motion of the particles is contributing to fluctuations. One possibility is “shot noise” in the PMT’s response. If much smaller τ’s could be sampled, you would expect a second rise in the curve from 0.924 to unity. Dividing each G by G0 = 0.924, then taking the natural log yields the following plot:

Next we calculate the wave number from (126). The wavelength λ in this equation is the wavelength in the sample whereas λ0 = 632.8 nm is the wavelength in vacuum (or in air). The refractive index of pure water is about n =1.33 so

= λ

λ0 = 476 nm n

Converting θsca = 178º into 3.11 radians, (126) leads to Q = 0.0264 nm–1 and the slope can then be converted into the intensityaverage diffusion coefficient

−slope nm 2 µm 2 D= = 7.72 × 106 = 7.72 sec sec 2Q 2 Finally we can convert D into particle radius using the Stokes-Einstein equation (43): = a

kT = 31.6 nm 6πµD

To calculate the above value we used a viscosity for water at 25ºC of 0.893 mPa-s. Doubling this, we obtain 63 nm, which is quite close to the nominal size quoted by the manufacturer of the particles. Polydispersity Even so-called monodisperse particles usually have some degree of polydispersity; a few percent standard deviation in size is typical. Small degrees of



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Intensity (Percent)

25 20 15 10 5 0 0.1

1

10

100

1000

10000

Size (d.nm)

Figure 11

polydispersity can be inferred from experimental data like that above using the method of cumulants. Taking the log of both sides of (127):

ln G ( Q, τ )  = − 2Q D τ  2

(128)

Γ

In the method of cumulants, we acknowledge that the sample exhibits more than one rate of exponential decay. (127) is generalized to

) G ( τ=

∞

∫ g (Γ) e

−Γτ

dτ

0

where g(Γ) represents the distribution (perhaps Gaussian) of decay rates. Then (128) is generalized to

ln [G ( τ )] = −Γτ +

µ 2 2 µ3 3 τ − τ + 2! 3!

(129)

where σD is the standard deviation in the diffusion coefficient D. The size distribution displayed by the Malvern instrument for the data in Example #1 is given by Figure 11. The instrument appears to assume a lognormal distribution. Notice the peak coincides with an average diameter of 63 nm. While the second cumulant provides a measure of the standard deviation of the distribution, what kind of average is Γ in (129)? The answer turns out to be the Z-average.♥ This is suggested by the label on the y-axis of Figure 11: the y-value represents the percent contribution of any size to the total intensity of scattering. At least for Rayleigh scattering, this would be proportional to the volume-squared times the number of particles of that size.

where Γ is the mean of g(Γ) and µ2, µ3 etc. are the second and higher moments of g(Γ). These moments would be zero if our sample is truly monodisperse (one size). A closer examination of Figure 10 shows that the data points at large delay times τ lie systematically above the black regression line representing (128). A somewhat better fit can be obtained by including the second term in (129). This is given by the red regression curve in Figure 10: Γ = −1.078 × 104 sec−1

and

µ2 = 1.118 × 106 sec−2 2!

The value of µ2 can be translated into a standard deviation in particle size using the following formula

µ 2 = 2σ2D Q 4


♥ G.D. Patterson et al., “Light scattering characterization of polystyrene latex with and without adsorbed polymer,” Colloids & Surfaces A 202, 9-21 (2002).


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This video contains many demonstrations of bubble, drop and particle motion caused by surface tension. In this chapter, we will analyze several of the phenomena demonstrated by this video.

All liquid molecules near the interface experience a net force normal to the interface pulling them back into the liquid. Each interfacial molecule then moves away from the interface, but conservation of mass causes another molecule to take its place. In the absence of other forces, the molecules of the liquid phase will rearrange themselves to minimize the interfacial area. Thus small droplets tend to be spherical because that geometry has the smallest area for a given volume.

Molecular Origin of Surface Tension

Surface Tension as an Energy

In thermodynamics, the Gibbs free energy is an extensive property of material whose value is proportional to the mass or volume of material. When the system consists of a mixture of two or more phases, the free energy also contains a contribution which is proportional to the surface area between the phases. This contribution is called surface tension (units are energy per unit area or force per unit length).

From a continuum point of view, the tendency to minimize interfacial area is achieved by assigning a contribution to the free energy of the air-water mixture which is proportional to the interfacial area. That contribution is called surface tension which we will denote by γ.

Surface Tension In class we will view the video Surface Tension in Fluid Mechanics which can be found at Hunt Library with the call number 530.427 S961. It can also be found online at YouTube.♠

Such a contribution arises from intermolecular forces. All molecules experience van der Waals attraction for one another. Consider a single water molecule buried deep in a pool of water. Such a molecule is represented by the blue dot in the lower right of the figure below. This interior water molecule is surrounded on all sides by other water molecules and is pulled equally in all directions (indicated by the four arrows). As a result it feels no net force. air water

For a liquid containing nL moles having a free energy of GL joules/mol in equilibrium with its vapor containing nV moles having a free energy of GV joules/mol, where the interface between liquid and vapor is A, the total free energy of the system is

= G n L G L + nV GV + Aγ Subject to any constraints imposed, G will be minimized by minimizing A.

Next, consider a second water molecule located next to the air/water interface. Such a molecule is represented by the blue dot in the upper left of the figure above. Owing to the very low density of air compared to water, there are virtually no molecules located above our interfacial water molecule. Consequently, the intermolecular forces acting on it are no longer balanced: it feels a net force pulling it normal to the interface into the pool.

♠ http://www.youtube.com/watch?v=MUlmkSnrAzM



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work goes into increasing the surface area of the soap film by 2l dx which increases the free energy by 2 γ l dx.♣ The factor of two arises because the soap film has two air-water interfaces: one each on the top and bottom. Assuming there are no irreversible losses of energy:

dW = F dx = γ dA = γ ( 2l dx ) = 2 γ l dx where γ is the surface tension. Dividing out the dx we have

F = 2γ l

(130)

This force must be continuously applied to overcome the thermodynamic tendency to reduce the interfacial area and keep the film from shinking in area. Solving for γ: To understand the consequences of surface tension, consider a soap film (colored area in sketch above) supported by a rectangular wire frame, in which the right side of the wire frame is a slide wire which can be moved either right or left. Suppose we apply a force F and drag the slide wire to the right a distance dx. The amount of work done is F dx. This

= γ

F force =[ ] 2l length

Values of the surface tension for various liquids are shown at several temperatures in Table 1. Strictly speaking, the term “surface tension” refers to the interfacial free energy between a liquid and its vapor. However, when the density of the gas phase is much

Table 1 Surface Tension for Common Liquids*

*Taken from CRC Handbook of Chemistry & Physics, 87th edition, 2006-7.

♣ The volume of liquid held in the film is a constant, so the film also thins a little as the area is increased.



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Table 2 Surface vs Interfacial Tensions*

*taken from Hunter, Vol. 1, page 233.

less than the liquid, replacing the vapor by a mixture of vapor and air usually doesn’t change the value of the interfacial free energy. Replacing the gas phase by second immiscible liquid does however change the interfacial free energy. The term interfacial tension is the generic term used for the interfacial free energy per unit area between two arbitrary (immiscible) phases. Table 2 shows the effect of replacing the gas phase in contact with various organic liquids by liquid water. At least in these examples, the resulting interfacial tension between liquids lies between the surface tensions for the two liquids.

Lecture #9 begins here Laplace Pressure One manifestation of surface tension is that it induces a hydrostatic pressure difference across a curved interface: the concave side of the interface has higher pressure. Recall the demonstration in Trefethen’s video in which he blows a soap bubble at the end of a pipe, then holds his finger over the end of the tube:


When he removed his finger and unblocks the tube, the soap bubble shrinks forcing air out through the tube. This demonstrates that at equilibrium, the pressure inside the bubble is greater than outside. Consider a spherical soap bubble of radius R composed of a liquid having surface tension γ. Consider a force balance about the upper hemisphere:


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The thick blue curve above represents a side view of the upper half of the soap bubble, represented as a hemisphere of radius R. The dotted rectangle is the side view of a cylinder of height R and radius R whose top and bottom surfaces are circular disks of radius R. The red arrows represent various forces that act on the system. The four downward-pointing arrows labelled γ2πR represent the force (distributed along the equator of the hemisphere) exerted by the lower half of the sphere on the upper half.

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which is surely measureable but not large. To get a pressure difference of 1 atm, we need a diameter of 2R = 2.88 µm. A sphere is a particular shape for the interface. More generally, two radii of curvature are needed to characterize a curved surface and the pressure difference is given by the Young-Laplace equation. This derivation of the Young-Laplace equation is adapted from Hunter, Vol 1.

Surface tension pulls downward with a force equal to the product of surface tension γ and the perimeter of the soap film 2πR [see (130)]. But we need to double this because the soap film has both an inner and out surface and both pull downward. Also acting downward is the hydrostatic pressure outside the bubble po. This exerts a downward force equal to the product of the pressure and the projected area of the sphere πR2. The total of the downward forces is downward:

2×2πRγ + πR2po

This is balanced by the upward force of hydrostatic pressure pi inside the bubble: πR2pi

upward:

Equating these two forces and solving for the pressure difference: soap bubble:

Consider a drop of liquid (see figure above) having surface tension γ in air but not necessarily a sphere. A spheroidal cap is bounded by a contour (the thick pink line) which is everywhere the same distance d from some arbitrary point X on the surface. The three distances denoted d might not look equal in the diagram owing to perspective.

4γ pi − po = R

If instead of a soap bubble, we had a liquid drop in air (or an air bubble immersed in liquid), we would not need to double the perimeter. Then the pressure difference turns out to be: liquid drop:

or air bubble:

2γ pi − po = R

(131)

2γ pi − po = R

Notice that it doesn’t matter whether the condensed phase is inside or outside, the pressure is always higher on the concave side of the interface (i.e. inside the sphere). Example: Calculate ∆p for a 1 mm air bubble submerged in water at 20ºC; at this temperature, the surface tension of water is 73 mN/m (see Table 2). Solution: Using (131):

 mN  2  73  2γ m  =  pi − po = = 292 Pa = 0.003 atm R 0.0005 m Copyright© 2016 by Dennis C. Prieve

In the sketch above, the edge of the spheroidal cap has been re-drawn and re-scaled as the pink curve labelled ACBDA. Contours AB and CD (which lie completely on the surface of the drop) both contain the point X and form a right angle at point X. In the limit that d→0, both AB and CD become circular arcs of radius r1 and r2, respectively. From the differential geometry of surfaces (Weatherburn, 1930), the directions of AB and CD (which are perpendicular) can be rotated around point X (keeping the two curves at right angles at X) such that r1 is a maximum (R1) and r2 is a minimum (R2). Furthermore the sum of the reciprocals of the two radii r1–1 + r2–1 is independent of the orientation of the two perpendicular axes AB and CD:


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1 1 1 1 + = + r1 r2 R1 R2 This sum of the reciprocals of the radii is called the mean curvature of the surface at point X, r1 and r2 are called radii of curvature and R1 and R2 are called the principal radii of curvature. In the same way that a force F was needed to overcome the thermodynamic tendency of the soap film to shink in (130), the remainder of the surface must exert a force to keep the area of the spherical cap from shinking. This force is exerted tangent to the surface as shown as the arrow at point A in the figure and its magnitude for a differential arc length ds (the analog of l) measured along ACBDA is

dF = γ ds The component of this force which is normal to the spherical cap at X (i.e. in the direction of the line XY) is

dFvertical = γ ds sin θ When d→0, the angle subtended by an arc of length d and radius r1 equals their ratio:

d lim θ = r1 d →0 and

Thus

lim sin θ = θ =

d →0

d r1

Spring, 2016

This vertical component of force on the cap is acting downward. But the drop is not accelerating; so another force of equal magnitude but opposite direction must be acting on the cap. This force is hydrostatic pressure: the pressure inside the drop is higher than the pressure outside by ∆P. The upward force due to this higher pressure is ∆P multiplied by the area of the disk (πd2): 1   1 ∆P πd 2 = γπd 2  +   R1 R2 

or

1   1 ∆P =γ  +   R1 R2 

which is called the Young-Laplace equation. The sum of the two radii reciprocals is called the curvature J of the surface:

= J

1 1 + R1 R2

Note that a flat surface has both radii equal to infinity: R1 = R2 = ∞ and J = 0. In other words, a flat surface has zero curvature. Curvature can have a profound effect on phase equilibria. For example, small water drops have a higher vapor pressure than bulk liquid at the same temperature. The reason is that the pressure inside the drop is higher, which increases its Gibbs free energy. Case #1: Equilibrium between bulk phases. First consider the usual problem of phase equilibrium of a one-component system (say pure water) between a liquid and its vapor at a flat interface:

dFvertical d = γ ds r1

This is the contribution to the vertical component of the force from ds near points A or B. Averaging similar contributions from points C or D yields dFvertical γd  1 1  γd  1 1  = + = + ds 2  r1 r2  2  R1 R2 

Since the sum of the reciprocals of the radii is independent of orientation, this average should apply for all points along ACBDA. We can integrate over the entire edge of the spherical cap. In the limit d→0, the edge is just a circle of radius d.

γd F= vertical 2

1   1 R +R  2   1

∫ edge of cap

ds

   2 πd

or

1   1 + Fvertical = γπd 2    R1 R2 


Thermal, mechanical and chemical equilibrium require: TV = TL = T L V P= P= P0 (T ) ∞ ∞

V L G= ∞ G= ∞ G0

(132)

According to Gibbs Phase Rule (P + F = C + 2), for a single component system (C = 1), there is only one degree of freedom (i.e. F = 1) for two phases (P = 2) to co-exist. Thus you can specify temperature or pressure, but not both. In other words, for a given temperature, there is only one pressure at which the


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 ∂G    =V  ∂P T

two phases co-exist. This is called the vapor pressure P0(T) which is plotted in the following graph:

where V is the molar volume. Changing the pressure of the vapor from the vapor pressure P0 (T ) = P∞V to

PrV causes the following change in the free energy of the vapor: GrV

PrV

ideal gas V law Pr

V G∞

P∞V

P∞V

= ∫ dG

V = ∫ V dP

∫

RT dP P

where the second equation substitutes the ideal-gas law for the molar volume of the vapor. Integrating V GrV − G∞ RT ln =

Notice that at 100ºC (the normal boiling point of water), the vapor pressure is 1 atm. Case #2: Equilibrium between small droplets and their vapor. Now consider the same component at the same temperature but this time the liquid is divided into equal-sized liquid drops in equilibrium with their vapor.

