Quick viewing(Text Mode)

Phase Equilibria: How to Calculate Oxygen Solubility in Cell

Phase Equilibria: How to Calculate Oxygen Solubility in Cell

BE.360J/10.449J SUPPLEMENTARY HANDOUT FALL 2002

PHASE EQUILIBRIA:HOW TO CALCULATE OXYGEN IN CELL CULTURE MEDIUM

1. Phase Equilibria Introduction A collection of molecules is in equilibrium with its surroundings if there is no net transfer of energy or matter across the boundaries to the environment. Individual molecules may constantly be zinging across the boundaries, or bumping into neighbors and transferring heat, due to thermal motion. But if we average over millions of molecules, the number of molecules leaving the volume of interest ("control volume") gets cancelled out by an equal number returning to the control volume, so the net motion across the boundary is zero. This situation is analogous to chemical reaction equilibrium -- when equilibrium is reached, the rate of the forward reaction is equal to the rate of the reverse reaction, and there is no net measurable change in the of reactants or products with time. In each case, the equilibrium is dynamic, meaning that forward and reverse reactions are constantly occurring. However, the overall measurable properties of the system do not change with time, and we consider the system at equilibrium1. Definitions and Measuring Equilibria (1) Consider first the situation where Phase I and Phase II are pure components A and B (phases are liquid, , or gas). Imagine that you put A & B together in a closed vial, shook the vial vigorously, then waited an "infinitely" long time, keeping the vial at constant temperature. At infinity, you should observe 2 separate phases (e.g., a layer of salt crystals at the bottom of a layer of ; oil layered on top of water, or liquid/gas). [Note -- if you do not observe 2 phases at infinite time, then you did not add enough of the component in the missing phase…i.e., you did not reach its solubility limit. Go back and add more until you see two phases at infinity…] You then carefully withdraw a small sample of Phase II and determine the concentration of A in it. This concentration is defined as the "solubility" of A in B -- it is the most A that will ever dissolve in B at that temperature. You also withdraw a small amount of Phase I and measure the concentration of B in it. This gives you the solubility of "B" in "A" at that temperature. How do you know you are at equilibrium? It is

1 [Note -- we reserve the term "equilibrium" for a system where no net transfer of stuff is occurring. We use "steady state" to refer to a system where there may be net transfer of mass or energy to designate that the rate of transfer does not change with time. Steady state of course also implies that concentration profiles or temperature profiles do not change with time. An equilibrium system is at steady state, but a steady state system is not always at equilibrium]

-1- not always easy to tell! Systems come to equilibrium over anywhere from a few moments to a few days. Equilibrium is determined by careful sampling, always keeping the vial at a constant temperature. When the no longer change with time (may require many many points to determine you have a flat line within statistical significance), you are at equilibrium.

If "A" (Phase I) is a liquid and "B" (Phase II) is a gas, the equilibrium concentration of A in B is usually reported as a vapor pressure of A, and is often designated as a "saturation" vapor pressure. Antoine's equation is a common correlation of the vapor pressures of various compounds as a function of temperature. Gasses (such as oxygen and nitrogen) are typically only sparingly soluble in liquids and their are typically reported as Henry's Law coefficients. Liquid/liquid solubilities are often reported as volume percents (vol%). [For example, when water and oil are mixed, a small amount of water dissolves in the oil, as in the homework problem]. The solubilities of in liquids are often reported as weight percents (wt%). There is no set, standard reporting structure, so just look at units carefully.

(2) Now consider that Phase I and Phase II are pure components A & B and a small amount of component "C" is added. Component C distributes throughout the two phases according to its relative solubility in each phase. Visualize a container filled with an oil/water mix (A&B), to which a brilliant blue hydrophilic dye ("C") is added. The water phase will appear deep blue, but the oil phase will appear only faintly blue. Hydrophilic means "water-loving," and the dye partitions into the aqueous phase. [Note that such dyed, oil/water-filled devices were a commercial fad not too long ago as "instant ocean" (usually, the "oceans" contained a tiny air bubble to allow for thermal expansion/contraction).] Akeyfactorin understanding phase equilibria is to realize that there are two possible routes to determining the concentrations of the dye (aka, component "C") in each of the two phases. Both methods give you a parameter known as the partition coefficient -- the relative solubility in one phase compared to the other, or ceqm = = I eqm K partition coefficient eqm ,wherecI denotes the concentration of C in phase I cII eqm when the system is at equilibrium, and cII denotes the concentration of C in phase II when the system is at equilibrium.

METHOD ONE: Conduct experiments such as those in part (a) where you determine the solubility of C in A and B respectively. In other words, you add C to A so that you have two phases, and measure the concentration of C in A to get its solubility, and do the same thing for C mixed with B. An alternative that saves on vials is to mix A, B, & C to get 3 phases (imagine adding enough dye to the "ocean" that it precipitates out) and measure the concentration of C in the A & B phases. The partition coefficient is then the ratio of the = solub ility limit so lubility limit solubilities of C in Phase I and Phase II respectively, or K cI / cII .A subtle assumption we make with this approach is that the solubilities of A&B in each other do not affect the solubility of C in either phase. This approach is generally used if you have tabulated solubilities of C in A and C in B, but no tabulated partition coefficient. So, if you are told for example that the solubility of the dye in water (phase I) is 100 mM and in the oil it is 0.1mM, the partition coefficient is 100/0.1 = 1000.

