Register to reply

Formula for calculating Electric potential of a solution from its ion concentration?

Share this thread:
dr_bin
#1
Aug30-08, 10:35 AM
P: 5
Hi everyone,

The subject of this topic is what I am looking for. For example if the concentration of a monovalent cation in a solution is 1 mol/L, then how much is the electric potential of this solution? (This solution just contains one kind of monovalent cation and not any anion)


Thanks in advance.
Phys.Org News Partner Physics news on Phys.org
Mapping the optimal route between two quantum states
Spin-based electronics: New material successfully tested
Verifying the future of quantum computing
Borek
#2
Aug30-08, 11:37 AM
Admin
Borek's Avatar
P: 23,397
Never heard about solutions containing just a cation. Are you sure about the question? I guess it is a chemical one, about the Nernst equation.
dr_bin
#3
Aug30-08, 12:45 PM
P: 5
Thank you Borek, the Nernst equation is perhaps what I am looking for.

Indeed I've never heard about solutions containing just a cation, too. I just assumed it.

Perhaps I should describe my idea as: [cation]-[anion]=1 mol/L. I think by biological means we can create a solution like that and isolate it. Because of the higher cation concentration this solution must have a positive electric potential and this potential is what I am trying to figure out. I'm not sure which one will be the determinant of this potential: the difference in ion concentrations( i.e, [cation]-[anion] ), or the quotient of them ( i.e, [cation]/[anion]).

I have viewed the Nernst equation and still have no idea of apply it in this task. Anyone has an idea?

Borek
#4
Aug30-08, 01:07 PM
Admin
Borek's Avatar
P: 23,397
Formula for calculating Electric potential of a solution from its ion concentration?

There is no such thing as charged solutions, or at least I have never heard about one. Each solution is electrically neutral. Potential is the effect of the redox reaction occuring in the solution. It is possible that charge is not uniformly distributed (double layer, cell membranes) but it doesn't mean solution is charged.
dr_bin
#5
Aug30-08, 01:29 PM
P: 5
We know that cell membranes have a "membrane potential" due to the uneven distribution of dissolved charged particles across the 2 sides of the membrane. So what happen if we withdraw the fluid of just 1 side of a biological membrane? Will we have a solution that is charged?
Borek
#6
Aug30-08, 01:48 PM
Admin
Borek's Avatar
P: 23,397
Try to separate such a solution, we will discuss it once you succeed. Apparently, if you take volume small enough, it may contain a single cation or single anion, it is still not the same as having charged solution. Solution and its properties are macroscopic.
atyy
#7
Aug31-08, 09:53 AM
Sci Advisor
P: 8,393
Quote Quote by dr_bin View Post
We know that cell membranes have a "membrane potential" due to the uneven distribution of dissolved charged particles across the 2 sides of the membrane. So what happen if we withdraw the fluid of just 1 side of a biological membrane? Will we have a solution that is charged?
I suppose the strict answer is yes. But not in practice.

Some points:

1) In a neuron not firing action potentials, there is excess negative charge inside the neuron. That means there is excess positive charge outside the neuron. Postive and negative charges attract, so the negative charge in the neuron is going to get as close to positive charge outside. Since negative charge cannot cross the lipid bilayer, the closest it can get to the outside is the inner surface of the lipid bilayer. This is why excess charge is associated with the "surface" of the membrane and not the "volume" of the cytoplasm. So for the strict answer, you must withdraw the volume *and* surface bits of the cytoplasm.

2) Strictly speaking, changing the electric potential of a cell membrane means putting more ions on one side of the membrane, which means changing the concentration of ions in the cytoplasm. However, the amount of excess charge needed to produce a biologically relevant change in potential is very small compared to a biologically relevant change in ion concentration, so we get an excellent approximation if we pretend that we can change electric potential without changing ion concentration. Now obviously if a neuron is very small, then it is easier to change the concentration of the cytoplasm. So the approximation can fail for small neurons (so I've heard).

3) You will see the Nernst equation being applied as if only one sort of ion were present. This is for changes in electric potential after the opening of a channel in the cell membrane that only allows one sort of ion to pass.
dr_bin
#8
Aug31-08, 10:52 PM
P: 5
Thank you, atyy.

Some points to your post:

1) As you have stated:
"excess charge is associated with the "surface" of the membrane and not the "volume" of the cytoplasm"
So I can infer that the concentration of charge is higher at the vicinity of the cell membrane than in the cytoplasm. Is it right? I don't know what method is used for measuring ion concentration of cells, but I guess it is either by withdrawal of some intracellular fluid for analysis or by introducing some kind of "probe" into the cell for a direct measurement. People have measured the ion concentration of biological fluid. For example the Na+ concentration of the intracellular and extracellular fluid of a resting neuron is 14mEq/L and 142mEq/L, respectively. As i have infered, the smaller the distance to the cell membrane, the higher is the ion concentration. So where in the cell did they obtain the measurement of 14mEq/L? And is it correct?