PrV

P∞V (T )

r PL , TL, GL PV, TV, GV A similar integration can be performed for the liquid. One difference is that the molar volume of a liquid is practically independent of pressure, so the resulting change in free energy is:

Because of the curved interface, the mechanical equilibrium requirement is different, making the pressure inside the liquid higher than that of the vapor: TV = TL = T

PrL

− PrV

2γ = r

GrV = GrL

(133) (134)

The pressures are clearly different in the two phases. In experiments, it is the pressure of the outer phase (vapor in this case) which can be directly measured. As in Case #1, for a given temperature, there is only one pressure (in the vapor) for which both liquid and vapor co-exist. As we will now see, the pressures of both phases will differ from the vapor pressure with a flat interface, even when the temperature is the same. The pressure affects the Gibbs free energy of either phase. The effect of pressure on the molar Gibbs free energy can be calculated from:


L L L V  GrL − G= ∞ V  Pr − P∞ (T ) 

According to (132) and (134), the left-hand sides of the above two equations are the same, so the righthand sides must also be equal: PrV RT ln = V L  PrL − P∞V (T )  P∞V (T )

Finally, we substitute PrL from (133): RT ln

PrV 2γ   = V L  PrV + − P∞V (T )  V ( ) r P∞ T      PL  r  

But we cannot easily solve this equation for PrV. However, it turns out that the increase in vapor pressure PrV − P∞V is always small compared to the Laplace pressure-difference 2γ/r, so we can neglect

PrV − P∞V on the right-hand side, leaving:


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RT ln

PrV P∞V

(T )

Spring, 2016

2 γV L r

≈

Solving for the new vapor pressure PrV: for drops:

 2 γV L  PrV = P∞V (T ) exp    rRT 

T = 100ºC (135)

≥ P∞V (T )

which is called Kelvin's equation for drops. Note that as r→∞, Kelvin’s equation correctly predicts no change in vapor pressure. But as r becomes smaller, the vapor pressure increases with a decrease in size. Example #1: Evaluate the vapor pressure of water in liquid droplets at 100°C as function of drop size. Solution: The surface tension of water at this temperature is γ = 58.9 mN/m♣ and the vapor pressure for flat interface is P∞V = P0 = 1 atm. The molar volume of liquid water can be calculated from its density ρ = 1 gm/cc and molecular weight M = 18 gm/gmol:

VL =

M = 18.0 cc/gmol ρ

Then Kelvin’s equation for drops becomes

 0.684 nm  PrV = P∞V (T ) exp   r   The corresponding pressure inside the liquid drop is higher by an amount equal to the Laplace pressure difference: L P= PrV + r

2γ r

Substituting in various values of r yields the graph shown below:

Thus the effect on vapor pressure PrV (the red curve) is not significant except for nanoparticles. Notice that the pressure in both phases increases as the droplets get smaller. Proof: Let’s now go back and test the main approximation made in the derivation of Kelvin’s equation (135) for drops: PrV − P∞V 

2γ r

(136)

In terms of the graph above, the left-hand side is given by the difference between the red curve and 1 atm whereas the right-hand side is given by the green curve. At least for very small drops ( 2r < 10–7 m), the green curve is clearly much larger — in agreement with the assumption (136). For larger drops, this graphical comparison is inconclusive because both quantities tend to vanish as drop size increases, making it impossible to see which vanishes faster. But in this limit of large drops, an analytic simplification of (135) is possible. Writing the argument of the exponential in (135) as x, then substituting the result into (136) gives

(

)

lim PrV − P∞V= lim ( e x − 1) P∞V ≈ xP∞V r →∞ x →0  x

Next, we substitute the argument of the exponential in (135) for x in (136):

(

)

ideal

2 γV L V gas law 2 γ V L 2γ  lim PrV − P∞V ≈ P∞ = V rRT r V r r →∞  0.001

where in the second equation, we have made use of the ideal gas law: P∞V V V = RT . At least for water at ♣ Vargaftik et al., J. Phys. Chem. Ref. Data, 12(3), 817820 (1983).


100ºC and 1 atm, molar volume of liquid V L is on


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the order of 0.001 of the vapor V V . Thus (136) is also satisfied for large drops. Kelvin’s equation for drops has a couple of important implications. Thomson’s Equation: Boiling Point of Drops 1) The main implication of Kelvin’s equation (135) for drops is that vapors need to be subcooled (at least transiently) a few degrees below their usual boiling point to begin condensation. Upon cooling a superheated vapor, the first liquid drops which form would be small. For these drops to be stable or grow, the temperature has to be lower than the normal boiling point. The boiling point for small droplets is given by Thomson’s equation:♦

Tr =

( )

T∞ P∞V

 2 γV L  exp  −   r ∆H vap      − 0.052 nm  r  

(137)

where ∆Hvap is the molar heat of vaporization (40.65 kJ/mol for water at 100ºC). Thomson’s equation is derived by combining Kelvin’s equation (135) with the Clausius-Clapeyron equation from thermodynamics. More precisely, Tr in (137) is the temperature of vapor and liquid at equilibrium when the liquid exists as droplets of radius r. Below is the boiling (or condensation) point of pure water (no air) for a constant 1 atm of pressure in the continuous outer vapor phase.

PV = 1 atm

In any real two-phase system, a range of droplet sizes can be expected: even if we somehow started with all of one size of droplet, random Brownian collisions of vapor molecules with liquid droplets will lead to small increases in size for some droplets; and water molecules are continuously escaping from liquid droplets, leading to small spontaneous decreases in size for other droplets. ♦ This equation is named after James J. Thomson, the elder brother of Lord Kelvin (William Thomson).


Spring, 2016

Suppose we hold the temperature and pressure fixed at (Tr, PrV ). According to our terminology, droplets of radius r are at equilibrium under these conditions. But what about larger or smaller droplets which exist at least transiently? According to the plot above, droplets smaller than r have a lower boiling point and are superheated; they can be expected to evaporate completely because any reduction in their size puts them further from equilibrium, increasing their rate of evaporation. Conversely, droplets larger than r have a higher boiling point (i.e. subcooled) and will grow by condensing more vapor. In the final equilibrium state, all of the vapor becomes condensed (assuming r p E: ♣ A capillary is a long tube having a small bore (from Latin capillus meaning hair). ♦ Radius must be much smaller than 2 γ ρg .


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pC – pE = ρgh

Spring, 2016

(141)

Since pC = pD according to (140), the left sides of (139) and (141) are equal, so the right-hand sides must also be equal:

2γ ρgh = r Solving for the capillary rise height h:

= h

2γ a2 = ρgr r

a≡

where

2γ ρg

(142) The expression (144) for capillary rise can be rewritten in terms of the capillary constant: (143)

= h

a2 cos θ r

is called the capillary constant or the capillary length, which is a property of the liquid having units of length.

Example: Compute the capillary rise of benzene in a clean glass capillary with l mm diameter. Assume cosθ = 1.

The derivation above assumes a hemispherical shape for the meniscus. More generally, the shape depends on the contact angle θ.

Solution: γ = 29 mN/m at 20°C and ρ = 0.879 g/cm3. Eq. (143) gives a = 0.259 cm and Eq. (144) gives h = 1.35 cm using r = 0.05cm and cosθ = 1. Curvature of an Arbitrary Contour

Contact angle is always measured through the liquid between the solid and gas phases as shown above. Assuming the interface is still spherical, its radius of curvature is R instead of the radius r of the capillary. From the geometry above, we see that r = R cosθ. Replacing r in the equation above by the new radius of curvature R = r/cosθ, obtain

= h

2γ cos θ ρgr

(144)

Our analysis of capillary rise in the previous section is only valid if the diameter of the capillary is very small (see footnote on page 62). In larger capillaries, gravity will distort the spherical shape of the meniscus. To analyze this distortion, we need to compare the Laplace pressure difference across a curved interface with the hydrostatic pressure difference along the meniscus due to gravity. To evaluate the Laplace pressure difference, we need to generalize (139) to nonspherical surfaces. In this section, we first consider how to calculate the curvature of a 2-D contour, and then the curvature of a surface in 3-D. Consider an arbitrary contour lying entirely in the xy plane. Let the contour be given by the function y = f (x).

Now cosθ is positive for 00. This corresponds to contact angles in the range 0 ≤ θ ≤ 90° (i.e.wetting surfaces). For contact angles larger than 90° (i.e. nonwetting surfaces), we expect h R. Repeat steps 2) to 4) until (172) is satisfied to the required tolerance.

This trial-and-error procedure to solve a boundaryvalue problem (i.e. the boundary conditions are specified at two different values of the independent variable r) is called the shooting method.♦ Special Case: contact angle θ = 0. For a given value of the capillary length a and capillary radius R, the solution for b when θ = 0 is summarized by the figure below, which is a plot of Sugden’s table (Table 2-6 on p67 of Berg): 1

R/b

a≡

2γ = 3.86 mm ρw g

Next we use the calculated value of R/a = 0.259 in the Sugden table to determine R/b = 0.979. Since R is known, we can then calculate b = 1.022 mm. The capillary rise height can now be calculated from (168). Substituting the known values of a and b, we obtain h = 14.60 mm. In the next example, we will turn the problem around. Instead of giving the surface tension and asking for the capillary rise, we give the measured rise for an unknown solution and ask for the surface tension. Example #2: A fluid whose density is 876.5 kg/m3 exhibits a capillary rise of 6.25 mm in a 2-mm diameter capillary tube. Find the surface tension, assuming the contact angle is 0º. Solution: Since the surface tension γ is unknown, both a and b are unknown. As an initial guess, we assume the meniscus is a hemisphere with the same radius as the capillary tube: b0 = R = 1 mm. Then solving (168) for a yields

a0 = hb0 = 2.50 mm

0.8

Having a trial value for a allows us use Sugden’s table to obtain a better estimate of b:

0.6

bi +1 =

0.4

0.2

Spring, 2016

0

1

2

R/a

Notice for R/a ≪ 1, that R/b = 1; in other words, when R≪a we have b = R. This is consistent with our early analysis of capillary rise on page 62.

Lecture #12 begins here Example #1: calculate the capillary rise height h for water (ρ = 1 g/cm3, γ = 73 mN/m) in a clean glass (θ = 0º) capillary having a radius R of 1 mm. Solution: First we calculate the capillary length from (158) (neglect the density of air):

R R f   ai 

(173)

where the function f (R/ai) = R/bi+1 is Sugden’s table given by the graph on the previous page. Next we can update our estimate of a again using (168):

ai +1 = hbi +1

(174)

Combining (173) and (174) yields a compact recursion formula:

ai +1 =

Rh R f   ai 

This rapidly converges in a few steps. Sequential values of a are given by the following table

♦ Guessing b is like taking aim at a target. Integrating over the range of values of the independent variable is like firing a shot and, finally, testing to see if the (172) is satisfied is like seeing if the target was hit.



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Once we have converged in a we can calculate the surface tension from (158):

a≡ or

2γ = 2.5621 mm ρw g γ = 28.2 mN/m



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Contact Angle Surface tension arises because molecules residing near the surface of a liquid possess a higher energy than molecules in the bulk of the liquid. This difference in energy results from the greater number of neighbors which bulk molecules have compared to surface molecules (recall Molecular Origin of Surface Tension on page 53).

Spring, 2016 • θ = 0° -- Liquid is said to “completely wet the solid” or to “spread on the solid”

Example: water on clean glass at 25°C has θ = 0. • θ < 90° -- Liquid is said to “partially wet the solid”

These same differences between surface molecules and bulk molecules applies to solids as well as liquids. Thus we might anticipate that solids have a kind of surface tension associated with them. One important difference is that solids cannot deform to minimize their surface energy. So the manifestation of surface energy of solids is more subtle. An important illustration of surface energy in solids is the contact angle.

• θ = 90º

• θ > 90° -- Liquid is said to be “partially nonwetting”

When a drop of fluid is placed on a solid (see figure above), the drop may assume a variety of shapes, depending not only on the volume of the drop and its liquid-vapor surface tension, but also on the nature of the solid. At equilibrium, the drop will form some angle with the solid. This is called the contact angle. This angle (denoted by θ in the figure below) is independent of the size of the drop, but it does depend on the particular solid and liquid used. Thus, the contact angle represents an equilibrium property of the solid/liquid interface.

Example: mercury on clean glass at 25°C has θ = 140° • θ = 180º — liquid is said to be “completely nonwetting”

As it turns out, the contact angle can be greatly altered by contamination, especially contamination of the solid surface. The reason for this becomes obvious as soon as we draw a connection between contact angle and surface energy (as we will do in the next section). Then any surface contamination might be explained through a change in the surface energy. Derivation of Young's Equation

By convention, contact angles are always measured through the liquid and 0 x0, respectively, where x0 denotes the location of the plane dividing the two phases.

where molar partial

Partial Proof: For a more complete derivation, see Berg §3.E. Here we will just reproduce the final steps in the derivation so you can see why Gibbs choice of a dividing surface is convenient.

for x > x0

Substituting this into the definition for Γ and splitting the integral into two:

= Γ

where G is Gibbs free energy, ♣ k is the phase and i (or j) denotes the component.

for x < x0

x0

 [=] J/mol   T , P,nkj ≠ i

∑ nis d µis + Ad γ =0 i

where nis is the number of moles of component i adsorbed to the surface (s) and µis is the chemical potential of those surface molecules. Solving for dγ:♦ −d γ=

nis ∑ A d µis= i  Γi

where

ns Γi ≡ i

∑ Γ i d µi

(189)

i

A

♣ recall G is defined as U + PV – TS = H – TS. ♦ We have dropped the superscript on µ because (at i equilibrium) components will distribute themselves between the phases such that µiα =µβi =µis =µi .


