1 2.Properties of Colloidal Dispersion Colloidal size : particle with linear dimension between 10-7 cm (10 AO) and 10-4 cm (1  ) nm particle weight/ particle size etc. Shapes of Colloids : linear, linear, spherical, rod, cylinder spiral sheet Shape & Size Determination

2 Molar Mass ( for polydispersed systems) Number averaged Weight averaged Viscosity averaged Surface averaged Volume averaged Second moment Third moment Radius of gyration Number averaged Molar Mass n

3 2.1 Colligative Propery In solution Vapor pressure lowering  P = ik p m Boiling point elevation  T b = ik b m Freezing point depression  T f = ik f m Osmotic pressure  = imRT (m = molality, i = van’t Hoff factor) In colloidal dispersion  Osmotic pressure

4 Osmosis Osmosis  the net movement of water across a partially permeableartially permeable membranemembrane from a region of high solvent potential to an area of low solvent potential, up a solute concentration gradient

5 Osmotic pressure  the hydrostatic pressure produced by a solution in a spacehydrostaticpressure divided by a semipermeable membrane due to a differential insemipermeable membrane the concentrations of solute For colloidal dispersion The osmotic pressure π can be calculated using the Macmillan & Mayer formula π = 1 [1 + Bc + B’c 2 + …] cRT M M M 2  =  gh

6 Molar Mass Determination Dilute dispersion h = RT [1 + Bc ] c  gM M Where c = g dm -3 or g/100 cm -3 M = M n = number-averaged molar mass B = constants depend on medium Intercept = RT  g n Slope = intercept x B/M n h (cm g -1 L) c c (g L -1 )

7 Estimating the molar volume From the Macmillan & Mayer formula : B = ½N A Vp where B = Virial coefficient N A = Avogadro # V p = excluded volume, the volume into which the center of a molecule can not penetrate which is approximately equals to 8 times of the molar volume Example/exercise : Atkins

8 Osmotic pressure on blood cells Donnan Equlibrium Donnan Equlibrium : activities product of ions inside = outside

Donnan equilibrium a NaCl, L = a NaCl, R (a Na+ ) L (a Cl- ) L = (a Na+ ) R (a Cl- ) R activity a =  C Where a = activity  = activity coefficient C = molar conc. log  = - kz 2

10 Reverse Osmosis  a separation process that uses pressure to force a solventsolvent through a membrane that retains the solute on one side and allowsmembranesolute the pure solvent to pass to the other sidesolvent Look for its application : drinking and waste water purifications, aquarium keeping, hydrogen production, car washing, food industry etc.water purifications Pressure

2.2 Kinetic property :  either the random movement of particles suspended in a fluid or the mathematical model used to describe such random movements, often called a Wiener process.Wiener process  The mathematical model of Brownian motion has several real-world applications. An often quoted example is stock market fluctuations.stock market = 2Dt D = diffusion coefficient Brownian Motion

2.2 Kinetic property : DiffusionDiffusion the random walk of an ensemble of particles from regions of high concentration to regions of lower concentration

Einstein Relation (kinetic theory) where D = Diffusion constant,Diffusion constant μ = mobility of the particles mobility k B = Boltzmann's constant,Boltzmann's constant and T = absolute temperature.absolute temperature The mobility μ is the ratio of the particle's terminal drift velocity to an applied force, μ = v d / F.

For spherical particles of radius r, the mobility μ is the inverse of the frictional coefficient f, therefore Stokes law givesStokes law f = 6  r where η is the viscosity of the medium.viscosity Thus the Einstein relation becomes This equation is also known as the Stokes-Einstein Relation. Diffusion of particles

Fick’s Law 1 st law J = -D 2 nd law 1 st Law 2 nd Law where Flux mole m -2 s -1  = molar concentration  = chemical potential D = diffusion coefficient

2.2 Kinetic property : ViscosityViscosity a measure of the resistance of a fluid to deform under shear stressresistancefluid shear stress where: is the frictional force, r is the Stokes radius of the particle,Stokes radius η is the fluid viscosity, and is the particle's velocity.

 = -  PR 4 t 8VL R P L  =  t  o  o t o  = viscosity of dispersion  o = viscosity of medium Unit:Poise (P) 1 P = 1 dyne s -1 cm -2 = 0.1 N s m -2 Viscosity Measurement

Viscometer

 [c P] liquid nitrogenliquid 77K0.158 acetone0.306 methanol0.544 benzene0.604 ethanol1.074 mercury1.526 nitrobenzene1.863 propanol1.945 sulfuric acid24.2 olive oil81 glycerol934 castor oil985  [c P] honey2,000–10,000 molasses5,000–10,000 molten glass10,000–1,000,000 chocolate syrup 10,000–25,000 chocolate chocolate * 45,000–130,000 ketchup ketchup * 50,000–100,000 peanut butter~250,000 shortening shortening * ~250,000   Intermolecular forces

Intermolecular forces intermolecular forces are forces that act between stable molecules or between functional groups of macromolecules.molecules macromolecules Intermolecular forces include momentary attractions between molecules, diatomic free elements, and individual atoms. These forces includes London Dispersion forcesLondon Dispersion forces, Dipole-dipole interactions Dipole-dipole interactions and Hydrogen bondingHydrogen bonding,.

Einstein Theory  =  o (1+2.5  )  = volume fraction of solvent replaced by solute molecule  = N A cV h MV where c = g cm -3 v h =hydrodynamic volume of solute  -1 =  sp = 2.5 ,  sp : specific viscosity  o

Einstein Theory  =  o (1+2.5  )  = N A cV h MV  -1 =  sp = 2.5 ,  sp : specific viscosity  o [  ] = lim  sp = 2.5 , [  ] : intrinsic viscosity c c [  ] = K(M v ) a Mark-Houwink equation K - types of dispersion a – shape & geometry of molecule

Assignment 2 1. At 25 o C D of Glucose = 6.81x m  s -1  of water = 8.937x10 3 P  of Glucose = 1.55 g cm -3 Use the Stokes law to calculate the molecular mass of glucose, suppose that glucose molecule has a spherical shape with radius r 5 points (3-5 students per group)

Assignment 2 2. Use the data below for Polystyrene in Toluene at 25 o C, calculate its molecular mass c/g cm  /10 -4 kg m -1 s Given : K and a in the Mark-Houwink equation equal 3.80x10 -5 dm -3 /g and 0.63, respectively (5 points) Due Date : 21 Aug 2009