--%>

Donnan Membrane Equilibria

The electric charge acquired by macromolecules affects the equilibrium set up across a semipermeable membrane.

Laboratory studies of macromolecule solutions as in osmotic pressure and dialysis studies confine the macromolecules to one compartment while allowing passage of small ions or solvent in or out compartment. Much of the transport occurring in cells and cell compartments in living systems can be similarly described. In all such cases, the equilibrium state that would be reached as a result of the net transport of the small ions can be markedly affected if the macromolecule carries a charge, as is generally the case.

Except at the isoionic pH, proteins and nucleic acids carry a charge as a result of a net gain or loss of protons. Additional charges are acquired by the binding of other species, e.g. the binding of Mg2+ ions by phosphate groups. Thus, macromolecules in laboratory or biological systems generally carry a charge. The overall electrical neutrality of the solution results from a corresponding opposite charge contributed by ions, called counterions, included in the remaining ionic make up of the solution.

Suppose such a macromolecule or, specifically, a protein solution is separated from pure water by a semipermeable membrane that allows passage of small ions but prohibits the passage of protein molecules. Such a situation could arise in an osmotic pressure study or in the dialysis of the protein solution. Suppose the protein carries a net negative charge and that Na+ ions are the counterions. The Na+ ions will tend to diffuse to the low concentration region of initially pure water. Electrical neutrality would be lost and this process prevented if it were not for the dissociation of water. This occurs, and H+ ions tend to accumulate on the proteins side of the membrane while the corresponding OH- ions accumulate, along with the buffered, pH charges will occur to upper the osmotic pressure or dialysis experiment.

In such ways are led to deal with the equilibrium between protein solutions, which are often themselves buffered, and buffer solutions. The complication arise can be illustrated by considering the simplest situation of the protein-sodium-ion solution separated by a semipermeable membrane from a sodium chloride solution.

Suppose the proteins species P carries a negative charge of -z. the neutrality of the solution is achieved by the presence of z positive charges, Na+ ions for example, for each protein concentration is cP, as the initial Na+ concentration in the protein compartment is zeP.

Species concentration in a Donnan-membrane equilibrium study:

368_donnan membrane.png 



Rearrangement leads to x, the concentration of chloride that develops in the protein compartment:

At large salt concentrations, the effect of the protein is overwhelmed and x = 1/2cs. The two compartments achieve equal salt concentrations. At large a protein concentration, however, the passage of salt into the protein compartment is prevented, even though this rejection of the chloride ion by a solution that contains none of that ion.

Donnan-membrane equilibrium calculated from the above equation for z = 1:

2230_donnan membrane1.png 

The effects of various concentrations of protein and electrolyte are shown in the table. Only at high concentration relative to the protein concentration is the effect of the confined charged protein small. Therefore many studies of proteins or other polyelectrolytes in solution are made at high electrolyte concentration and at a pH near the isoionic point.  

   Related Questions in Chemistry

  • Q : What is Elevation in boiling point? The

    The boiling of a liquid may be defused by the temperature at which its vapour pressure which is equal to atmospheric pressure. The effect of addition in a non-volatile solute on the boiling point shown and its solution containing non-volatile solute with tempe

  • Q : Cons of eating organic foods Illustrate

    Illustrate the cons of eating organic foods?

  • Q : Hybridization Atomic orbitals can be

    Atomic orbitals can be combined, in a process called hybridization, to describe the bonding in polyatomic molecules. Descriptions of the bonding in CH4 can be used to illustrate the valence bond procedure. We must arrive a

  • Q : Explain the process of adsorption in

    The process of adsorption can occurs in solutions also. This implies that the solid surfaces can also adsorb solutes from solutions. Some clarifying examples are listed below: (i) When an aqueous solution of ethano

  • Q : Molarity of sodium hydroxide Can

    Can someone please help me in getting through this problem. Determine the molarity of a solution having 5g of sodium hydroxide in 250ml  solution is: (i) 0.5  (ii) 1.0  (iii) 2.0   (d) 0.1Answer: The right answer i

  • Q : Composition of the vapour Choose the

    Choose the right answer from following. An ideal solution was obtained by mixing methanol and ethanol. If the partial vapour pressure of methanol and ethanol are 2.619KPa and 4.556KPa respectively, the composition of the vapour (in terms of mole fraction) will be: (

  • Q : What are isotonic and hypotonic

    The two solutions which are having equivalent osmotic pressure are called isotonic solutions. The isotonic solutions at the same temperature also have same molar concentration. If we have solutions having different osmotic pressures then the solution having different

  • Q : Problem on equilibrium composition The

    The catalytic dehydrogenation of 1-butene to 1,3-butadiene, C4H8(g) = C4H6(g)+H2(g) is carried out at 900 K and 1 atm.

    Q : How to calculate solutions ionic

    Transference numbers and molar conductors can be used to calculate ionic mobilities. This tables under is giving the transference numbers for positive ions at 25 degree C and the values obtained by extrapolation to infinite dilution:

    Q : Problem on molarity-normality-molality

    Can someone please help me in getting through this problem. The solution ofAl2(SO4)3 d = 1.253gm/m comprise 22% salt by weight. The molarity, normality and molality of the solution is: (1) 0.805 M, 4.83 N, 0.825 M (2)