USMLE Forums banner

1 - 12 of 12 Posts

·
Registered
Joined
·
2 Posts
Discussion Starter #1
Hi everyone!
I don't know what would happen to the RBC in this situation. Hope you help me guys!!! Confuses me a lot when they say NaCl and Urea (Just NaCl is okay, but...)
 

·
Registered
Joined
·
86 Posts
Hi everyone!
I don't know what would happen to the RBC in this situation. Hope you help me guys!!! Confuses me a lot when they say NaCl and Urea (Just NaCl is okay, but...)
Normal serum osmolality is about 295 mOsmol/kg H2O.

If those are two separate solutions:

In 200mM NaCl RBCs would swell due to influx of water to water down the hyperosmolar solution inside the cells and to concentrate the solution outside the cells.

In 600mM urea RBCs would shrink due to efflux of water to water down the hyperosmolar solution outside the cells and to concentrate the hypoosmolar solution inside the cells.

If this is a mixed solution you can not say anything based on this information. You need to know the ratio of the volumes of solutions mixed. If you mixed 1 liter of NaCl solution with 1 liter of urea solution you would have a different osmolarity as opposed to mixing 100 liters of NaCl solution with 1 liter of urea solution.
 

·
Registered
Joined
·
10 Posts
Normal serum osmolality is about 295 mOsmol/kg H2O.

If those are two separate solutions:

In 200mM NaCl RBCs would swell due to influx of water to water down the hyperosmolar solution inside the cells and to concentrate the solution outside the cells.

In 600mM urea RBCs would shrink due to efflux of water to water down the hyperosmolar solution outside the cells and to concentrate the hypoosmolar solution inside the cells.

If this is a mixed solution you can not say anything based on this information. You need to know the ratio of the volumes of solutions mixed. If you mixed 1 liter of NaCl solution with 1 liter of urea solution you would have a different osmolarity as opposed to mixing 100 liters of NaCl solution with 1 liter of urea solution.
Sorry, but that is not correct.

Whether water moves in or out of the cell, depends on the "effective osmolarity" of the solution in which it is being kept. The cell maintains itself at roughly 300 mOsm (notice the unit). The ECF osmolarity is calculated mainly by sodium, it's;

(2 X [Na] + 10) Osm (again, notice the unit)

"Effective Osmolarity" is simply how concentrated the solution is, but it is only because of those solutes that do not cross the cell membrane. Urea easily crosses the cell membrane, so when a cell is placed in such a solution, the urea will quickly distribute evenly between the cell and the solution, thus no change in effective osmolarity, ergo no water movement, which means NO change in cell size.

about the NaCl.. yup this is osmotically active (actually the one most quoted too).. already mentioned the formula to calculate ECF osmolarity. but the mistake Nodo made was in not noticing the units. it says 300 mM, not osm. which means that its telling you how much Na is there in moles.. the osmolarity needs to be calculated,

2 X [Na] + 10
2 X [200] + 10
~410 osm (now we have the right unit)

So in truth, this is a rather concentrated solution, and water will move out of the cell, which will end up shrinking.

To summarize,

RBC in 200 mM NaCl ---> RBC shrinks
RBC in 600 mM urea ---> no change

This got rather lengthy, sorry about that. But i hope it helps!
 

·
Registered
Joined
·
86 Posts
Sorry, but that is not correct.

Whether water moves in or out of the cell, depends on the "effective osmolarity" of the solution in which it is being kept. The cell maintains itself at roughly 300 mOsm (notice the unit). The ECF osmolarity is calculated mainly by sodium, it's;

(2 X [Na] + 10) Osm (again, notice the unit)

"Effective Osmolarity" is simply how concentrated the solution is, but it is only because of those solutes that do not cross the cell membrane. Urea easily crosses the cell membrane, so when a cell is placed in such a solution, the urea will quickly distribute evenly between the cell and the solution, thus no change in effective osmolarity, ergo no water movement, which means NO change in cell size.

about the NaCl.. yup this is osmotically active (actually the one most quoted too).. already mentioned the formula to calculate ECF osmolarity. but the mistake Nodo made was in not noticing the units. it says 300 mM, not osm. which means that its telling you how much Na is there in moles.. the osmolarity needs to be calculated,

2 X [Na] + 10
2 X [200] + 10
~410 osm (now we have the right unit)

So in truth, this is a rather concentrated solution, and water will move out of the cell, which will end up shrinking.