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Special Case #1: Two components For binary solutions, we have only the solvent and the solute. Then the sum in (189) contains only two terms: −d γ = Γ solvent d µ solvent + Γ solute d µ solute    0

Choosing the dividing surface such that Γsolvent = 0 and solving the remainder for Γsolute leads to (188) which is called Gibbs adsorption equation. For a thermodynamically ideal solution:

( d µi )T , P

= RTd ln Ci

(190)

 ∂γ     ∂ ln Ci T , P

(191)

Using this in (188):

Γi =−

1 RT

Figure 12: Aqueous solutions of normal alcohols at 25ºC.♥

Note that if surface tension is decreased by the addition of the solute, the solute must be adsorbed at the interface (since Γ > 0). This is the case for most nonelectrolytes and surfactants. On the other hand, addition of a simple electrolyte like NaCl increases surface tension,♠ which implies negative adsorption (Γ CSDS, the first term in the above sum can be neglected, leaving excess NaCl:

∂γ 1   Γ SDS = −  RT  ∂ ln CSDS T , P

which differs from (195) by a factor of two but now resembles (191) for nonelectrolytes. Langmuir Isotherm Irving Langmuir (1918) derived a relationship between the amount of gas adsorbed on a solid surface (denoted by Γ moles per area) and the concentration of the adsorbing component in the gas phase (denoted by C moles per volume). He assumed there existed a finite number of surface sites at which a molecule could adsorb (denoted by Γmax moles per area). He further assumed that the kinetics of adsorption were first-order in C and in the number of available sites Γmax– Γ while the rate of desorption was first-order in the number of occupied sites Γ:

dΓ = ka C ( Γ max − Γ ) −  dt 

net adsorption rate

adsorption rate

 C  Γ (C )  Csat → Γ max 1 

as

C →0 Csat

as

C →∞ Csat

Eliminating Γ between Gibbs adsorption equation for ideal solutions (191) and Langmuir’s isotherm (199):

−

Γ max C 1 dγ = RT d ln C Csat + C

Multiplying both sides by –RT dlnC:

kd Γ 

desorption rate

d γ = − RT Γ max

where ka and kd are rate constants for adsorption and desorption, respectively. At equilibrium, the rate dΓ/dt must vanish. Solving for Γ at equilibrium: ka C Γ max C Γ max = Γ = kd + ka C kd +C ka

= − RT Γ max

or

dC Csat + C

d ( Csat + C ) Csat + C

= − RT Γ max d ln ( Csat + C )

where Csat is treated as a constant in the second equation above. Integrating from C = 0 and γ = γ0 (pure water) to C = C and γ = γ:

The ratio of rate constants kd/ka must have units of concentration so we denote this ratio as Csat. Making this substitution leaves

Γ max C Γ (C ) = Csat + C

where the red curve is (199). The other two lines (brown and blue) are the two asymptotes for large and small concentrations:

(199)

γ0

∫ γ

0

d γ = − RT Γ max ∫ d ln ( Csat + C ) C C

= RT Γ max ∫ d ln ( Csat + C ) 0

which is called the Langmuir adsorption isotherm. While this equation was derived for gases, it works well also for adsorption of solutes.


C +C  γ 0 −= γ RT Γ max ln  sat   Csat 

(200)


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which is called Szyszkowski’s equation.♥

Lecture #14 begins here Let’s compare (200) to the data for alcohols shown in Figure 12 on page 85. We have already seen in the example on page 85 that a straight line on these semi-log coordinates implies that the surface excess is independent of C. In terms of the Langmuir isotherm, this situation corresponds to saturation, or to C ≫ Csat. Dropping Csat relative to C in the numerator of the logarithm in (200) leads to

 C  γ 0 −= γ RT Γ max ln    Csat 

Now the log of the quotient equals the difference between the logs:

γ 0 −= γ RT Γ max ( ln C − ln Csat ) with a little rearrangement leads to γ=

for C≫Csat:

( γ 0 + RT Γmax ln Csat ) − RT Γ max ln C

which is a linear relation between γ and lnC having a negative slope. Thus Szyszkowski’s equation (200) is consistent with the linear portions of the curves in Figure 12 having negative slope. At the other extreme of C ≪ Csat, the log in (200) can be approximated using a Taylor series expansion:

C C +C   ln  sat ln 1 + =  Csat   Csat

 C CMC

= C [ monomer ] + N [ micelles ]

Spring, 2016

length of either chain, you can achieve a wide variety of properties. They are commercially available as the Brij® series (ICI America). Another variation is H(CH2)n−C6H4−(OCH2CH2)mOH alkane−benzene ring−PEO available as the Triton X® series (Union Carbide). One particular example is Triton X-100:

where N is the aggregation number (i.e. the number of surfactant molecules per micelle).

monomer

CMC

N

-1

CMC

s elle mic

C

Returning to Figure 13 on page 88: the reason for the plateau above the CMC is that C which appears in (199) and (200) is the monomer concentration – not the total concentration. Since the monomer concentration is constant above the CMC, so Γ and γ are constant above the CMC. Surfactant Types SDS is said to be an anionic surfactant. When it is dissolved in water, it dissociates into a sodium ion and a negatively charged (i.e. anionic) tail:

which has a branched alkane chain. Another popular commercial nonionic surfactant is the Pluronics® line (BASF). Pluronics are triblock co-polymers: HO(CH2CH2O)m/2-(CH2CH3CHO)n(CH2CH2O)m/2H PEO−PPO−PEO where the polypropylene oxide (PPO) block is slightly more hydrophobic than the PEO block, so in water the PPO block typically adsorbs, leaving the two PEO tails protruding from the surface. Commercial surfactant which slightly more complicated molecular structure are the Span® series (Croda International) which are generically known as sorbitan esters:

H(CH2)12SO4–Na+ → H(CH2)12SO4– + Na+ Another anionic surfactants is sodium stearate: H(CH2)17COO–Na+ An example of a cationic surfactant is hexadecyltrimethylammonium bromide (CTAB):♠ H(CH2)16N(CH3)3+Br– Of course, the polar group need not be ionic. Perhaps the most common nonionic surfactants are the polyethylene oxides, the simplest of which are H(CH2)n−(OCH2CH2)mOH

which is sorbitan (the 4-C ring plus the 2 attached C’s), an ester linkage (the –O– ) and a fatty acid (the R-carboxyl group). The R of a fatty acid is an alkyl chain which is the nonpolar portion of the molecule. Replacing the three –OH groups of the Spans with polyethoxy groups –O(CH2CH2O)xH produces another series called the Tween® series (Croda International):

alkane−PEO which consists of a nonpolar alkane chain connected to a polar polyethylene oxide chain. By varying the ♠ “Hexadecyl” is also called “cetyl” – hence the C in CTAB, which stands for CetylTrimethylAmmonium Bromide.



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Other shapes include cylinders, lamellar bilayers and vesicles. Vesicles are also bilayer structures. Perhaps the most common occurance of this bilayer structure is in biological systems.

Lipid bilayers of phosphoglycerides are a very important part of nearly all biological membranes -from cell walls to the lining of the gut. Their function in biology is to make transport of solutes through the membrane highly selective. Phospholipids are an biologically important class of surfactants. They have two tails instead of one. They also tend to form bilayers which are the basis of all cell membranes. One of them (dipalmitoyl lecithin) is a lung surfactant which keeps the air sacs open in the lung by reducing surface tension; without them capillarity would make breathing almost impossible. Micelle Shape Reference: Berg §3.I Spheres are not the only shapes which micelles can take.


Packing Parameter The shape of the micelle depends largely on the architecture of the surfactant molecules: in particular, the size the headgroup relative to the size of the tail. This is quantified in a dimensionless number called the packing parameter or packing factor:

P≡

at v = a0  c a0

where v is the volume of the micelle core per surfactant molecule, a0 is the cross-sectional area of the headgroup and c is the length of the tail. If we think of the tail as a cylinder then the ratio of its volume to its length would equal its cross-sectional area of the cylinder. Thus the packing factor can also


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be thought of as the ratio of the cross-section of the tail at to the area of the headgroup a0.

Spring, 2016

volume of the tail of one surfactant molecule yields the number of surfactant molecules (i.e. the aggregation number):

4π 3 R na = 3 v Dividing the surface area of the spherical micelle by the area per headgroup also yields the aggregation number:

4πR 2 na = a0 Eliminating na between these two equations: The area occupied by the headgroup depends on the balance between hydrophobic attraction between tail segments near the outer edge of the core and hydrophilic repulsion between the headgroups themselves (as suggested by the figure above). For surfactants having alkane tails, the fully extended length and volume can be estimated from known structural properties of alkane molecules: normal alkanes have a C-C bond length of 0.154 nm and a bond angle of 109.5º. The terminal methyl group has a radius of 0.21 nm while the half length of the bond between the alkane and the polar head is 0.06 nm.

v R = a0 3 Dividing both sides by the length of the tail yields the packing factor: R ≤1

v R 1 lc 1 P≡ = ≤ a0  c  c 3 3 Now the radius of the core must be less than the extended length of the hydrocarbon tail (otherwise you would have a vacuum at the center); in other words R/c ≤ 1.

Using this information we can deduce a formula for the extended length of an n-carbon alkane chain:

= c [ 0.21 + 0.126 ( n − 1) + 0.06] nm = ( 0.144 + 0.126n ) nm This is close to what Tanford (1980) recommends:

=  c ( 0.15 + 0.1265n ) nm and

v =( 27.4 + 26.9n ) × 10−3 nm3

Example: show that the packing factor must be less than 1/3 in order to get spherical micelles. Solution: let R denote the radius of the core of the micelle. Dividing the volume of the core by the



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Larger values of the packing factor leads to other shapes for micelles as shown in table 8.2 (at right).

Solubilization Surfactant solutions above the CMC can solubilize otherwise insoluble material. This is dramatically illustrated in the photos below♠ which show a dye (bromothymol blue) in dodecane with or without a surfactant.

A

♠ courtesy of Ben Yezer (PhD thesis, CMU, 2016).


B

C

In vial A we have the dye only (no surfactant) in dodecane. The dye is virtually insoluble in pure dodecane so it collects at the bottom as the red precipitate. The remaining liquid is colorless owing to the absence of any dissolved dye (the red color at


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the meniscus is just the reflection of the precipitate at the bottom). Vial B contains the same dye in dodecane, but this time we have also dissolved a commercial surfactant: sodium dioctysulfosuccinate (also known as aerosol OT or AOT).♣ AOT is soluble in dodecane but the solutions are colorless.

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The different colors of Vials B and C is likely to inverted AOT micelles being acidic while OLOA micelles are basic. HLB Number Reference: Berg §9.B.2

Notice that the solution with AOT and dye has a yellow-orange color; this indicates that the dye has been solubilized in the surfactant solution. There is still some precipitate at the bottom because we have added more dye than can be solubilized.

With multiple series of commercial surfactants available (see Surfactant Types on p92) and each series having multiple variants, it becomes challenging to decide which surfactant to use for a particular application. One useful guide is the HLB number, which stands for hydrophile-lipophile balance.

Vial C contains the same dye in dodecane along with a different commercial surfactant: polyisobutylene succinimide (also known as OLOA).♥ Once again, the intense blue color indicates the dye has been solubilized in the OLOA-dodecane micelles.

For nonionic surfactants, the HLB number is related to the fraction of the total molecular weight of the molecule which is polar (i.e. hydrophilic). More precise, it is calculated from

Both AOT and OLOA in dodecane are known to form micelles above a CMC. Because dodecane is very nonpolar, these are inverted micelles having the hydrophobic tail on the outside and the polar head on the inside:

inverted micelle

= HLB

M.W. of polar part 100% × total M.W. 5

Thus HLB’s for nonionic surfactants vary between 0 and 20, with HLB=10 having equal amounts of polar and nonpolar character. The HLB’s for the Spans and Tweens are shown below (along with the HLB’s needed for some typical applications):

normal micelle

Thus the core of the micelle is water-like. Since bromothymol blue is soluble in water, it can be solubilized in these inverted micelles. The color of the dye reflects the pH (at least in aqueous solutions). The relationship between color of the aqueous dye solution and pH is indicated by the following chart:

For ionic surfactants, the HLB is calculated from

HLB = 7.0 + ∑ ( Group Numbers ) where the group numbers are given in the table below:

♣ AOT is widely used in pharmaceuticals as a stool softener. ♥ OLOA stands for Oronite Lubricating Oil Additive. It is added to motor oil to control sludge formation in internal combustion engines. Oronite is the division of Chevron which markets this product.



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Lecture #15 begins here Flocculation There is an irresistible tendency for agglomeration to occur in most lyophobic systems.♥ interfacial tension -- a contribution to the free energy of a two-phase system that is proportional to the interfacial area

R1

R1

+

R2

Since interfacial tensions are always positive, the free energy of any such system can be reduced by reducing the area. For example, consider two spherical oil droplets which coalesce together to form one larger sphere: V2 = 2V1 → then

R2 = 21 3 = 1.26 R1

A2 = 2−1 3 = 0.79 2 A1

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The main challenge in formulating chemical products which are two-phase mixtures (e.g. printing ink or house paint) is to obtain a useful shelf-life for the product by slowing coalescence or flocculation. This act of “slowing” is often referred to as “imparting stability to the dispersion.” stabilize — to slow agglomeration by imparting repulsion between particles Much of the remainder of this course will deal with flocculation and how to slow it down. We begin our story of stability of dispersions by predicting how fast flocculation occurs if no stabilization is present. Kinetics of Rapid Brownian Flocculation In a stagnant fluid, particles undergo Brownian motion, which from time to time results in a collision between two particles. Let's suppose that every collision between two singlets results in the formation of a doublet — i.e. the particles stick together for some reason. Then the rate of flocculation is simply the rate of Brownian collisions. Reference: Kruyt, Colloid Science, Vol. 1, Elsevier, Amsterdam, 1952. pages 278-82

Thus there is a 21% reduction in area as a result of coalescence of two drops. This represents the driving force for for coalescence of emulsions. Attractive Forces Since solid/liquid interfaces also possess interfacial tension, a similar thermodynamic argument might be made for agglomeration of solid particles (also known as “flocculation”). However the rigidity of solid particles prevents them from coalescing or fusing together when they come in contact. So no significant area change occurs on flocculation. Yet in a microscope, the particles can clearly be seen to stick together. This is evidence for some long-range attractive force. This attraction is a macroscopic manifestation of the same intermolecular van der Waals forces which give rise to interfacial tension. Later we will see that van der Waals force between identical particles is always attractive and the force can have a range comparable to the particle size. ♥An analogy between this irresistible tendency of colloidal particles to flocculate and the inevitability of death of living organisms was made by Theo Overbeek (the “O” in DLVO theory), whose once morbidly described flocculation as “the decay of the corpse of the colloid.”


Smoluchowski♣ (1916) calculated this rate by computing the rate of diffusion of particles to a sphere fixed at the center of the coordinate system. The maximum driving force for diffusion is established by setting the concentration of particles to zero on the collision sphere having radius 2a:♦ at r=2a:

c=0

(201)

♣ Marian Smoluchowski (1872-1917): Polish physicist who modelled the kinetics of flocculation and the electrophoresis of charged colloidal particles. http://www.cyfronet.krakow.pl/~ncadamcz/MS.htm ♦ While particles do not actually “disappear” when they arrive at r=2a, this “perfect sink” condition leads mathematically to the diffusion-limited rate, which is what is desired.