2)
the amount of excess charge needed to produce a biologically relevant change in potential is very small compared to a biologically relevant change in ion concentration, so we get an excellent approximation if we pretend that we can change electric potential without changing ion concentration
Actually this is what makes me started this topic. Textbooks also said like you but it was just a qualitative statement and I would like to check this statement quantitatively: How many ions need to pass the channel in order to change the local membrane potential from -90mV to about+35mV in an action potential?

3)
Quote Quote by atyy
You will see the Nernst equation being applied as if only one sort of ion were present
I think we can use the Goldman-Hodgkin-Katz equation in case of there are several different ions.
Borek
#9
Sep1-08, 03:10 AM
Admin
Borek's Avatar
P: 23,397
Quote Quote by dr_bin View Post
How many ions need to pass the channel in order to change the local membrane potential from -90mV to about+35mV in an action potential?
You may treat the membrane as a condenser, that'll move the question from the biology domain to pure physics.
atyy
#10
Sep1-08, 06:12 AM
Sci Advisor
P: 8,393
Quote Quote by dr_bin View Post
Actually this is what makes me started this topic. Textbooks also said like you but it was just a qualitative statement and I would like to check this statement quantitatively: How many ions need to pass the channel in order to change the local membrane potential from -90mV to about+35mV in an action potential?
OK, let me see if I can answer your second question first, since that's easier. Just in case I get stuck or give you the wrong answer, let me refer you to "Foundations of Cellular Neurophysiology" by Johnston and Wu, where they do the calculation in Chapter 1 or 2.

Anyway, we follow Borek's suggestion and treat the cell membrane as a capacitor. Why can we do this? A capacitor is two conducting plates separated by an insulator. In our case, the conducting plates are the ionic solutions inside and outside the cell, and the insulator is the cell membrane.

The equation for a capacitor is Q=CV.

An approximate value for the capacitance per membrane area is 1 uF/cm2 ("Biophysics of Computation" by Christof Koch). Let's say a cell is about 30 um = 0.003 cm in diameter, so its surface area is about 9X10-6 cm2. So C=9X10-12F.

And we want the change in membrane potential to be V=35+90mV=125mV=100mV (approx).

So the charge difference needed is
Q=CV=9X10-12F X 100mV = 900 X 10-15 coulombs = 1X10-12 coulombs (approx)

The charge on one ion is about 10-19 coulombs, which means about 107 ions will have to be moved.

Now the volume of a 30 um cell will be about 27000 um3. A 1 molar solution of singly charged ions about contains 1023 ions in 1 L. A 10 mM solution contains 1021 ions in 1 L, and 1021 X (27000 um3/ 1 L) = 3 X 1010 ions in 27000 um3.

So our error is 107/1010=10-3. That means if the solution was originally 10 mM, by changing the potential by 100 mV, the solution will become about 10.01 mM or 9.99 mM, which is close enough to 10 for most purposes.
atyy
#11
Sep1-08, 06:59 AM
Sci Advisor
P: 8,393
Quote Quote by dr_bin View Post
So I can infer that the concentration of charge is higher at the vicinity of the cell membrane than in the cytoplasm. Is it right?
Yes that's right.

Quote Quote by dr_bin View Post
I don't know what method is used for measuring ion concentration of cells, but I guess it is either by withdrawal of some intracellular fluid for analysis or by introducing some kind of "probe" into the cell for a direct measurement. People have measured the ion concentration of biological fluid. For example the Na+ concentration of the intracellular and extracellular fluid of a resting neuron is 14mEq/L and 142mEq/L, respectively. As i have infered, the smaller the distance to the cell membrane, the higher is the ion concentration. So where in the cell did they obtain the measurement of 14mEq/L? And is it correct?
The 140 mM NaCl for the extracellular fluid is the concentration of "Normal Saline" which people use in hospitals everyday. So we would know if this was drastically wrong! There are several other solutions they use in hospitals like Ringer's, but let's not discuss that here. This value does not include the solution near the cell membrane. In general, you have to take special care if you want to extract the surface charge (Like Borek, I actually don't know if that's technologically possible - we need a precision chemist or physicist to answer that). But as you can see, even if they had gotten the surface charge, the value would be say 141 mM or 139 mM instead of 140 mM. The variation in this value between healthy individuals is already greater than that!

The intracellular solution in a healthy neuron is about 14 mM NaCl. There are also other ions like potassium in the intracellular solution so that its osmolarity is the same as the extracellular solution. I don't know how people got the 14 mM value, although I use this number everyday! But I have reasons for thinking this number is correct. We can detect, without disrupting the cell membrane, the pattern of action potentials that occur in an auditory cortex neuron when you play a tone to an animal. We can also use a method called whole cell recording which records "intracellularly", and which we can use to change the intracellular solution of the neuron. When we replace the true intracellular solution of the neuron with artificial intracellular solution containing about 14 mM NaCl, the pattern of action potentials caused by playing a tone doesn't change. We can also do experiments where we use some abnormal artificial intracellular solution and disrupt the pattern of action potentials. The famous experiments of Hodgkin and Huxley (and others such as Cole) used very similar reasoning 60 years ago in the squid giant axon to figure out that the action potential was caused by Na rushing into the neuron.