To summarize,

RBC in 200 mM NaCl ---> RBC shrinks
RBC in 600 mM urea ---> no change

This got rather lengthy, sorry about that. But i hope it helps!
Totally agree on the urea part. My mistake.

However, mM means a millimolar, not millimol. A capital M is a unit for molarity which is a notation for mol/liter. So 200mM for NaCl actually define concentration, not the total amount of solute.

Now on to the mOsm business. You never use mOsm because it is the measure of the total amount, rather you use mOsm/kg of water which is the measure of concentration. Osmole is a term used to describe the amount of solute in moles that contributes to the osmotic pressure. In the case of NaCl moles=osmoles.

Molarity is a ratio of number of moles of solute and the volume of solution in liters (mol/liter). Since 1 kg of water is roughly 1 liter and since in the case of NaCl mOsm=mmol, in this case no conversions are necessary(mol=osmole, kg=liter).

200mM (or 200mOsm/kg of H20) solution will be hypoosmolar for RBCs and they will swell.

This part does not concern the question of RBC swelling.

The formula that you used is used to approximate serum osmolarity. It is not to be used anywhere else. It is used to estimate serum osmolarity by using sodium concentration alone. It takes into account that Na is by far the most highly concentrated cation and the fact that serum carries a net zero charge meaning that there needs to be the same concentration of anions (Cl, HCO3) in the serum. Hence, 2 times the sodium concentration. You get the osmolarity created by sodium and exactly the same amount of osmolarity created by anions, which you should add together. And then you have that plus 10 because there are other things in serum contributing to osmolarity. In this example the only solute in the solution is NaCl and only NaCl will contribute to osmolarity, so there is no need to use any formulas, concentration of NaCl is equal to the osmolarity of the solution.

When you use the formula be careful to use concentration of sodium, if you are saying that 200mM depicts the number of moles of sodium in the solution, you can not use that number in that formula.

Great point on urea, though. 100% correct on that one.
 

·
Registered
Joined
·
10 Posts
Totally agree on the urea part. My mistake.

However, mM means a millimolar, not millimol. A capital M is a unit for molarity which is a notation for mol/liter. So 200mM for NaCl actually define concentration, not the total amount of solute.

Now on to the mOsm business. You never use mOsm because it is the measure of the total amount, rather you use mOsm/kg of water which is the measure of concentration. Osmole is a term used to describe the amount of solute in moles that contributes to the osmotic pressure. In the case of NaCl moles=osmoles.

Molarity is a ratio of number of moles of solute and the volume of solution in liters (mol/liter). Since 1 kg of water is roughly 1 liter and since in the case of NaCl mOsm=mmol, in this case no conversions are necessary(mol=osmole, kg=liter).

200mM (or 200mOsm/kg of H20) solution will be hypoosmolar for RBCs and they will swell.

This part does not concern the question of RBC swelling.

The formula that you used is used to approximate serum osmolarity. It is not to be used anywhere else. It is used to estimate serum osmolarity by using sodium concentration alone. It takes into account that Na is by far the most highly concentrated cation and the fact that serum carries a net zero charge meaning that there needs to be the same concentration of anions (Cl, HCO3) in the serum. Hence, 2 times the sodium concentration. You get the osmolarity created by sodium and exactly the same amount of osmolarity created by anions, which you should add together. And then you have that plus 10 because there are other things in serum contributing to osmolarity. In this example the only solute in the solution is NaCl and only NaCl will contribute to osmolarity, so there is no need to use any formulas, concentration of NaCl is equal to the osmolarity of the solution.