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while specifying a nonzero bulk concentration for particles: as r→∞:

c → c∞

(202)

where c∞ is the number concentration of particles in the flocculating suspension. In spherical coordinates, if c(r,t) is the local concentration of particles ([=] particles/m3), then the radial flux is given by Fick's law as: dc ] particles −D [= J r ( r, t ) = dr m 2 -sec

(203)

Conservation of particles requires

∂c + ∂t

.J ∇ 

(

1 ∂ 2 r Jr r 2 ∂r

Thus we can also compute how the relative distance between the two changes with time:

)

=

After some initial transient, flocculation becomes quasi-steady and the time derivative can be neglected, leaving:

(

)

= 2D1t = 2D2t

= 0

d 2 r Jr = 0 dr

To show this, consider two Brownian particles which at time zero are located at the origin. At a later time t we know their mean square locations from Einstein's equation:

(204)

Substituting Fick's law for the radial flux:

= - 2 + But =0 because either particle is equally likely to be on either side of the origin. In other words, the probability of x2x1 being positive or negative is the same so that the average is zero. This leaves:

d  2 dc  r =0 dr  dr 

= + = 2(D1+D2)t

which can be easily integrated to obtain the general solution for the concentration profile. Integrating

For identical particles D1=D2 and the relative Brownian motion between two moving particles is twice as large as that when one is held stationary (D1=0).

( r ) c∞ 1 − 2a  c= r   The rate of collisions with a stationary target sphere is

( −4πr 2 J r ) r =2a = 8πDac∞

(205)

Although we have evaluated the left-hand side at r=2a (where the diffusing particles accumulate), the same result is obtained for any r since r2Jr is independent of r according to (204). If the target sphere is allowed to also undergo B.M. with the same diffusion coefficient as the other particles, what will be the collision rate? It turns out to be twice as large.♣

Generalizing (205), we replace D by 2D. For a single moving target particle: no. of collisions/time = 16πDac∞

(206)

To get the rate of collision per unit volume, we multiply by the number of targets per unit volume, c∞: no. collision/vol-time = 16πDac∞2 = b11 (207) This determines the rate at which singlets combine to form doublets. Not surprisingly then, flocculation is a second-order reaction (i.e. the rate is proportional to the square of the concentration). The resulting doublet may combine with a singlet or another doublet to form a triplet or quadruplet

♣ See 2014 Exam #1, Prob. 1d.



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Note that the above equation is exact when i=j. It's also approximately true if ai and aj are not very different, as can be seen in the following graph:

Quadruplets can also be formed by combining a triplet and a singlet: After substituting (210), (209) becomes:

bij = Of course, all of the above might then further combine to form larger aggregates. To describe the kinetics of aggregation we have to be able to predict the of collisions between each type of aggregate and any other. Smoluchowski computed the rate of these collisions by treating any size aggregate as a sphere. To compute the rate of collisions between spheres of radii ai and aj, we start with (207) and replace: 2a →ai+aj 2D →Di+Dj

8kT ci c j 3µ

(211)

With the rate of each type of encounter known, we can now try to predict how the concentration of any size of aggregate will evolve with time. A mass balance on any species gives:

{

rate of − rate of } {formation } {depletion}

rate of = accum

Let's take a particular species, say quintuplets (k=5). Particles of size 5 are formed by the combination of 1+4 or 2+3, where particles of size 5 are depleted by combining with particles of any size:

c∞2 →cicj

dc5 = b14 + b23 - (b51+b52+...+b5∞) dt

where ai+aj is known as the collision diameter of the pair. Thus (207) becomes:

The corresponding mass balance for species of any size is:

bij = 4π (Di+Dj)(ai+aj)cicj The the diffusion coefficient of any size sphere is inversely proportional to its radius. Bigger spheres have less mobility and a smaller diffusion coefficient:

= Di m= i kT

kT 6πµai

(208)

which is called the Stokes-Einstein equation [recall (42)a]. Substituting it into the expression for the collision rate, we have:

bij =

2kT 3µ

1 1   +  a + a j ci c j  ai a j  i  

(

)

(209)

The main approximation made by Smoluchowski is:

 1 1  +  ai a j 

  ai + a j ≈ 4  

(

)


(210)

k −1 ∞ dck 1 i = for k≥2:= bij − ∑ bik ∑ 2 dt=i 1 =i 1

(212)

j = k −i

The factor of 1/2 in front of the sum is to avoid double counting the same type of collision (for example, i,j = 1,k-1 and k-1,1). Substituting (211) into (212), we obtain a set of ordinary differential equations in the concentrations:   k −1 ∞  i=  dck 8kT   1 =  2 ∑ ci c j − ck ∑ ci  dt = 3µ  i 1 =i 1   j = k −i   

Now this set of O.D.E.'s can be solved analytically once we have specified some initial conditions. Let's suppose we start with all singlets of some known concentration: for k=1:

c1(0) = co


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ck(0) = 0

The particular solution to this initial-value problem is given by: ck ( t ) =

(

( tτ)

1+ t

k −1

τ

)

c k +1 0

(213)

where τ is called the flocculation time:

 0.163 s  3  µ m 3µ  = τ ≈ c0 4kTc0

(214)

when co is the initial number of singlets per unit volume. The number was computed using the viscosity of water (0.893 cp) at room temperature (298K). The total concentration of particles, irrespective of their size can also be easily computed: c≡

∞

∑ ck

k =1

c0 = t 1+ τ

(215)

This time dependence is characteristic of a secondorder reaction. τ represents the half life of the particle concentration: the total particle concentration, c, is reduced by flocculation to half its initial value at t=τ.

Experimental Verification: the Coulter Counter Since the availability of monodisperse latexes in the 1950's, there have been many experimental measurements of flocculation rates. One such study is that of Swift & Friedlander [J. Colloid Sci. 19, 621 (1964)], whose used a Coulter counter to measure the concentration of flocs of various sizes at different times. In the late 1940’s and early 1950’s, Wallace H. Coulter developed the Coulter Counter to rapidly count and size blood cells. A schematic of a modern version, now sold commercially by BeckmanCoulter,♣ is shown in Fig. 5-69 of Berg.

EXAMPLE: compute the flocculation time for a suspension of polystyrene latex particles having a diameter of 1 micron and a particle concentration of 2×108 particles/cm3. (These are the conditions of the Swift & Friedlander experiment described below.) Solution: (214) yields

τ = 815 sec

Notice that the value of the diffusion coefficient was not needed although we are describing the rate of Brownian flocculation. This is because in the theory leading to (214), it was the product of the diffusion coefficient and the collision diameter which determines the rate. Since the collision diameter equals the particle diameter and the diffusion coefficient is inversely proportional to diameter, the diameter cancels out. For the dilute concentration of this example, Brownian flocculation is quite slow which allows us to measure

Fig. 5-69. Schematic of a Coulter Counter. The glass tube is immersed in the beaker containing the particles to be counted. By withdrawing a plunger in the tube, fluid containing cells is drawn through a small aperture. The particle suspension must be sufficiently dilute so that only a single particle resides in the aperature at any time. ♣ http://www.beckmancoulter.com.



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While particles are being drawn through the aperature, the electrical current produced by applying a small voltage between two electrodes — one on each side of the aperature — is monitored. While an insulating particle resides in the aperature, the current decreases because the particle is blocking part of the path for current. The amount of the decrease in current is proportional to the volume of the particle (independent of its shape). Thus the number of “blips” of current represents the number of particles in the volume drawn into the glass tube; the magnitude of each blip give the size.

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force arises between particles which retards the aggregation of particles. Two common forces are: electrostatic repulsion between like charges and steric repulsion arising from adsorbed usually nonionic polymers. Nearly all particles, bubbles or drops in water are charged. We know this because when you apply an electric field across water containing a particle, bubble or drop, you can observe it to move. An example is the red blood cell experiment below.

Some are Swift & Friendland’s♠ results are plotted below.

The surface of the blood cell or other particle acquires charge by a number of mechanisms:

Eq. (215) can be linearized by taking the reciprocal of both sides: 1 1 1 = + t c ( t ) c0 c0 τ

—COOH → —COO– + H+ —NH2 + H+ → —NH3+

The slope of this plot should be

µm 1 1 = = 6.1 c0 τ  0.163 s s  3  µ m  

 adsorption of ions (e.g. ionic surfactants like SDS and hydrolyzed metal cations like AlOH+2)  dissolution of ions from ionic solids (e.g. AgI)  dissociation of acidic or basic sites on interface:

3

Two particles with like charge will experience electrostatic repulsion. If large enough, this repulsion can stabilize a dispersion. Another type of repulsion is steric repulsion:

which should be independent of particle concentration or size. The experimentally determined slope is low by a factor of two, suggesting that the ‘rapid’ rate of flocculation measured is slower than expected. Later we will show that this occurs because the diffusion coefficient of particles is reduced when the two particles are close to one another.

Lecture #16 begins here Repulsive Forces Stabilizing a dispersion usually means formulating the mixture so that some long-range repulsive ♠ D.L. Swift and S.K. Friedlander, “The Coagulation of Hydrosols by Brownian Motion and Laminar Shear Flow,” J. Colloid Interface Sci. 19, 621-647 (1964).


The white circles in the schematic above represent bare colloidal particles and the layer of thickness δ represents macromolecules attached to the surface with their soluble tails protruding from the surface.


When two particles approach one another with a separation distance S0 < 2δ, they experience mutual repulsion owing to two mechanisms: 1) a higher osmotic pressure in the region of overlap of the adsorbed layers (arising from an increase in the polymer segment density), and 2) a decrease in entropy owing to a reduction in the degrees of freedom for the polymer chains. While the polymer chains might be covalently attached to sites on the particle surface, more often the polymers are physically adsorbed. By choosing a diblock polymer, with one block soluble in the solvent and the other block insoluble, the insoluble block (called the “anchor”) adsorbs on the surface while the soluble block (called the “stabilizing moiety”) extends into the solvent. Some typical diblock polymers are listed below:

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1000

Stability Ratio, W

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CCC 10

1 0.01

0.1

1

NaCl Concentration (mol/dm3)

Stability ratio is the ratio of the Smoluchowski rate of flocculation to the actual rate:

W≡

Smoluchowski's rate actual rate

(216)

which is usually greater than unity. Notice that above a certain NaCl concentration called the critical coagulation concentration or CCC, the stability ratio becomes independent of salt concentration. At low salt concentrations, the stability ratio can easily reach 1000: meaning the rate is only 0.1% of the rate predicted by Smoluchowski’s equation. It has long be recognized that electrostatic repulsion between like charged particles can reduce the rate of flocculation to practically zero — thus causing the stability ratio to rise. Slow Brownian Flocculation Repulsive forces between the particles can profoundly slow the rate of flocculation. This can be seen in the sketch below showing experimental data taken by Robert Ofoli (PhD Thesis, CMU, 1994) on sulfonated (negatively charged) polystyrene latex particles having a diameter of 0.154 µm using smallangle light scattering:

When W  1, the colloid is said to be kinetically stable. This means that the rate of flocculation is so slow that the colloid has a useful shelf-life, which is almost as good as if it were thermodynamically stable (which might correspond to W = ∞). Now let’s refine the theory to include the effects of interparticle interactions. Forces of interaction between particles affect the rate of aggregation by causing the particles to migrate in the force field at some terminal velocity. This represents another transport mechanism which acts in parallel with diffusion. The Russian scientist N.A. Fuchs (1934) modified Smoluchowski’s analysis to include migration in the force field. (203) becomes: dc Jr = −D + mFc  dr  migration diffusion

where m is the hydrodynamic mobility (terminal velocity per unit applied force) and F is the interparticle force (F>0 for repulsion).



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The mobility m is the reciprocal of the friction coefficient f, which in turn is related to the diffusion coefficient D by (41). The force F can be written as the derivative of the potential energy φ; thus we make the following substitutions:

m=

D kT

and

F= −

dφ dr

Then the flux becomes dc D  d φ  −D + − Jr = c dr kT  dr   dc c d φ  = −D  +   dr kT dr 

(217)

Substituting (217) into the continuity equation (204): d 2 d   dc c d φ   + r J r = −r 2 D  0  = dr dr   dr kT dr  

(

)

Boundary conditions are provided by (201) and (202) as r®∞:

c ® c∞

at r=2a:

c=0

Fig. 7-10 from Berg. Jar tests for a sol of arsenic trisulfide (As2S3), a yellow pigment. The numbers on each couvette denote the concentration (in mM) of a 1-1 electrolyte. Clearly the CCC lies between 60 and 70 mM. The jars can now be refilled with different electrolyte concentrations (between 60 and 70) and the experiments repeated to narrow the interval in which the CCC lies. Below is a table of CCC values determined for this same sol.

The particular solution leads to the following expression for the stability ratio defined by (216):

exp φ ( r ) kT  dr r2 2a ∞

W = 2a ∫

(218)

If we substitute φ(r) = 0, this reduces to W = 1, which means the rate of flocculation is just what Smoluchowski calculated. On the other hand, if φ(r) >> kT (strong repulsion), then the exponential in the integrand can become quite large, causing W >> 1. Effect of Electrolyte Concentration Determining the CCC by measuring the kinetics of flocculation (as Ofoli did above) requires considerable care, patience and a small-angle light scattering apparatus (including a laminar-flow hood to avoid the spurious effects of dust). A much easier technique to see qualitatively the effect of electrolyte concentration and to quantify the CCC is the jar test. Jars filled with the colloid are prepared with the same colloid concentration but different electrolyte concentrations. After thoroughly mixing, they are allowed to settle for an hour or so. Six jars (actually they are spectrophotometer couvettes), marked with their electrolyte concentration (in mM) are shown below:


“Potential-Determining” vs. “Indifferent” In trying to understand what determines the concentration required to cause rapid flocculation, it is important to distinguish two general classes of electrolytes. Some ions directly impact the surface charge density (or the related surface potential). These are called potential-determining ions — an ion species whose concentration determines the surface charge density There is often one concentration of a “potential determining” ion at which the surface charge density vanishes. Then rapid flocculation will ensue. At either higher or lower concentration of a potentialdetermining ion, a nonzero charge (either positive or negative) will occur and the dispersion might be stable (slow flocculation).