-2- METHOD TWO: Simply measure the concentration of C in each phase after the system is at equilibrium, even if C is well below its solubility limit in the two phases. The of C has to be the same in each phase, and thus the ratio of concentrations in the two phases is always the partition coefficient. For example, you add a small amount of the dye to the oil/water mix -- say, just enough to get a visible blue color in the water -- and find that the concentration in the water (phase I) is 20 mM. You then measure the concentration in the oil phase and find that it is 0.02 mM. Then K = 20/0.02 = 1000. You then decide to add more dye to get a deeper blue, and find that the concentration in the water phase (after the system came to equilibrium) is now 45 mM. You measure the concentration in the oil phase and find that it is 0.045mM. The partition coefficient is 45/0.045 = 1000. You could take a similar approach to determine the Henry's law coefficient for oxygen in water -- add water and a nitrogen/oxygen mix to the vial, and measure the final equilibrium concentrations in each phase as the nitrogen-oxygen ratio is varied.

2. Mass Transfer Across Interfaces

Mass transfer is inherently a non-equilibrium process, characterized by the net motion of molecules down a concentration gradient. We study mass transfer at the continuum scale (i.e., anything we do is averaged over zillions of molecules), and the concentration profile of the diffusing solute is continuous within a single phase. Often, mass transfer occurs across the interface between two phases -- liquid/liquid (e.g., oil/water), liquid/gas (e.g., air/water), liquid/solid (e.g., water/polymer film), or gas/solid (e.g., air/polymer film). In this class, we will consider only diffusion of dilute solutes and "slow" mass transfer rates. The nature of the interface is not disrupted due to convective flow of material across the interface under these conditions. At the interface between two phases, the chemical potential of the diffusing solute in phase I must equal that in phase II. We imagine that the interface between the two phases is an infinitely thin plane and that the phases are in equilibrium in this plane. There is typically a discontinuity in concentration at the interface. Here at the interface, the concentration of the diffusing solute A in Phase I is related to that in Phase II via a partition coefficient, Henry's law coefficient, vapor pressure, etc. Note that the concentration of A at the interface is often well below its solubility limit, but that partition coefficients, etc., are simply the relative concentrations in each phase at equilibrium.

-3- OXYGEN SOLUBILITY IN CELL CULTURE MEDIA and CELLULAR OXYGEN CONSUMPTION RATES

The solubility of gasses in liquids is correlated in terms of “Henry’s Law,” which says the concentration of gas dissolved in the liquid is linearly proportional to the partial pressure (or concentration) of the gas in the vapor phase:

Hcliq =Pgas Values of H for oxygen dissolved in select liquids: H (atm-L/mmol) Pure water (15°C) 0.65 Pure water (25°C) 0.79 Pure water (35°C) 0.92 1M NaCl (25°C) 1.12 Cell culture medium (37°C) 1.08

For cell culture medium in equilibrium with pure air at 37°C, the concentration of dissolved oxygen can be calculated as:

CO2 = (0.21 atm)/ H = (0.21 atm)/(1.08atm-L/mmol) = 0.19 mmol/L

Cell culture media typically use carbon dioxide at 5-10% in the gas phase as a buffer, and the air is usually humidified to prevent evaporation. The vapor pressure of water in air at 37°C is

47 mm Hg (0.06 atm). Correcting for 5% CO2 and water in the gas phase (air is 21% oxygen):

CO2 = (0.21 atm)(1 – 0.06 – 0.05)/ H = (0.187 atm)/(1.08atm-L/mmol) = 0.17 mmol/L oxygen solubilities in Blood and Plasma: Table 41 (p 90) in Fornier:

arterial: PO2 = 95 mm Hg, plasma = .130 mM, as oxyhemoglobin = 8.5 mM

venous: PO2 = 40 mmHg, plasma = 0.054 mM, as oxyhemoglobin = 5.82 mM

Typical specific oxygen consumption rates for mammalian cells are: Rat hepatocyte 8.4- 11 x 10-8 mol/cm3/sec. (cell diam = 25 microns = 4.3 - 5.7 10-8 mol/cm3 cells-sec.)

-4- Islets: Unstimulated 2.6 10-8 mol/cm3 cells-sec Stimulated: 4.6 10-8 mol/cm3 cells-sec. (beta cell diameter ~12 microns)

9 Hybridoma (transformed lymphocyte, ~10 microns diameter): 0.3-0.4 mmol O2/10 viable cells/hr (Miller et al., 1988)

Given the oxygen consumption rate of mammalian cells – oxygen is rapidly consumed and needs to be replenished continuously.

-5-