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CHEM 305 Viscometry When a macromolecule moves in solution (e.g. of water), it induces net motions of the individual solvent molecules, i.e. the solvent molecules will feel a force. - neglect Brownian motion.

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Transcript of Viscometry - University of British ColumbiaMeasuring viscosity Ostwald viscometer Wilhelm Ostwald...

CHEM 305

Viscometry

When a macromolecule moves in solution (e.g. of water), it induces netmotions of the individual solvent molecules, i.e. the solvent moleculeswill feel a force.

- neglect Brownian motion.

CHEM 305

To describe this force, let us consider two sheets of fluid, of area A

F

A

dvdx

h

F = Aηdvdx

η = viscosity

CHEM 305

To measure viscosity, the capillary effect is used. The force for the movementof solvent depends on the hydrostatic pressure, i.e.

Fup = Pπa2

For a small cylindrical sheet, at a radialdistance x, the differential force will be

dFup = 2Pπx dx

If the fluid is flowing through the capillaryat a steady state, this force must bebalanced by a frictional force, i.e.

Fdown = -A η

= - 2πxlη

where the negative sign indicates that it is in the direction opposite to the applied force.

ax

dvdx

dx

l

dvdx

CHEM 305

The net force on the sheet due to fluid motion is the differential force felt by thetwo sides of the sheet, i.e.

dFdown = - 2πlη d[x(dv/dx)] dxdx

These two differential forces (up and down) are equal, therefore

2Pπx dx = - 2πlη d[x(dv/dx)] dxdx

Px = - lη d[x(dv/dx)] dx

Integrating this equation once gives

½ Px2 + c1 = -ηlx (dv/dx)

and again ¼ Px2 + c1 ln x + c2 = -ηlv

where c1 and c2 are integration constants.

CHEM 305

The integration constants can be obtained by looking at the boundary conditions.

1) At x=0

¼ Px2 + c1 ln x + c2 = -ηlv

Therefore, c1 = 0.

2) At x=a

¼ Pa2 + c2 = -ηlv

Therefore, c2 = - ¼ Pa2.

Thus we can write,

v = P (a2 – x2) (flow velocity).4ηl

-∞ cannot be infinite!!!

= 0

CHEM 305

Unfortunately, flow velocity is not easily measured – better to use the volume rateof flow, which is defined as

dV = ∫ 2πxv dxdt

= πP ∫ (a2-x2) x dx2ηl

dV = πPa4 Poiseuille’s lawdt 8ηl

0

a

0

a

CHEM 305

Measuring viscosity

Ostwald viscometer

Wilhelm Ostwald(1853-1932)

Nobel Laureate 1909(for his work in catalysis,chemical balance, and

Reaction rates)

CHEM 305

h1

h2

l

a

A fluid of density ρ is allowed to fall from heighth1 to h2, in a determined time t. The hydrostaticpressure felt by the solution is given by ρgh.Using the equation for the volume rate of flow,

we can determine the time required for the totalvolume V to flow by integrating. The result is

t = 8ηl ∫ dV/h.πgρa4

The integral is a constant for a given apparatus,which is determined by measuring the time ittakes for a solution of known density to fall fromh1 to h2. Typically one uses the pure solvent in which the macromolecule will be studiedsubsequently.

h

dV = πP ∫ (a2-x2) x dxdt 2ηl 0

a

h1

h2

CHEM 305

1) Large volume of solution is required.

2) Shearing forces generated by the flow gradient are large.

Shear stress S = F/A = η (dv/dx)

-can cause distortions in the coildistribution of flexible molecules, which in turn means that the viscositycan be altered.

The average shear stress in a capillary viscometer can be determined byusing the equation:

where we know that c1 = 0.

½ Px2 + c1 = -ηlx (dv/dx)

CHEM 305

This allows us to write

Sx = η (dv/dx) = - Px2l

for a cylindrical sheet of fluid with radius x. To obtain the average shear stress,we need to integrate the expression over all sheets, i.e.

<S> = ∫ 2πxl dx Sx∫ 2πxl dx

= 2πl (-P) ∫ x2 dx = -P a3

2l (2l)(3)2πl ∫ x dx a2/2

<S> = - Pa3l

0

a

0

a

shear stress depends onthe height of the capillary

Assumption: that the pressure remains constant during capillary viscosity measurement – not the case!

CHEM 305

To minimize shear stress, we can use a different type of viscometer, namely

ω

The relative viscosities of anytwo solutions is given by

η2 = ω2η1 ω1

The shear can be altered by changing thestrength of the applied magnetic field. Theshear stress is 104 less that in an Ostwaldviscometer.