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The second class of ions are indifferent electrolytes — a salt, acid or base whose ions are not “potential determining” These ions are not directly involved the surface charging mechanism and the surface charge density is often insensitive to their concentration. Nonetheless, if their concentration is high enough (above the CCC reported in Table 7-2), they cause Debye screening of the electrostatic repulsion, which also leads to rapid flocculation. PDI Example #1: H+ and OH– The most common example of a potentialdetermining ion is H+ (or OH–) for solid surfaces bearing acidic or basic sites. The surface pH will determine the degree of dissociation of acidic or basic sites, thereby determining the surface charge. Suppose the surface contains both carboxyl and amine groups. Then the following surface reactions might occur

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concentrations: [COO–], [COOH], [NH2] and [NH3+]. Once the solution is obtained, the surface charge density can be calculated from

(

= σ e  NH 3+  − COO − 

)

(223)

Increasing [H+] (or decreasing the pH) will shift both of these reactions: it simultaneously decreases the number of negative carboxylate (–COO-) groups and increases the number of positive amines (–NH3+). Example: a red blood cell has both acidic and basic sites on its surface with the following density and dissociation constant: acidic:

Na = 2×1017 sites/m2 and Ka = 10–3 M

basic:

Nb = 1×1017 sites/m2 and Kb = 10–10 M

Plot the degree of dissociation and net surface charge density versus pH. Solution: For each reaction, we define the degree of dissociation (denoted α) as the fraction of the total sites which are charged:

αa ≡

COO −  Na

and

 NH +  3  αb ≡  Nb

These definitions can be used to express the number density of charged sites: The “mass-action” law for these reactions leads to the the following relations between the concentrations at equilibrium:

COO −   H +  = Ka [COOH ] and

[ NH 2 ]  H +  = K  NH +  3  

(219)

COO −  = α N a a

and

 NH +  = α N b b  3

With the help of (221) and (222) we can also express the number density of uncharged sites:

[COOH ]= (1 − α a ) N a and

[ NH 2 ]= (1 − αb ) Nb

Then (219) and (220) can be written as b

(220)

where [COO–], [COOH], [NH2] and [NH3+] have units of number of sites per unit area, [H+], Ka and Kb have units of number of ions per unit volume. The total number of sites (whether dissociated or not) remains fixed as the degree of dissociation changes: COO −  + [COOH ] = (221) Na

Nb [ NH 2 ] +  NH 3+  =

α a  H +  = Ka (1 − α a )

and

(1 − αb )  H +  = K αb

b

Solving these for the α’s:

Ka αa =  H +  + K a

and

 H +  αb =  H +  + K

b

For the particular values of the K’s given and various pH’s, the α's are given in the following plot:

(222)

Treating Ka, Kb, Na and Nb as known, then for any given value of [H+], (219), (220), (221) and (222) constitute four equations in the four unknown surface



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determining ion at which the surface charge vanishes In the example above, the PZC = pKa. This is a quirk of this particular example. More generally, the PZC for surfaces having acidic and basic sites can depend on both pKa and pKb as well as both Na and Nb.

where

pH = − log10  H + 

and

pKi = − log10 ( Ki )

provided both [H+] and Ki are expressed in mol/dm3. Finally we calculate the net surface charge density σ from (223):

σ= e ( αb Nb − α a N a )

Comment: In the example above, we have focussed on pH – or the corresponding value of [H+] – as what determines the charge (or the associated potential)♣ on the surface. Thus we say that H+ is the “potentialdetermining ion” in this case. We could just as easily say that OH– is the potential-determining ion. This is because the concentrations of the two ions are directly related by the dissociation equilibria of water: Kw

H 2O ( l )  H + ( aq ) + OH − ( aq )

and the mass-action law requires

 H +  OH −     = K′ w H O [ 2 ]

which is shown in the following plot:

e Nb e Na e Na e Nb

Since the degree of dissociation of water is always extremely small (even at pH = 0, [H+] = 1 M which is small compared to [H2O] = 55 M), the mass action law for water dissociation is usually written more simply as −  H +  OH =   

Kw [ H 2O ] K w′ ≡=

10−14 M 2

Taking minus the log10 of both sides: Comment #1: While this example (and the next one) helps to understand the qualitative changes in surface charge with pH, the results should not be taken too literally. We have in fact ignored a number of complications including nonideality of electrolyte solutions and physical adsorption of ions – both of which are often important. Comment #2: keep in mind, the results are given for a particular surface pH (which is the pH of the solution immediately next to the charged surface), which is only equal to the bulk pH when the surface is uncharged. In Charge Regulation on page 142 we will relate the two pH’s. If (as in this example) both acidic and basic sites are present, there is often one pH (or [H+]) at which the total surface charge is zero. This concentration is called: point of zero charge (PZC) — concentration (or –log10 of concentration) of a potentialCopyright© 2016 by Dennis C. Prieve

pH + pOH = 14 Because of this constraint relating the two concentrations, either H+ or OH– can be considered to be the potential-determining ion in this example. PDI Example #2: Oxides The charge-forming sites on the surface of many oxides (e.g. silica, alumina and titania) are usually hydroxides -MOH (where M is the metal atom: Si, Al or Ti) which can either accept or donate a proton to become positively or negatively charged:

♣ In the Special Case #1: Small Potentials beginning below on page 134, we will derive the relationship between surface charge and surface potential. At least near the PZC, the two are directly proportional and have the same sign.


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basic:

Kb = 10–10 M

Plot the degree of dissociation and net surface charge density versus pH. For this reason –MOH is said to be amphoteric. As in Example #1, H+ and OH– are the potentialdetermining ions. The “mass-action” law for these reactions leads to the the following relations between the concentrations at equilibrium:

 MO −   H +  = Ka [ MOH ]

[ MOH ]  H + 

and

 MOH +  2 

Solution: The fraction of total sites which exist in each of the three states can be calculated directly from (227) to (229):

(224)

= Kb

(225)

As in (219) and (220), we have defined the equilibrium constants Ka and Kb so they both have units of

concentration.♥ The total number of surface sites (positive, negative and neutral) is denoted as N:

 MO −  + [ MOH ] +  MOH +  = (226) 2 N  Treating Ka, Kb and N as known, then for any given value of [H+], (224), (225) and (226) constitute three equations in the three unknown surface concentrations [MO–], [MOH] and [MOH2+] whose solution is

 MO −  K a Kb = 2 N  H +  +  H +  K + K K b a b

[ MOH ] N

and

=

 H +  Kb 2

 H +  +  H +  K + K K b a b

(227)

By plotting the above data on log-log axes, we are able to see that at pH = pKa = 3, we have [MO-] = [MOH] as predicted by (224). However at this low pH virtually all of the sites exist as MOH2+. Similarly, at pH = pKb = 10, we have [MOH] = [MOH2+] as predicted by (225) but virtually all of the sites exist as MO-. Knowing the number of each type of site, we can calculate the net surface charge from (230):

(228)

2  MOH +   H +  2  = (229) 2 N  H +  +  H +  K + K K b a b

Once the surface concentrations are obtained, the surface charge density can be calculated from

(

= σ e  MOH 2+  −  MO − 

)

(230)

Example: an oxide has amphoteric sites on its surface with the following density and dissociation constants: N = 2×1017 sites/m2 acidic:

Ka =

10–3

M

The most interesting result is the pH at which the surface charge equals zero or [MO-] =[MOH2+]. Equating (227) to (229) and solving for [H+] yields

 H + 

PZC

= K a Kb

or taking minus the log10 of both sides: ♥ This is not a universally accepted convention.



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= pH PZC

pK a + pKb = 6.5 2

Lyklema lists some PZC’s for oxides in Appendix 3 of Volume 2:♦ Oxide pHPZC silica 2.1 – 3.5 titania 2.9 – 6.6 alumina 8.1 – 9.7 A range is given above, because the PZC depends on the crystal structure of the particular sample as well as how the sample is prepared. In addition to Lyklema’s appendix, there is an entire book on this subject: M. Kosmulski, Surface Charging and Points of Zero Charge, Vol. 145 of Surfactant Science Series, CRC Press, 2009. PDI Example #3: Ions from Ionic Solids

Although either ion can be chosen to be “potential determining”, let’s pick Ag+. Thus the point of zero charge is [Ag+]PZC = 3.0x10-6M or

pAgPZC = 5.52

Colloidal titration further reveals that [Ag+] < 3.0x10-6M → negative charge on solid

The dependence of surface potential ψ0 on [Ag+] can be quantified using the principles of phase equilibria. At equilibrium, the electrochemical potential of silver ions in the solid must equal the electrochemical potential of silver ions in solution Ag +

whose equilibrium is characterized by the solubility product. For AgI, this product is:♠ (231)

If there are no other ions present, then dissolution will produce equal concentrations of the two ions and dissolution will stop when the concentrations become K sp = 8.7×10-9 M

On the other hand, if the solution is prepared by mixing AgNO3 and NaI (for example),♣ [Ag+] does not have to equal [I-], although the product of these concentrations must still satisfy (231) if solid AgI is present. Colloidal titrations (see p487 of Berg) can reveal the sign of surface charge density (or surface potential). To have no charge on the AgI crystal ♦ J. Lyklema, Fundamentals of Interface and Colloid Science, Vol. II, Academic Press, 1995. ♠ source of values for solubility product and PZC for AgI is H&R, 3rd ed., p503. See also Berg, p458. ♣ Both AgNO and NaI are quite soluble in water, whereas 3 AgI is only sparingly soluble.


[Ag+][I-] = (3.0x10-6M)(2.5x10-11M) = Ksp

µs

K sp

AgI ( s )  Ag + ( aq ) + I − ( aq )

[Ag+] = [I-] =

surface, it turns out that we need to have a greater concentration of silver ions compared to the halide ions. In particular, the concentrations which lead to zero charge and also satisfy the solubility product are:

[Ag+] > 3.0x10-6M → positive charge on solid

Another important class of solids having potential determining ions are ionic crystals; for example, silver halides like AgI which is the light-sensitive component of photographic film. Dissolution equilibrium can be thought of as the following reaction:

Ksp ≡ [Ag+][I-] = 7.5x10-17M2

Spring, 2016

= µ aq + Ag

Expanding the electrochemical potentials contributions from chemical and electric parts:

into

µi0, s + zi eψ 0 = µi0,aq + kT ln ci ( x )  + zi eψ ( x ) (232)   kT ln ci∞

where the terms having a superscript “0” are the chemical potentials at some reference state for the material and x is the distance of the ion from the surface of the solid. While ci(x) and ψ(x) are not directly measurible, the sum of these two xdependent functions must be independent of x at equilibrium (recall that the criteria for phase equilibrium within a single phase is spatially uniform electrochemical potential). If we select the uniform potential of the bulk solution [i.e. ψ(∞)] far from the charged interface as our reference state for potential, then ψ(∞)=0 and the sum of the two terms in (232) becomes kTlnci∞, where ci∞ is the bulk concentration of component i, which is known. When ci∞ = ci,pzc the surface potential vanishes (ψ0 = 0) and (232) becomes

µi0, s = µi0,aq + kT ln ci, pzc

(233)

Subtracting (233) from (232), then solving for the surface potential, yields


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109

c kT ψ 0 = ln i zi e ci , pzc

(234)

which is called Nernst’s equation. Note that choosing ci = ci,pzc yields ψ0 = 0 at which the colloidal dispersion should be unstable. Choosing ci < ci,pzc or ci > ci,pzc yields either a positive or a negative ψ0, respectively. If large enough in absolute magnitude, repulsion of like-charged particles might be strong enough to stabilize the dispersion.

Spring, 2016

Schulze-Hardy Rule At least as early as 1882,♦ H. Schulze observed that when other salts (not composed of potentialdetermining ions) are added to an electrostatically stabilized dispersion of particles, they too can cause flocculation. Unlike potential determining ions, which destabilize only near a particular concentration, indifferent electrolyte cause rapid flocculation for all concentrations above some minimum value called the

Example: Taking the PZC to be pAg = 5.52 calculate the surface potential of AgI crystals as a function of the [Ag+].

critical coagulation concentration (CCC) — minimum concentration of “indifferent” electrolyte required to destabilize a dispersion

Solution: Converting the natural log appearing in (234) into log-base 10 and substitute zi = +1:

The CCC was measured for a various indifferent electrolytes (see Table 7-2) and was found to be about the same for all monovalent salts (KCl, NaCl, etc.). However, divalent salts have a much lower CCC, while trivalent salt have a even lower CCC:

 Ag +  kT  2.303 log  ψ0 = PZC e +  Ag    = 2.303

kT e

CCC = 100:1.6:0.05 for z = 1:2:3

 + +  log  Ag  − log  Ag  

(

kT pAg PZC − pAg = 2.303 e  

)

PZC

  

59.1 mV

which is plotted below:

Unlike the two previous examples, Nernst equation yields the surface potential rather than the surface charge density. Near the PZC, the two quantities are directly proportional and have the same sign, but the proportionality constant depends the concentration of indifferent electrolyte which is usually also present.

This dependence on the valence of the salt (more specifically, on the charge number of the counterions, for unsymmetric electrolytes) is called the: Schulze-Hardy Rule -- the CCC depends strongly on the charge number of the counterion This was observed before 1900. It was generally assumed that even indifferent electrolytes adsorbed on the surface and reduced the charge, thereby destabilizing the sol. Then makes sense in that, the higher the valance, the less electrolyte is needed to neutralize the surface charge. This was called the Freundlich adsorption theory after H. Freundlich who suggested it in 1903. if ions adsorbed:

CCC ∝ z-1

observed:

CCC ∝ z-6

We now know that the CCC is associated with Debye screening of the electrostatic interactions between particles (due to the compression of the double layer by added salt) rather than changes in the charge on the surfaces. By the end of the 1930's, Freundlich's adsorption theory was abandoned by colloid scientists in favor of Debye screening.

♦ H. Schulze, J. Prakt. Chem. 25:431 (1882) and 27:320 (1883); W.B. Hardy, Proc. Royal Soc., 66:110 (1900)



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110

Spring, 2016

whose projection onto the xy plane lies within one collision diameter of the center of our target, will eventually collide with our target.

