1 Chapter 7 Continued: Entropic Forces at Work (Jan. 10, 2011) Mechanisms that produce entropic forces: in a gas: the pressure in a cell: the osmotic pressure in macromolecular solutions: depletion in ionic interactions: electrostatic forces between macromolecules hydrofobia

2 Semipermeable membrane: colloidal particles carry (letward) momentum => membrane delivers an equal and opposite counter force, effectively directed rightward, dragging water molecules along! Pressure difference due to the transfer of momentum is given by the Van het Hoff relation (‘Gas law’): ∆p = k B T∆c. MAKE YOUR TURN 7C Note: without a pressure gradient across the pore => no net flow through the pore Wat veroorzaakt osmotische druk?

3 General case: spontaneous and driven osmosis combined Two situations in which an active pump keeps the pressure gradient across the membrane the same ( __ ) or even larger (---). Driven osmosis: due to pressure difference ∆p Spontaneous osmosis: due to a concentration difference ∆c (VhHoff) now there is a pressure difference => flow through the pore H 2 O flow rightward (osmotic machine) H 2 0 flow leftward (reverse osmosis)

4 Flow through the pore: Poiseuille flow (Ch. 5): Q=[m 3 /s] Flux: flow per area per second: If ∆c=0: j v = -L p ∆p If ∆p=0: j v = L p ∆c k B T (L p is the ‘filtration coefficient’) Combine driven+spontaneous osmosis:

5 Entropic force by electrostatic interactions: (now we account for the interactions between the molecules) Macroscopic bodies are electrically neutral. Why is that? 2 mm Water dropHow much work is needed to ionize 1% of the water molecules? Ions will start to form a positively charged shell around the drop Number of charges: !!! Make YOUR TURN 7D

6 In the cell: Thermal motion can make large neutral macromolecules to split into charged subunits! Take an acid macromolecule like DNA  A negatively charged macro-ion is left behind  + a surrounding cloud of positive cations. DNA Equilibrium between the urge to increase entropy, and the cost by having to overcome the electric field (potential energy) Looked at from far away the E-field of the macromolecule is neutral, but from nearby (nm scale) it’s not! ‘diffuse charge layer/electric double layer’

7 Two types of forces work on macromolecules: depletion drives them together electrostatic repulsion moves them apart Therefore the macromolecules are not all closely packed together Moreover: these forces act at very short range (order of nm) these forces have a strong geometric dependency (stereo-specificity) This ensures that certain molecules can find each other and stick together. What are the properties of the surrounding cation-cloud?

8 negative surface charge density (C/m 2 ) positive volume charge density (C/m 3 ) Gauss law: (close to the surface) (away from the surface)

9 When charge distributions have a complex geometry: use ‘Mean Field Approximation’: each ion senses on average the same electric field at x Question: what is c + (x)? (the cation distribution in average potential V(x)) - Far away from the surface: c + (x) -> 0 - Ions move independent of each other through the field V(x): Boltzmann distribution: But what is V(x)?

10 From:and: We also had the Boltzmann charge concentration: we get the Poisson equation: (I) and note that for the charge density: (II)

11 To solve (I) and (II) we first introduce the Bjerrum length in water: “How closely can two equal charges be brought together with an energy of k B T” At room temp, monovalent charges in H 2 O: 0.71 nm and the normalized potential: Rewriting in these new variables yields from (I) and (II): The Poisson-Boltzmann equation How do we solve this P-B equation? Boundary conditions! Your Turn 7E

12 Boundary conditions for V(x): (close to the surface) We further adopt: yielding: Moreover: (no charge, pure H 2 O) YOUR TURN 7F The solution is a function of the form V(x) = B ln (1 + x/x 0 ) (see Book, page 268), which gives: with The “Gouy-Chapman layer”

13 An interesting case: two negative macro ions with diffuse charge clouds, approaching each other: When D<2x 0 the ions start to sense each other. The cations are forced to stay between the ions (neutrality), which creates an osmotic pressure! We can again use the Poisseuille-Boltzmann eqn., but now with different boundary conditions! (V(x) symmetric re. centre) Now try a solution of the form: V(x)=Aln cos(  x): Question 7-i: verify that A=2 and

14 Second boundary condition: (close to the surface, x=±D) Question 7-ii: Verify that now So that: Grafic solution for 

15 The osmotic force between the macro ions is: k B T times the concentration difference: Comparison of theory with experiment Voor de durf-al: YOUR TURN 7G

16 Werkcollege van : Your Turns 7C, 7D, 7E en 7F Opgave 7.4