CHEM 305

Effect of solute on viscosity

The equations presented up to this stage all relate to the solvent. Ifwe now include a solute, we have the complicated task of computinghow a particle distorts the flow lines of a solution containing avelocity gradient. We start by calculating the energy per unit time neededto maintain the shear in the parallel plate system

F

A

dvdx

h

F = Aηdvdx

Energy = F vb = Ahη dv 2

t dx

This allows us to definethe viscosity of the rateof energy dissipation perunit volume (Ah) at unitshear (dv/dx = 1),

dE α ηdt

CHEM 305

Einstein showed that the rate of energy dissipation in a dilute macromolecularsolution is defined by

dE = dE (1 + νφ)dt dt

where φ is the fraction of the solution volume occupied by macromolecules andν is a numerical factor related to the shape (like the Perrin factor, but not thesame value – ν = 2.5 for a sphere). Given that dE/dt is proportional to theviscosity, we can write

ηr = ηsolution = 1 + νφη0

where ηr is the relative viscosity and η0 refers to the pure solvent. We can nowdefine a specific viscosity as:

ηsp = ηr – 1 = νφ

What does ηsp mean physically?

solution solvent

CHEM 305

We can further define an intrinsic viscosity as

[η] = lim ηsp = lim νφc2 0 c2 c2 0 c2

Let us now rewrite φ in terms of a hydrated volume of the solute, Vh. Recall φis the volume fraction occupied by the solute molecules, i.e.

φ = Vh NA c2M2

Therefore, for a spherical solute,

ηsp = 2.5 (Vh NA c2) and [η] = 2.5 (Vh NA) in cm3.g-1M2 M2

or putting in the definition for the hydrated volume

[η] = 2.5 (V2 + δ1V1)

independent of molecularweight!!!!!!!!!!

CHEM 305

Ribonuclease A

Lysozyme

Bovine serum albumin

Hemoglobin

Bushy stunt virus

-all near spherical macromolecules- range of possible values for [η]: for DNA [η]=5000;

for tropomyosin [η]=52

ab

CHEM 305

Effect of solute shape on viscosity

As alluded to on the previous slide, the shape of the macromolecule has alarge effect on the measured viscosity (DNA and tropomyosin – which arerod-like particles – vs. spherical particles). If the shape can be modelled asa rigid ellipsoid, then the intrinsic viscosity is defined as:

The factor ν is a Simha factor and is defined as:

ν = (a/b)2 + (a/b)2 + 14 for prolate ellipsoids5[ln(a/b) – 0.5] 15[ln(2a/b) – 1.5] 15

ν = 16 a for oblate ellipsoids15 b tan-1(a/b)

[η] = ν (Vh NA) = ν (V2 + δ1V1) M2

ab

CHEM 305

Using viscosity to estimate molecular weight

By combining viscosity with sedimentation or diffusion measurements, itshould be possible to obtain a good estimate of molecular weight of a biomolecule, while eliminating most shape effects.

Starting from the friction coefficient:

f = 6πη 3Vh F4π

And combining it with the sedimentation coefficient:

6π 3 F = 4π

s = M2 [1 - ρV2]NA f

⅓ M2 [1 - ρV2]Vh

1/3ηNA s

CHEM 305

[η] = ν (Vh NA) M2

Using the definition for intrinsic viscosity:

We can divide the equation above (after taking the cube root) by the equationon the previous slide to yield:

NA1/3 ν1/3 = [η] M2

Vh6π 3 F

NA1/3 ν1/3 = [η]1/3ηNAs

(162π2)1/3F M22/3(1 – V2ρ)

M2 [1 - ρV2]Vh

1/3ηNA s

β’

Scheraga-Mandelkernequation

shape factor

CHEM 305 Ref: Cantor and Schimmel, p. 652

Not very sensitiveto shape

CHEM 305

The parameter β’ is not very sensitive to shape but since M2 α (β’)-3/2, themolecular weight derived from the Scheraga-Mandelkern equation should beaccurate to within 10%.

For most practical applications, the intrinsic viscosity is used to determinethe molecular weight by solving the equation:

[η] = k Ma

where k and a are constants specific to the system.

E.g. DNA (rod-like macromolecules)

[η] α M1.8

whereas for coils

[η] α M0.5 to M1.0

CHEM 305

Summary: Transport processes

- Diffusion

- Electrophoresis

-Sedimentation: - sedimentation velocity- equilibrium ultracentrifugation

- Viscosities

All of these methods can be used to yield information on the molecular weightof a biomolecule.

E.g. Lysozyme Method

Chemical structureSedimentation and diffusionSedimentation equilibriumViscosity (Scheraga-Mandelkern)

Molecular weight

14 211 g. mol -114 10014 50012 400

Sedimentation and diffusion

s = M [1 - ρV2] D = kT = RTNA f f NAf

M = sRTD [1 - ρV2]

Svedberg equation

f /fmin = [ ]1/3 F(V2 + δ1V1)V2

Sedimentation equilibrium

c2(x) = c2(x0) exp {[M2(1-V2ρ) ω2/2RT] (x2-x02)}

lnc 2

x

slope α M2

CHEM 305

[η] = ν (Vh NA) M2

Viscosity

NA1/3 ν1/3 = [η]1/3ηNAs

(162π2)1/3F M22/3(1 – V2ρ)

β’

Scheraga-Mandelkernequation

shape factor