Lecture #17 begins here Rapid Flocculation by Shear Reference: H.R. Kruyt, Colloid Science, Vol. 1, Elsevier, Amsterdam, 1952. pages 289-92. Besides diffusion, another familiar mechanism for transport is convection. Suppose now, that the particles are completely non-Brownian, but that the fluid in which the particles are dispersed moves. Of course, if the velocity of the fluid were uniform everywhere, the relative position of the particles would remain unchanged -- no flocculation would occur. Thus it is necessary for the fluid velocity to vary from point to point so that the particles can move relative to one another. The first analysis of this problem was again done by Smoluchowski (1916):

End View (xy plane) In other words, if particle center is located at: (x,y,z) and yz>a:

φ A =− φ A =−

Aa 12 h

16 A  a  9  h 

6

Notice that the attractive energy decays like h–1 when h ≪ a and like h–6 when h ≫ a . The more rapid decay at large separations has the same exponent as between two atoms [see (256)]. The transition between these two values of the exponent is gradual as is shown in the following plot

m

where dM is the mass of the volume element dV, m is the mass of one atom and ρ/m is the no. of atoms/vol.

ρ φ A =−Cvdw   m

2

1

∫ ∫ r 6 dV1dV2

V1 V2

after factoring the constants Cvdw, ρ and m out of the integrals. Then, after performing the integration over the two volumes for the case of two spheres of radius a, we obtain:

A 1 1 s( s + 2)  φ A (h) = −  + + 2ln  2 12  s( s + 2) ( s + 1) ( s + 1)2  where

s=

h 2a

2

and

 πρ  A =   Cvdw [ = ] J m


(257)

This change in exponent is due to geometry only (retardation is not included in this result).


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The results for several other geometries are summarized in the following table (taken from Fig. 4.4.3 in Hunter, Foundations, Vol I).

Instead of treating every atom as a discrete and independent oscillator, Lifshitz theory treats the three materials as continua. Using Lifshitz theory, the interaction energy per unit area between two half spaces, composed of materials 1 and 2, separated by thickness l of material 3, is A132 ( l ) φ A (l ) = − 12πl 2

This is also the expression we would get from Hamaker's superposition theory for this geometry. The main difference in Lifshitz's theory is that A132 depends on properties of the continuum — rather than properties of the individual atoms — and A132 will also depend on the separation distance l. Here are the general expressions which can be used to evaluate the Hamaker constant: ∞

∞ 3 A132 = − kT ∑ ′ ∫ x ln 1 − ∆13 ∆ 23e- x  2 n =0

(

+ ln 1 − ∆13 ∆ 23e

= ∆ jk

(

rn

-x

) dx

Hamaker’s superposition of atomic or molecular interactions gives the correct dependence on the geometry but the proportionality constant A is not accurately predicted by (257) – even when the two bodies are separated by vacuum. When another media separates the bodies, their interaction is more complicated: the net interaction might even be repulsive. While repulsive interactions can be predicted in the framework of Hamaker’s superposition theory, the assumption of pairwise additivity of intermolecular forces is inherently flawed: the interaction involves the simultaneous interaction between many molecules. Lifshitz's Continuum Theory Reference: §4.7 of Hunter, Foundations of Colloid Science, Vol. I, Clarendon Press, Oxford, 1987.


2 sk=

ξn =

(258)

ε j sk − ε k s j sk − s j = ∆ jk ε j sk + ε k s j sk + s j 2

Figure 14

)

 2ξ l  x +  n  ( ε k − ε3 )  c  2

2πnkT h

rn=

2ξ n l ε 3 c

ε k = ε k ( iξ n )

where c = 3.00×108 m/s is the speed of light in vacuum and h = 1.0545×10–34 J-s is Planck’s constant divided by 2π. The prime (' ) on the sum denotes that the first term is to be multiplied by 1/2. The material properties enter this equation through εk(iξn). εkε0 is the electric permittivity of the material (k=1,2,3), where ε0 = 8.854×10–12 F/m is the electric permittivity of vacuum. εk itself is dimensionless in these equations♣ and is sometimes called the dielectric constant of the material. ♣ Elsewhere in these Notes ε has units of F/m and is called the permittivity rather than the dielectric constant.


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εk is evaluated at a purely imaginary frequency iξn. This might appear to be quite artificial, but it is a consequence of extending the definition of functions of a real quantity (e.g. the real-valued frequency ω) to complex values of the argument. In mathematics, this extension of the definition of a function is called analytic continuation. Cauchy’s Residue Theorem and the Kramers-Kronig Relation are two powerful theorems which are employed to evaluate contour integrals on the complex plane of frequencies. The resulting εk(iξn) turns out [see (271) on page 128] to be a real-valued function which is very well behaved (decays monotonically and continuously to zero as ξn → ∞). By contrast the same function evaluated at purely real frequencies is highly singular and nonmonotonic in the vicinity of each frequency where the material absorbs light.

Below♥ are results showing how the Hamaker constant depends on separation distance for two polystyrene half-spaces, separated by water, with 0.1 M NaCl or without added salt.

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Spring, 2016

∞

∞ 3 0 0 -x A132 ( 0 ) = − kT ∑' ∫ x ln 1 − ∆13 ∆ 23e dx ♣ 2 n =0 0   

(

(

)

)

0 0 0 0 − Li3 ∆13 ∆ 23 ≈−∆13 ∆ 23

ε j ( iξ n ) − ε k ( iξ n ) ∆ 0jk ( ξn ) = ε j ( iξ n ) + ε k ( iξ n )

where

Since the ∆0’s are now constants (i.e. they no longer depend on dummy integration variable x), we can evaluate the integral once and for all. When 0 0 ∆13 ∆ 23 ≪1, we can approximate the logarithm:

(

)

0 0 -x 0 0 -x ln 1 − ∆13 ∆ 23e ≈ −∆13 ∆ 23e

0 0 ∆ 23 outside the integral, we obtain Factoring ∆13 ∞

∫

(

1  

∞

)

0 0 -x 0 0 x ln 1 − ∆13 ∆ 23e dx ≈ −∆13 ∆ 23

0

∫

xe- x dx

0 0 0 = −∆13 ∆ 23

The following simple result is a good approximation (max error is 20%):

3 ∞ 0 0 A132 ( 0 ) ≈ kT ∑'∆13 ∆ 23 2 n =0 The table below summarizes the contributions of various terms to the sum

Figure 15: Hamaker constant for two polystyrene halfspaces (separated either by pure water or a solution of 0.1 M NaCl) calculated from Lifshitz theory (258).

The reduction in Hamaker constant with separation distance is the manifestation of retardation. The greater decay to zero in salt water occurs because of Debye screening of the zero-frequency term. Owing to the complexity of these expressions, it’s hard to see how A depends on the properties of the materials. The expressions simplify somewhat if we look at the unretarded limit: in other words, if we let l → 0. Then rn = 0 and sk = x leading to ∆ jk = 0.

Material Properties and Frequency Dependence Reference: §2.2 of Hunter, Foundations of Colloid Science, Vol. I, Clarendon Press, Oxford, 1987.

Then (258) for the Hamaker constant A simplifies to

The only material property which appears in the previous section is the dielectric permittivity ε, which is the continuum property (of the medium separating

♥ taken from Prieve & Russel, J. Colloid Interface Sci.125, 1 (1988). A reprint is available on Blackboard.

♣ Li (z) is the polylog function with index n. Li (z) is the n 3 trilog function. See http://en.wikipedia.org/wiki/Polylogarithm.



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125

the point charges) which appears in Coulomb’s law [see (76) on page 30]. This property reflects how the medium becomes polarized by an electric field. The analogous property for atoms and molecules is the polarizability α. ε and α for the same material are related by the Clausius Mossoti relation (see page 30): α=

3ε0 ε − ε0 N ε + 2ε 0

Spring, 2016 The general solution of this equation for ω ≠ ωo is given by: x(t) = Asin(ωot) + Bcos(ωot) + Ccos(ωt) (261) where

qE0 C (ω) = m(ωo2 − ω2 )

(78)

It turns out that the ability to polarize a material by applying an electric field depends on the frequency of that electric field. The largest polarization occurs at zero frequency. As the frequency becomes very high, the molecules don’t have time to respond. The following simple harmonic oscillator model provides insights into the frequency dependence.

k m

ωo =

and

(262)

is called the natural frequency (a.k.a. resonance frequency). This natural frequency is the frequency with which the electron would oscillate back and forth if there was no electric field forcing it. This is the frequency of oscillation which would result if you initially displaced the spring away from equilibrium and then just let it go without touching the spring again.

Harmonic Oscillator Model The direct proportionality between the dipole moment of a H atom and the electric field generating it [see (74) on page 30] is very much analogous to a linear Hookean spring whose displacement from its equilibrium position is proportional to the applied force. Let’s now explore this analogy a little further to see what it says about the dynamics for an oscillating force.

qE = 0

x(t)

t

q F = qE

∆t =

x(t) Let

k = spring const restoring force = kx

Suppose the electron has mass m. Then the dynamics of the electron are governed by Newton's law of the motion: ma = ΣF m

d2x = qE − kx dt 2

(259)

(260) into (259) gives: m

d2x dt 2

+= kx qE0 cos(ωt )


Now let’s return to the response described by (261) when a sinusoidal force is applied with a frequency different from the natural one. Let's focus attention on the particular solution, because the values of the integration constants A and B will depend on the initial conditions, whereas the particular solution is more universal: it is independent of the initial conditions. Indeed, there exists one set of initial conditions for which A=B=0. Special Case: ω=0 (E = E0) In this case the particular solution is a constant:

Now let's try to predict the response of the electron if the electric field oscillates with time: E(t) = E0cos(ωt)

2π ωo

(260)

qE0 qE0 = k mωo2

C= (0)

We can use this as a gauge to judge the amplitude of the response as a function of frequency. (262) can be rewritten as

C ( ω) C (0)

=

ωo2 ωo2 − ω2

(263)


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bg Cb0g

Cω

Spring, 2016

The dipole moment induced by the oscillating electric field is proportional to the displacement x(t) of the electron cloud:

ω = ωo

p(t) = qx(t) = αE(t)

1

ω

(265)

Substituting (261) with A=B=0:

q2 α ( ω) = m(ωo2 -ω2 )

(266)

Damped Harmonic Oscillator Model Notice that as the frequency increases from zero to the natural frequency the amplitude of the response increases to infinity, then reverses sign. Above the natural frequency, the amplitude decays to zero as the frequency becomes much greater than the natural value. Driving a system at it's natural frequency is clearly special and mathematically singular. Let's see if we can understand what is happening. It turns out that if ω=ω0, (261) is not the general solution. Instead, the general solution is:

There is one additional feature we can add to this simple-harmonic oscillator model to make it more realistic: friction. Any real mechanical oscillator would have irreversible losses of energy owing to viscous drag on the object moving back and forth. Likewise, an oscillating electon cloud around a hydrogen atom radiates energy in all direction as scattered light: this radiated energy also represents an irreversible loss of energy. We can model the effect of such losses by adding a viscous drag force to (259) :

d2x dx m = qE ( t ) − kx  − f 2  dt dt  spring  external

x(t) = Asin(ω0t) + Bcos(ω0t) + Dt sin(ω0t) (264) D = qE0ω0/2k

where

inertia

x(t)

f = 6πµa

where

t

viscous drag

restoring force

force

A particular solution (not the general solution) to this refined model for E(t) = E0 cosωt is: x(t) = C(ω) cos ωt + S(ω) sin ωt

( )  qE where C ( ω) = 0 k C 0

Notice that this particular solution oscillates, but the amplitude grows with time without bound. There is a simple physical explanation. Suppose the system is already oscillating at its natural frequency when we start to drive it at the same frequency. Even without any driving force, the system would oscillate with constant amplitude. With a oscillating driving force of the same frequency, we pull the electron further from the center with each cycle than it would otherwise move. In effect, the electron cloud is absorbing energy. Indeed this natural frequency corresponds to an absorption peak on a spectrogram. Of course, atoms and molecules generally possess many electrons and they also display multiple absorption peaks in their spectra.


(267)

and

qE S ( ω) = 0 k

(ω

− ω2 ω02

− ω2

)

2 0

(ω

2 0

(268)

)

2

2

 f  +   ω2 m

f ωω02 m

(ω

2 0

− ω2

)

2

2

 f  +   ω2 m

Notice that for f = 0, this equation for C(ω) reduces to (263), which was singular at ω = ω0. But when f ≠ 0, neither of these expressions is singular. The plot below was prepared by choosing f /m = 0.3ω0. C(ω) and S(ω) have both been normalized by C(0) = qE0/k.


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Spring, 2016

θ ei= cos θ + i sin θ

1 4

C( ω) 2

0

0

S( ω)

−2

where θ is any real number (either positive or negative). Notice that by applying Euler’s formula, we find that the real part of (269) is just the cosine we had before. In effect, all we have done is to add an imaginary contribution [-i sinωt], which we will drop by taking the real part of the final answer. But the mathematical operations needed to obtain the final answer are much easier. The particular solution to (267) becomes:♠

−4 0.1

1

ω

C(w) S(w) undamped

= x (t ) 10

ω 0

(

)

re

The red and blue curves in the figure above show how the normalized C and S depend on ω for this one particular value of f. Notice that damping significantly attenuates C compared to its singular behavior predicted from (263) (shown above as black curve). Another very important difference compared to (261) is the appearance of the sine term in (268). The sum of a sine and a cosine of the same frequency produces a cosine wave with a different amplitude and shifted in phase, as is illustrated in the following figure. 2

(270)

iθ

If we took the real parts of (269) and (270), we would obtain (260) and (268).♦ The real advantage of using the complex exponential form of the time dependence in (269) is that 1) differentiation w.r.t. t is easier and 2) the application of trig identities to calculate the phase angle is replaced by a conversion of the final complex number into polar form. Once r and θ are calculated, taking the real part of (270) is easy: = x ( t ) Re {re−i

} r cos ( ωt + θ ) =

( ωt +θ )

which should be compared to (269). Clearly θ is the phase angle for oscillations in x(t) relative to oscillations in E(t). Complex Polarizability and Permittivity Using (265) and (270) we can deduce the complex polarizability

1

q2 α ( ω) = m ω02 − ω2 − iωf

0

(

−1

−2

qE0 ( ) = e−iωt re−i ωt +θ 2 2 m ω0 − ω − iωf  

0

1

2

cos(wt) cos(wt) + 0.3sin(wt) 0 cos(wt) + 0.5sin(wt) cos(wt) + sin(wt)

3

ωt

♠A positive exponent can also be used in (269) in place of the negative exponent; then the complex conjugate of (270) is obtained (i is everwhere replaced by –i).

π

All of this behavior can be represented more compactly using complex numbers. For example, instead of representing the driving force by a real function [E(t) = E0 cosωt], we represent it as an complex exponential function E(t) = E0 exp(–iωt)

♦Note that ω in (270) is the natural frequency for 0 undamped oscillations. In other words ω0 is given by the equation following (262). This is not the natural frequency for damped oscillations, which is smaller and given by 4km − f 2 ωdamped = 0 4m 2

(269)

The real and imaginary parts of complex exponentials can be deduced with the help of Euler’s formula: Copyright© 2016 by Dennis C. Prieve

)

But this natural frequency for undamped oscillations only shows up as part of the transient solution before steady oscillations commence.