Now, the intracellular solution of any neuron is not as well known as the extracellular composition. If we knew it, then we would be able to calculate the membrane potential of the neuron. Or if we knew the membrane potential of the neuron, we would be able to infer the intracellular solution. We know that the inside of a neuron at rest is about -70 mV relative to the outside of 0 mV. But when we make this measurement we usually disrupt the cell membrane, and so we don't actually know if it is -90 mV or -50 mV, and that can make a difference in some cases, as in this paper by Gulledge and Stuart:

----------
Gulledge AT, Stuart GJ. Excitatory actions of GABA in the cortex. Neuron. 2003 Jan 23;37(2):299-309.

Little is known about how GABAergic inputs interact with excitatory inputs under conditions that maintain physiological concentrations of intracellular anions. Using extracellular and gramicidin perforated-patch recording, we show that somatic and dendritic GABA responses in mature cortical pyramidal neurons are depolarizing from rest and can facilitate action potential generation when combined with proximal excitatory input. Dendritic GABA responses were excitatory regardless of timing, whereas somatic GABA responses were inhibitory when coincident with excitatory input but excitatory at earlier times. These excitatory actions of GABA occur even though the GABA reversal potential is below action potential threshold and largely uniform across the somato-dendritic axis, and arise when GABAergic inputs are temporally or spatially isolated from concurrent excitation. Our findings demonstrate that under certain circumstances GABA will have an excitatory role in synaptic integration in the cortex.
----------
atyy
#12
Sep1-08, 09:34 AM
Sci Advisor
P: 8,393
Quote Quote by atyy View Post
That means if the solution was originally 10 mM, by changing the potential by 100 mV, the solution will become about 10.01 mM or 9.99 mM, which is close enough to 10 for most purposes.
I should say what we mean by "close enough to 10 mM for most purposes". The values you gave of 14 mM and 140 mM for Na ion concentration are what you put into the Nernst equation. If the concentration changes, then you should use the new concentrations of say 14.01 mM and 139.99 mM instead. You will find that the new values don't change the potential calculated using the Nernst equation by much. As for "most purposes", it would be wise to check this each time you use the equation (in research, not in an exam).

Although I don't know how to extract the surface charge, people do address questions of exactly how close to the surface the charge is. This is important if you have a protein in the membrane, and some part of the protein is sensitive to a particular ion. For example, in http://www.jbc.org/cgi/content/full/270/34/19936, they mention "distance of the effective calcium activation site from the channel".
dr_bin
#13
Sep1-08, 11:22 AM
P: 5
Very excellent! Thank you very much atyy and Borek!

My primary concern is solved. Considering the cell as a capacitor, that's what I didn't think about. Even if I thought about it, I still didn't have the "capacitance" to calculate.

Your calculation of the number of ions is reasonable. But I still don't understand this statement:
Quote Quote by atyy
But as you can see, even if they had gotten the surface charge, the value would be say 141 mM or 139 mM instead of 140 mM.
How can we know that the higher concentration of charge at the cell membrane is just 1 mM? It seems that you apply the previous calculation in this statement, and I can hardly find any relevance here.

Sorry If I'm a bit annoying.

Thank you again for your interesting posts, and for your enthusiasm!
atyy
#14
Sep1-08, 07:07 PM
Sci Advisor
P: 8,393
Quote Quote by dr_bin View Post
How can we know that the higher concentration of charge at the cell membrane is just 1 mM? It seems that you apply the previous calculation in this statement, and I can hardly find any relevance here.
Yes, I applied the previous calculation here. We don't know the higher concentration at the cell membrane. The point is that we use the calculation to estimate how big our lack of knowledge is.
DaleSpam
#15
Sep1-08, 08:04 PM
Mentor
P: 16,987
Quote Quote by dr_bin View Post
We know that cell membranes have a "membrane potential" due to the uneven distribution of dissolved charged particles across the 2 sides of the membrane. So what happen if we withdraw the fluid of just 1 side of a biological membrane? Will we have a solution that is charged?
Let's say that you start with a membrane between two solutions, one is distilled water and one is a NaCl solution. Let's further say that the membrane is only permeable to Na+. Due to the concentration gradient some of the Na+ will cross the membrane. Thus the "pure" side will have an overall positive charge and the "salt" side will have an overall negative charge. Through simple electrostatic attraction the negative charges will attract the positive charges and will accumulate on the membrane. Because of how strong this attraction is, the bulk solutions will still remain neutral and all of the charge will reside on the membrane surface (as you would expect in a conductor).


Register to reply

Related Discussions
Calculating the electric field from the potential Introductory Physics Homework 5
Calculating electric potential Introductory Physics Homework 6
Calculating the Electric Potential from an Electric Field problem Introductory Physics Homework 1
Calculating the electric potential engery of the system Introductory Physics Homework 18
Calculating Electric Potential in A Uniform Electric Field Introductory Physics Homework 4