--%>

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 : Acid Solutions Choose the right answer

    Choose the right answer from following. Volume of water needed to mix with 10 ml 10N NHO3 to get 0.1 N HNO3: (a) 1000 ml (b) 990 ml (c) 1010 ml (d) 10 ml

  • 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 : Sedimentation and Velocity The first

    The first method begins with a well defined layer, or boundary, of solution near the center of rotation and tracks the movement of this layer to the outside of the cell as a function of time. Such a method is termed a sedimentary velocity experiment. A

  • Q : Problem on making solution Select the

    Select the right answer of the question. The weight of H2C2O42H2O required to prepare 500ml of 0.2N solution is : (a) 126g (b) 12.6g (c) 63g (d) 6.3g

  • Q : Soluation of Ideal Gas Law problems

    Explain the method, how do you solve Ideal Gas Law problems?

  • Q : Explain various chemicals associated

    During processing of food, several chemicals are added to it to augment its shelf life and to make it more attractive as well. Main types of food addi

  • Q : Mole fraction of hydrogen Give me

    Give me answer of this question. In a mixture of 1 gm H2 and 8 gm O2 , the mole fraction of hydrogen is: (a) 0.667 (b) 0.5 (c) 0.33 (d) None of these

  • Q : Molality of a glucose solution What

    What will be the molality of a solution containing 18g of glucose (having mol. wt. = 180) dissolved in 500g of water: (i) 1m  (ii) 0.5m  (iii) 0.2m  (iv) 2m

  • Q : Unit of molality Select the right

    Select the right answer of the question. The unit of molality is: (a) Mole per litre (b) Mole per kilogram (c) Per mole per litre (d) Mole litre

  • Q : Determining highest normality What is

    What is the correct answer. Which of the given solutions contains highest normality: (i) 8 gm of KOH/litre (ii) N phosphoric acid (iii) 6 gm of NaOH /100 ml (iv) 0.5M H2SO4