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Using the mathematical principles of analytic continuation, the definition of this function can be extended to complex values of its argument ω. In particular, notice that for purely imaginary frequencies (i.e. ω = iξ), the polarizability is a purely real function which decays monotonically with an increase in imaginary frequency iξ: 2

q α ( iξ ) = 2 m ω0 + ξ2 + ξf

(

)

In particular, the singularity at ω = ω0 has disappeared (provide ξ is a positive real number). The plot below compares α(ω) and α(iω) for the undamped case (i.e. f=0). 3

1

2

1

0

Spring, 2016

addition, condensed media (e.g. liquid water) are not sufficiently dilute so that this approximation to the Clausius Mossoti relation applies. Nonetheless, the permittivity spectra of nearly any material can be fit to a semi-empirical function having the following form:

gj ε ( iξ ) g h 1 0 + 0 +∑ =+ (272) ε0 ξ2 f0 + ξ j ξ2 + f j ξ + ω2j where the first term following unity is the contribution from free electrons (present only in metals), the second term in the contribution from the orientation of a permanent dipole (present only for polar liquids like water) and the remaining terms are contributions from absorptions peaks j in the IR or UV. For example, Parsegian & Weiss♦ fit experimental results for water using the permanent-dipole term, 5 IR peaks and 6 UV peaks (a total of 35 fitting constants). Their results for ε(iξ)/ε0 is shown as the ×’s in the figure below:

0

−1

−2 0.1

ω

1

ω

alpha(w) alpha( i*w)

10

0

For this reason, imaginary frequencies offer a convenient alternative for curve fitting of experimental results, which will be summarized below. The Kramers-Kronig relationship of analytic-continuation theory allows experimental data taken at real frequencies to be mapped into purely imaginary frequencies.♣ For gases (which have ε/ε0 ≈ 1), the Clausius Mossoti relation (78) gives

Nq 2

ε ( iξ ) − ε 0 = m ωo2 + ξ2 + ξf

(

)

(271)

These expressions for α(ω) and ε(ω) (which display a single absorption peak at ω = ω0) were developed for a simple atom like H. More generally, atoms and molecules possess many electrons and they also display multiple absorption peaks in their spectra. In ♣Further information can be found in §2.2 and §4.7.2 of Hunter, Vol. I.


where

2πnkT ξn = h

and h = 1.0545×10–34 J-s is Planck’s constant divided by 2π. The pluses and circles also show comparable data for tetradecane and polystyrene. To obtain the complete spectra requires combining data from several experiments covering various subsets of the spectra (microwave, IR, visible or UV). The solid curves are simpler approximations to the spectra which are discussed in the next section.

♦J. Colloid Interface Sci. 81, 285 (1981).


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A plot of n2–1 vs (n2–1)ω2 is called a Cauchy plot.

Cauchy Plot Full characterization of the permittivity spectra ε(iξ) requires a lot of experimental data and effort. This has generally hindered the use of Lifshitz theory for van der Waals forces. A glimmer of hope was offered in the work of Hough & White♠ who suggested that an adequate representation of the permittivity spectra can be obtained from a single undamped (i.e. f=0) absorption peak in the UV. In other words, (272) can be approximated by

−2 The slope equals ωUV while the intercept equals C. Cauchy plots for polystyrene (triangles), tetradecane (squares) and water (circles) are shown below (the units on the x-axis are 1030 Hz2):

ε ( iξ ) gUV = 1+ 2 2 ε0 ξ + ωUV In the above expression, we are still evaluating ε at a purely imaginary frequency iξ (ξ itself is real). To get the corresponding expression for real frequencies, we replace ξ by –iω; then the argument iξ becomes ω and ξ2 becomes –ω2. In the absence of absorption, ε/ε0 = n2, where n is the fractive index of the material [recall (76)]. In terms of purely real frequencies, this corresponds to

Using this simplified characterization of the spectra, the solid curves of ε(iξn) vs n were obtained (see figure just before last one).

gUV ε ( ω) = n 2 ( ω) = 1 + 2 ε0 ωUV − ω2 Note that as long as ω is in the visible portion of the spectra (i.e. ωvis < ωUV), this function is well behaved. Subtracting 1 from both sides:

gUV 2( ) n= ω −1

gUV = 2 ωUV − ω2

1−

2 ωUV

ω2 2 ωUV

Multiplying both sides by the denominator of the right-hand side: 2   gUV  n 2 ( ω) − 1 1 − ω = ≡C   ω2  ω2 UV  UV 

This relationship can be rearranged to obtain n 2 ( ω) − 1 = C +  n 2 ( ω) − 1

where

C=

ω2 2 ωUV

gUV 2 ωUV

♠ Adv. Colloid Interface Sci. 14, 3 (1980).



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Theory vs. Experiment Below are direct force measurements between crossed cylinders of mica (about 1 cm in diameter) in air.♥

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The dotted line was calculated using a Hamaker constant of

= A 2.2 × 10−20 J The agreement is quite good for separations below about 7 nm. At larger separations, retardation causes the force to decay more rapidly with further increases in separation than the theory predicts. Retardation becomes more evident at much larger separations where the forces are weaker and harder to measure.

Figure 16: van der Waals interactions measured between crossed mica cylinders in air measured with the surfaceforces apparatus (SFA).

The dotted line shows the corresponding theory for nonretarded van der Waals forces. The form of this equation can be deduced from the entry in the Figure 14 for crossed cylinders of radius Rc separated by a gap equal to d:

R φA (d ) = −A c 6d

(273)

Differentiating to get the force:

dφ(h) AR − = − c F= dd 6d 2 and dividing by the radius:

F A = − Rc 6d 2 ♥ Israelachvili & Adams, JCS Faraday Trans I, 78, 975 (1978).


Figure 17: Potential energy profiles measured for a 6 µm polystyrene sphere and polystyrene plate, separated by an aqueous solution having different concentrations of NaCl and measured with total internal reflection microscopy (TIRM).♣

Figure 17 shows the severely retarded van der Waals interaction between a 6 µm sphere of polystyrene interacting with a glass microscope slide, separated by dilute aqueous salt solutions having various concentrations. The upturn in energy for separations less than 50 nm represents double-layer repulsion which is not present in Figure 16 because the cylinders are uncharged in air. The solid curve is the common envelope of the attractive portion of the four curves; this envelope represents van der Waals attraction. An empirical equation used to fit this solid curve is

♣ Bevan & Prieve, Langmuir 15, 7925 (1999).


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 −d  φA (d ) = −3kT exp    40 nm  To give you an idea how weak these van der Waals forces are: replacing the 1-cm radius of the cylinders Rc in (273) by the 3 µm radius of the sphere in Figure 17, the potential energy evaluated from (273) at d = 300 nm is –10 kT so the nonretarded interaction is “off the scale” for Figure 17. The two dashed curves represent two different approximations (for the geometry), both of which use Lifshitz theory to predict the retarded interaction across half-spaces. Deviations between theory and experiment for separations below about 70 nm are due to roughness on both surfaces (which is not a problem in the previous figure employing mica which is atomically smooth).



132

+

-+ -+ -+ -+ - -+

Whereas van der Waals attraction between droplets or particles is responsible for the thermodynamic tendency of a dispersion to be unstable and flocculate, electrostatic repulsion between particles of like sign can reduce the rate of flocculation to near zero and impart kinetic stability to the colloidal state. The interaction between charged bodies across water is profoundly different than across vacuum or a simple dielectric material. This profound difference is due to the presence of mobile charges (called “ions”) in water, as we will now describe.

Now let’s bring two such particles together:

-+ + -+ + - -+

Q2 r

Coulomb’s law (see page 30 or 115) states that two point charges Q1 and Q2, separated by a distance r, experience an electrostatic force given by

F=

1 Q1Q2 4πε r 2

Electrostatic interactions across water are more complicated than interactions across simple dielectrics like air or oil because water always contains ions which are also charged. In response to the charge on the interacting particles, these ions are rearranged: oppositely charged ions (called “counterions”) are attracted to the particle while ions of the same sign of charge (called “co-ions”) are repelled: +

-+ -+ -+ -+ - -+

-+ + -+ + - -+

r

+

-+ + -+ + - -+

+

Because each particle plus its cloud is electrically neutral, no electrostatic force is felt between two such particles in water until they get close enough for their clouds to overlap. Then we will see that the electrostatic repulsion decays exponentially with r :

Electrostatic Forces in Water vs. Coulomb’s Law

Q1

+

The Electric Double Layer

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-+ -+ -+ -+ - -+

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F ∝ e-κr where 1/κ is the thickness of the charge cloud, also known as the Debye length. The resulting repulsion is called the double-layer force. The resulting severe weakening of the electrostatic forces is called Debye screening: the cloud essentially hides the bare charge borne by the particles until they get very close together.

Lecture #19 begins here Gouy-Chapman Model of Double Layer Let's try to quantify this description. In particular, we would like to know how thick this cloud of counterions is. Because of the charge on the interface (the negative charges in the schematic above), ions in solution (the blue region) feel an electrostatic force. The force per unit charge is called the electric field:

)[ ] E ( x=

force N J m V = = = charge C C m

+

At equilibrium, a diffuse cloud of counterions surrounds each charged particle (see above). Moreover, the magnitude of charge in this diffuse cloud (in blue) equals the magnitude of charge born on the surface of the particle (in yellow). Since the diffuse cloud’s charge is always of opposite sign to the particle itself, viewed from a distance, the particle together with its cloud appears electrically neutral. The two layers of charge collectively form the double layer — the discrete layer of fixed charges at the interface plus the diffuse layer of counterions in the electrolyte solution next to the surface


Like gravity, this vector field is conservative, thus there must exist a scalar potential — which we will denote by ψ(x) — such that: E = –∇ψ

(274)

As in the analogous equation relating the gravitational force to the gravitational potential, the minus sign is included by convention. ψ is called the electrostatic potential.

ψ ( x)[= ]

energy [ = ] volt ≡ joul charge coul


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Boltzmann’s Equation:

which is called the Nernst-Planck equation.

So how do ions distribute themselves next to a charged surfaces? Let’s restrict attention to a uniformly charged plane surface of infinite area.

Next, we need boundary conditions. Far from the surface (x→∞), any effect of charges at x=0 will decay to zero, leaving the ion concentrations equal to their value ci∞ in the bulk of the solution (assumed to be of uniform composition). For this to happen, the electrostatic potential must tend to some constant value (say ψ→0 as x→∞):

uniformly charged interface

as x→∞:

x

Then any electric field resulting from the surface charge will act purely normal to the surface and we expect the electrostatic potential and the ion concentration to depend only on the normal distance from the charged surface (say x). The charge on the surface attracts oppositely charged ions and repels like charged ions. The flux of ions will have contributions from diffusion and electromigration of ions driven by the electric field:

dc J ix = − Di i + mi Fix ci dx

(275)

where Di is the diffusion coefficient of ion species i, ci is the number of ions per unit volume, mi is the hydrodynamic mobility of ion species i and Fix is the electrical force acting on ion i induced by the local electric field Ex. Recalling the electric field is force per unit charge, we can calculate the force on any ion by multiplying the electric field by the charge on that ion:

dψ Fix = zi eE x = − zi e dx

(276)

where zi is the number of elemental charges (including sign) born by any ion and e = 1.60217733×10–19 C is the quantity (absolute magnitude) of charge on one proton or one electron. For example, zi is +1 for Na+ and –1 for Cl–. The hydrodynamic mobility mi is the reciprocal of the friction coefficient fi, which in turn is related to the diffusion coefficient Di by (41) (see page 19): Di = mikT

(277)

ci → ci∞ and ψ → 0

(279)

By taking ψ=0 far from the charged surface, we are choosing this location as the arbitrary reference state for potential. If no adsorption or reaction of ions occurs at the wall (x=0) and the wall is otherwise impenetrable to ions, the flux Jix must vanish at the wall. Then another appropriate boundary condition is Jix = 0 and ψ = ψ0

at x=0:

(280)

where the value of the surface potential ψ0 is sometimes known;♦ even if ψ0 is unknown, this boundary condition then serves as its definition. Continuity of species mass requires

∂ci + ∇ . Ji = 0 ∂t For steady state, the partial time derivative is zero. For one-dimensional diffusion in the x-direction, this yields

dJ ix =0 dx

or

Jix = const = 0

where this constant value of the flux is determined by the boundary condition (280). Then (278) becomes

z e dψ   dc 0= − Di  i + ci i  kT dx   dx After dividing out the Di, we have a separable, firstorder ODE: ci ( x )

∫

ci∞

ψ( x )

dci ze = − i ∫ dψ ci kT 0

After applying boundary condition (279), we have  z eψ ( x )  ci ( x ) ci∞ exp  − i =  kT  

(281)

Substituting (276) and (277) into (275) yields

z e dψ   dc J ix = − Di  i + ci i  kT dx   dx


(278)

♦ For example, in the case of potential-determining ions like Ag+ or I–, it is given by Nernst equation (234).


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which is called Boltzmann’s equation. So if I knew the ψ(x), I could calculate the concentration profile. But alas ψ(x) is still unknown. I need another equation. The extra equation is provided by Coulomb’s law of electrostatics. If I knew the distribution of charges, I could calculate the force on them using Coulomb’s law.