When you use the formula be careful to use concentration of sodium, if you are saying that 200mM depicts the number of moles of sodium in the solution, you can not use that number in that formula.

Great point on urea, though. 100% correct on that one.
The formulae are correct, but you're interpreting them wrong. I didn't want to go into the molarity description so omitted that part.

But since you mentioned it, here's what you're missing, A mole of NaCl gives not one, but 2 osmoles i.e. a Na+ and a Cl- . Which is why that formula of concentration even in serum holds here as well. No matter where you dissolve 1 mM Nacl, you'll get a conc of 2 mosm (per volume of that solution)

So, yeah.. 200 mM Nacl (ie 200 mmoles of NaCl/L) will equal ~400 mosmoles/L

So the cell goes: O -> o

In case i m missing something here, plz do correct me.
Thanks!
 

·
Registered
Joined
·
161 Posts
Wayfarer is right and Nodo is wrong.
NaCl forms 2 ions in water. Na+ and Cl- respectively.
Hence you multiply 200 x 2 which gives you 400.
This is explained in Kaplan.

When converting mM to mOsm for Step 1 purposes, multiply mM by the number of ions formed by the same substance in the solvent which is usually water.

Also beware of substances like urea in the question, which do not contribute to osmolality.

This is all you need to know.
 

·
Registered
Joined
·
86 Posts
The formulae are correct, but you're interpreting them wrong. I didn't want to go into the molarity description so omitted that part.

But since you mentioned it, here's what you're missing, A mole of NaCl gives not one, but 2 osmoles i.e. a Na+ and a Cl- . Which is why that formula of concentration even in serum holds here as well. No matter where you dissolve 1 mM Nacl, you'll get a conc of 2 mosm (per volume of that solution)

So, yeah.. 200 mM Nacl (ie 200 mmoles of NaCl/L) will equal ~400 mosmoles/L

So the cell goes: O -> o

In case i m missing something here, plz do correct me.
Thanks!
OK. Two comments lol.

1. You are correct again in saying that 200mM of NaCl actually gives an osmolarity of 400. I had to convert mM to mOsm/L by multiplying by two.

2. I still do not understand the use of the formula. I might be wrong on this one as well but let me explain where I stand. The way I understand the formula, and what I said in one of my previous posts, is that the formula you mentioned is used to approximate serum osmolarity by using sodium concentration alone. Again, I could be wrong but I just do not understand the usage of that formula in this situation. All you need to do in this case is convert mM to mOsm/L (2x molarity of NaCl).
 

·
Registered
Joined
·
86 Posts
That is exactly what he said.
The formula is for ECF osmolality which he mentioned as well.
I guess I will rephrase my question. What is the logic behind using a formula for calculating ECF osmolality when we are not dealing with ECF?
 

·
Registered
Joined
·
4 Posts
when you are talking about whether the rbc,s will shrink or swell in a given solution so whatever the concentration of any ions dissolved in that solution would be percieved as an ECF thats bcz thats out of the cell thats what Extracelluler fluid is so we will simply use the equations as we use for ECF.hope it helps
 

·
Registered
Joined
·
86 Posts
when you are talking about whether the rbc,s will shrink or swell in a given solution so whatever the concentration of any ions dissolved in that solution would be percieved as an ECF thats bcz thats out of the cell thats what Extracelluler fluid is so we will simply use the equations as we use for ECF.hope it helps
Technically you are right. However, the formula is only true for ECF present in the body. ECF in the body has many components in addition to sodium and chloride. The solution in this question only has Na and Cl. Therefore the osmolarity in the body ECF is approximately equal to 2x[Na] + 10 but in this question it is equal to [Na] + [Cl]. Plus 10 is for the other components of ECF in the body, there is no plus 10 in this question, it only contains sodium and chloride.

I made many mistakes in this thread, but I still do not understand why we need to use the ECF formula in this question. Maybe I am wrong again.
 
1 - 12 of 12 Posts
Top