In continuum mechanics – instead of specifying the discrete location in 3-D space of every ion – we smear out the discrete charges and specify the distribution using a continuous variable called the space charge density. Just like we define the mass of material per volume to be the density of the fluid — or we define ci(x,t) to be the local concentration of ions of species i — we can define: ρe(x,t) = local space charge density

(charge/volume) For a continuous medium having a uniform isotropic permittivity, Coulomb’s law can be written either as Gauss’s law (in differential form)♥

ρe ∇.E = ε

(282)

or – if we substitute (274) – it becomes Poisson’s equation:♦

ρe ε

(283)

where ε is the electric permittivity of the medium between the ions (e.g. the water): = ε0 8.854 × 10−12

ρe ( x ) = ∑ zi eci ( x ) i

=

C

2

F =[ ] 2 m N-m

is the permittivity of vacuum, and

ε= 78ε0 is the permittivity of water. The charge density ρe of the fluid arises from the charge born by each ion. Adding up the charges from each ion species i, then substituting Boltzmann’s equation (281) gives ♥ The more general form (for nonuniform, nonisotropic materials) is given by Maxwell’s equation (55). ♦ For a derivation of Gauss’s and Poisson’s equations from Coulomb’s law, see Electrostatics of Continuous Media.pdf at Blackboard.


 zi eψ ( x )   kT 

∑ zi eci∞ exp − i

(284)

Assuming ψ = ψ(x), (283) and (284) yield d 2ψ

Poisson-Boltzmann Equation

∇2ψ = −

Spring, 2016

dx

2

 z eψ ( x )  e = − ∑ zi ci∞ exp  − i  ε i kT  

(285)

which is called the Poisson-Boltzmann Equation. It is a second-order, nonlinear, ordinary differential equation for ψ(x). The Poisson-Boltzmann equation (285) represents one equation in one unknown. Boundary conditions are given by (279) and (280). Once we have the particular solution for ψ(x), we can calculate the ion concentrations profiles from Boltzmann’s equation (281). Owing to the nonlinearity (represented by the exponentials of the dependent variable ψ) of the Poisson-Boltzmann equation, no general solution exists (even in the simplest case of 1-D). However particular solutions cases can be obtained analytically in a couple of special, as we will now show. Special Case #1: Small Potentials If the potentials are small enough, then we can approximate the exponential using the first couple of terms of its Taylor series expansion:

lim eu =1 + u +

u →0

1 2 1 3 u + u + 2! 3!

Replacing u by –zieψ/kT:  z eψ ( x )  z eψ ( x ) exp  − i 1− i + ... = kT  kT 

Then Boltzmann’s equation (281) for ion concentrations can also be approximated:  zi eψ ( x )  ci ( x ) = + ... ci∞ 1 − kT   ( ) eψ x = ci∞ − zi ci∞ + ... kT

(286)

Truncating the series after the second term, (284) yields   eψ ( x ) zi ci e ∑ zi ci∞ − zi2 ci∞  = ρe e ∑ = ∑ kT i  i  i       2I  0 


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Now the first sum vanishes because the bulk solution must be electrically neutral. This leaves

2 Ie2 ρe ( x ) = − ψ ( x) kT where

I≡

1 ∑ zi2 ci∞ 2 i

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which is plotted in the graph below using zi = ±1.

(287) (288)

is called the ionic strength of the bulk solution. In the case of a symmetric univalent electrolyte (e.g. NaCl), the ionic strength is also the bulk concentration of salt. Substituting (287) into (283) yields d 2ψ dx 2

=

2e 2 I ψ = κ2 ψ εkT 

(289)

κ2

The remaining collection of constants is lumped into κ2 where κ is called the Debye parameter: κ≡

2e2 I [ =] m −1 εkT

(290)

(289) is called the linearized Poisson-Boltzmann equation. Integrating this ordinary differential equation subject to (279) and (280):

Substituting (291) into (287) gives the linearized form of the space charge density:

2 Ie2 2 Ie2 ρe ( x ) = − ψ ( x) = − ψ 0 e−κx (292) kT kT Integrating this over the entire fluid yields the net charge per unit area born by the diffuse cloud: σd ≡

ψ = 0 at x=∞ ψ(x) = ψ0 exp(–κx)


∞

ψ = ψ0 at x=0

yields

In the sketch above, we have exaggerated the changes in ion concentrations which must be small compared to respective bulk concentration of that ion (in order to linearize the mathematical problem).

0

(291)

which is sketched in the graph below for some arbitrary (but small) positive value of the surface potential ψ0.

    ∞ −κx  ψ 0 ∫ e dx   0    κ −1   ψ0  

∫ ρe ( x ) dx

2 Ie 2 = − kT 2 Ie 2 = − κkT

(293)

Global electroneutrality requires that the total charge of the solid and the fluid much vanish: ∞

σ + ∫ ρe ( x ) dx = 0

(294)

0   σd

σ = surface charge density (charge/area) σd = charge density (charge/area)

Substituting (291) into Boltzmann’s equation (281) and again linearizing the exponentials as in (286), yields

of diffuse cloud Substituting (293) into the above equation yields an equation for the surface charge density

eψ 0 −κx   c= e  i ( x ) ci∞  1 − zi kT  



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∞

σ = − ∫ ρe ( x ) dx = 0

2 Ie2 ψ 0 = κεψ 0 κkT 

(295)

The third equation was obtained by substituting the definition (290) of the Debye parameter κ. Notice that the surface charge density σ and the surface potential ψ0 have the same sign (since all the other symbols in the equation are positive). Of course the electrolyte solution must have a net charge σd opposite in sign to that born by the surface itself (σ). This is guaranteed by electroneutrality (294). Based on the form of (291) or (292), κ is clearly associated with the thickness of the diffuse cloud. Indeed κ–1 is called the Debye length: for

c∞ = 10–1 M

κ–1 = 10–6 m = 1µm

for

c∞ = 10–7 M

the fixed charges located at x=0 from the counterions located in the diffuse cloud. Special Case #2: Symmetric Binary Electrolyte

κε

κ–1 = 10–9 m = 1nm

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Based on the graphs of ψ(x) or of ci(x) shown above, the thickness of the counterion cloud is a couple of Debye lengths.

Let’s relax the small-potential assumption. We can still integrate the nonlinear Poisson-Boltzmann equation if there are only two different species of ions and they both have the same magnitude of charge (e.g. NaCl is 1-1, BaSO4 is 2-2); then we say the solution is a symmetric binary electrolyte: z+ = –z– = z c+ ∞ = c– ∞ = c∞

(297)

Then Boltzmann's equation (281) can still be used to calculate the charge density:  z eψ   z eψ  = ρe ( x ) z+ ec+∞ exp  − +  + z− ec−∞ exp  − −   kT   kT    zeψ   zeψ   = zec∞ exp  −  − exp  + kT    kT    

or

 zeψ  ♣ ρe ( x ) = −2 zec∞ sinh    kT 

(298)

Double-Layer Capacitance In (295), the surface charge per unit area σ is directly proportional to the surface potential ψ0. The proportionality constant κε has units of electrical capacitance per unit area (Farad/m2). A Farad is defined as one Coulomb per Volt. (295) can be written as

Cdbl ≡

σ ≈ κε ψ0

Actually the formula κε for capacitance of the double-layer is remarkably analogous to the formula for calculating the electrical capacitance of a parallelplate capacitor. If the separation distance between the two plates is d and the material between the plates is a dielectric having permittivity ε, then the capacitance per unit area of the parallel-plate device is given by

ε d

Replacing the plate separation d by the Debye length κ–1 yields the formula for the double-layer capacitance. In other words, the Debye length is analogous to the plate-separation. Indeed, the Debye length can be thought of as the characteristic distance separating


d 2ψ dx 2

=

2 zec∞  zeψ  sinh   ε  kT 

The argument of the sinh must be dimensionless, so let’s use this combination of variables to define a new dimensionless potential

(296)

when ψ0> κ–1). Overall electroneutrality requires


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σ0 + σd   fixed charges diffuse layer in a plane of charge

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= 0

All of the fixed charges on the interface at x=0 must be neutralized by an excess of counterions in the diffuse layer. Then the net charge in V (the interior of the cylinder) is zero and (306) and (307) reduce to

Dwx − Dsx = ε w Ewx − ε s Esx = 0  0

where εw and εs are the permittivities of the water and solid phase, respectively. Recall that one of the boundary conditions imposed on the PoissonBoltzmann equation leading to the profile displayed in Fig. 18 was ψ = 0 as x→∞. This corresponds to Ewx = –dψ/dx = 0 and the equation above further requires Esx = 0 inside the solid phase. A second feature in Fig. 18 is the discontinuity in the slope of (i.e. a “kink” in) the potential profile at x=0. This too is a consequence of Gauss’s law (306). If the length of the cylinder shown in Fig. 18 is just large enough to straddle the interface, then (306) and (307) reduce to

Dx x 0 = + − Dx = x 0− or

= σ

ε w E x x 0+= − ε s E x x 0− =   

=σ

Fig. 19. Potential profile with a Stern layer but no adsorption of counterions. We customarily assign the ion’s net charge to the exact center of the ion, although in reality the electron in the outer shell is distributed uniformly over the outer surface of the ion. If the center of the counterion (which is the dominate ion species in the diffuse layer) can only get within one of its radii δ of the wall, then there must exist a layer (called the Stern layer) within which there are no ions and no space charge. The thickness of this layer is denoted as δ in Fig. 19. Within this layer, ρe = 0 and Poisson’s equation (283) yields

0

where x=0– denotes the left side of the interface and x=0+ denotes the right side. We included only the charges on the interface σ (not the diffuse cloud σd) in the volume integral. Thus any net charge at an interface leads to a discontinuity in the normal component of the displacement vector across the interface. Stern Layer Reference: Lyklema, Vol. II, §3.6c. Stern Refinement #1 Stern’s first refinement was to take into account an ion’s finite size by observing that the center of an ion can only get so close to the charged plane at x=0:

d 2ψ dx 2

=0

Integration reveals that dψ/dx = const and ψ(x) must be linear in the Stern layer (as shown in Fig. 19). This causes the potential to drop somewhat (in absolute value) before we reach the inner edge of the diffuse cloud which is now located at x = δ. Smaller potentials inside the diffuse cloud lead to smaller concentrations of ions. We are now less likely to predict absurdly high concentrations if ψ0 remains fixed. One feature of Fig. 19 is the absence of a “kink” at x = δ. Once again, this is a consequence of Gauss’s law (306), which applied to a “pillbox” (i.e. the very short cylinder) straddling x = δ leads to

Dx or

− Dx x = = ε w Ex x = − ε w Ex x = =0 δ+ δ− δ+ δ− x= Ex

= Ex x = x= δ+ δ−

In the following example, we show how this accounting for finite ion size can relieve the absurdity seen in Example #1 on page 138.



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Example #2: Add a Stern layer to the previous example. Assume hydrated ions have a radius of 0.25 nm and keep the surface potential ψ0 the same as in Example #1. Calculate the (maximum) concentration of counterions next to the surface. Solution: The electrostatic potential profiles with or without a Stern layer are compared in the following plot:

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Comment #1: By keeping the surface potential ψ0 fixed we are in effect assuming that the solid is composed of Nernstian potential-determining ions (e.g. AgI). Then the surface potential satisfies Nernst’s equation (234): if the composition of the bulk solution is the same, the surface potential remains the same. Comment #2: The partial relief (from absurdity) seen in this example is a consequence of a smaller potential drop across the diffuse part of the doublelayer (i.e. x>δ). This smaller potential drop also means the diffuse cloud contain fewer charges: ψδ = 117 mV corresponds to σ = 0.179 C/m2 (instead of 0.641 C/m2) calculated from the Gouy-Chapman equation (304). σ = 0.179 C/m2 also represents the new surface charge density born by the solid. Comment #3: Instead of keeping the surface potential constant, If we had kept the surface charge density fixed (e.g. sulfonated polystyrene latex having covalently attached strongly acidic surface sites), the comparison with and without a Stern layer would look very different:

The blue curve is ψ(x) from Example #1, calculated from (302) using σ = 0.641 C/m2 which corresponds to ψ0 = 182 mV. The red curve for x>δ was also calculated from (302) but x is replaced by x–δ and ψ0 is replaced by ψδ = 117 mV: for x>δ:

 Ψ ( x)   Ψδ tanh   = tanh  4  4  

 −κ( x −δ ) e 

The linear portion of the red curve for 0 7 lie below both of these asymptotes. Analytical Approximation for Large Particles

When the zeta potentials are large, κR has to be very large indeed in order for Smoluchowski’s formula (393) for be accurate. The curve corresponding to κR =100 in Fig. 7 deviates from the

straight line corresponding to (393) for ζ greater than about 3. For ζ greater than about 6, U E actually goes through a maximum.

Fig. 8: Solution to Example on this page.

So how large does κR have to be to use (393)? O’Brien has suggested the following criterion:

 zeζ  κR  cosh    2kT 

(394)

which means that the minimum value of κR needed to

use (393) grows exponentially with ζ , regardless of sign (cosh is an even function). O’Brien & Hunter♦ developed an improved approximation for electrophoretic mobility of large particles in z-z electrolytes: for ζ >0:

 ζ ln 2 (1 − e− zζ ) 6 − 3 z  ζ − 2 U E = κR − zζ 2 2 e 2+ 3m 1+ 2 z

(395)

♦ R.W. O’Brien & R.J. Hunter, Canadian J. Chem. 59, 1878 (1981).


where

m ≡

2εRgasT z 12.86 = (396) 3µ Λ 0 Λ0

is the dimensionless drag coefficient for the counterions (assumed to be anions),♣ Λ0 is the ionic conductance of counterions at infinite dilution (conductivity divided by molar concentration) and Rgas is the universal gas constant (k times Avogadro’s number). The second equation above applies for aqueous solutions at 25ºC provided that the units of Λ0 are Scm2/mol. Example: compare predictions of (395) for U E

versus ζ in solutions of KCl having κR = 150 with corresponding results of Fig. 7. ♣ (395) is (6.121) in Berg. In his solution to Chapt. 6, Prob. 4, Berg instead uses the harmonic mean of the conductances of the cation and anion. This is not consistent with O’Brien & Hunter,♦ who use the counterion conductance only.


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Solution: Since (395) is only valid for positive zetas, the counterion is the anion of the salt. The ionic conductance of Cl– at 25ºC is♠

Λ0 = 76.31

S-cm 2 mol

(KCl was probably chosen for Fig. 7 because its two ions have nearly identical values of the conductance.) (396) gives m = 0.169. (395) gives the blue curve in Fig. 8. Notice that the blue curve has a maximum mobility of 7.15 which occurs at ζ = 6.8 which is very close to the maximum of the curve in Fig. 7 for κR = 150.

The mobilities calculated from approximation (395) for this and other κR values are compared in Fig. 9 with the exact numerical results from Fig. 7. While the comparison for κR = 150 is quite close, the comparisons for κR = 50 and 30 show greater deviation. Streaming Current So far in our discussion of electrokinetic phenomena, we have considered the motion of either fluid or solid that results from the application of an electric field: Electrophoresis and Electro-osmosis: E → U Now we would like to consider the inverse situation in which forced flow across a charged interface generates an electric field: Streaming Potential and Sedimentation Potential:

Fig. 9: Comparison of predictions of (395) (shown as dashed lines) with exact results from Fig. 7. Figure copied from O’Brien & Hunter (1981).

U → E Consider the pressure-driven flow through a large circular capillary. κ–1

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