Biochemistry More on Osmosis

Hello and welcome back to Educator.com and welcome back to Biochemistry.0000

Today, we are going to talk a little bit more about osmosis.0004

The last lesson, we introduced this notion of osmotic pressure, and we said that when you separate a solution of something from the pure solvent by a semi-permeable membrane, something that allows the water molecules to pass but nothing else; that the water will spontaneously and naturally pass from the pure solvent into the solution to dilute it.0008

And, the extent to which it actually does so, we could actually assign a number to it, we can measure it and we call that the osmotic pressure.0031

The process is osmosis and the number itself, we call the osmotic pressure.0040

Today, I just want to say a little more about it, maybe do a couple of more examples, define a few terms just to round out.0045

This is actually a very, very important concept primarily because cells are surrounded by this cell membrane and that cell membrane is a semi-permeable membrane.0053

You have the solution inside the cell, and then you have the solution outside the cell.0067

It's very, very important that the osmotic pressure inside and outside the cell, for the most part, stay the same, otherwise, you are going to have problems.0071

OK.0082

Let's jump on in, get started.0084

OK.0086

Once again, we said - well, actually you know what, that's fine - we said that if you separate the solution from a pure solvent, you can measure this thing called osmotic pressure.0088

Well, you can take two different solutions that have nothing to do with each other, let's say a lactic acid solution and a glucose solution, and you can also separate those by a semi-permeable membrane and the same thing will happen.0103

Now, you have one solution of a given osmotic pressure, another solution of another given osmotic pressure and the same thing is going to happen.0119

Solvent will start to flow in the direction of the one that has the higher osmotic pressure in order to bring its pressure down, and then the other one, it's going to raise its pressure.0125

What we are trying to do is get to a point where the two solutions have the same osmotic pressure; in other words, have the same number of particles per liter, that way it's in equilibrium.0137

So, let's draw that out.0148

Let's go ahead and draw our little U tube here, something like this.0152

And then of course, we have our semi-permeable membrane, and let's say over here we have a glucose solution.0162

I'll use Xs for the solute, and then let's say over here we have our - let's just use a salt - magnesium chloride, so we have a magnesium chloride solution, and that's this one, and I'll just use dots for this.0173

OK.0192

Let's say that the osmotic pressure of this one happens to be 2.6 atm, and let's say that the osmotic pressure of this one, π is equal to 1.2 atm.0194

In this particular case, you have this magnesium chloride solution, and all this osmotic pressure means it has a greater concentration of solute particles.0206

There are more free solute particles floating around in this solution than magnesium chloride, than there are in the glucose0216

So, what's going to happen is water, solvent is going to spontaneously flow across the membrane in - let me use red here, it's going to flow in, oops, let me get red, make sure I get...OK, there we go - it's going to flow in this direction.0222

It's going to flow out of the solution for glucose and it's going to flow into this solution for magnesium chloride.0238

Well, as water depletes from this, the concentration of the glucose solution is going to rise.0244

OK, starts here.0252

Well, this one, because water is flowing into it, you are actually diluting this, so its concentration is going to come down.0254

Water will keep flowing until they achieve the same concentration; until they achieve the same osmotic pressure.0261

And now, at this point it has reached its equilibrium and is perfectly happy.0267

So again, you can separate two different solutions, it doesn't have to be a solution and pure solvent.0271

Water will still spontaneously flow in the direction of the solution that has the more number of particles in it in order to dilute it.0275

It's always going to be that way- that is osmosis.0286

OK.0290

Let's go ahead and do an example here.0294

Let's go ahead and keep this red ink.0299

The numerical measure of osmotic pressure is used to determine the molar mass or the molecular weight of proteins, and this is going to be our first example.0304

Example 1: OK, 0.001g of a newly purified protein, so you've gone ahead and run through the process of protein purification, you've purified it, everything is good, and now, you need to measure its molar mass.0329

So, 0.001g of this protein was dissolved in enough water to make 1.00mL of solution.0354

So, we have this protein solution.0372

Now, the osmotic pressure of this solution was measured.0374

Sol'n is a shorthand for solution.0383

It was measured to be 1.22 Torr.0391

Torr is the unit of pressure.0399

The Torricelli is the same as millimeters of mercury, so there is 760 Torr in 1 atm.0400

We happen to measure it and it is 1.22 Torr at 25°C.0406

Now, again, when we measure osmotic pressure, we're measuring the solution against a pure solvent.0411

This is a measure, and the standard by which we measure it is from pure solvent.0417

That's what's happening here.0422

OK.0425

Now, the question is: "What is the molar mass of this protein?".0426

OK.0430

What is the molar mass of this protein?0431

OK, it's a great problem.0440

Well, again, take a look at what it is that you want- molar mass.0442

What is the definition of molar mass?0445

It's grams of whatever per mole.0447

Let's write down the definition of molar mass, so we know what it is that we're actually looking for.0454

Molar mass is equal to the grams of the protein divided by the moles of the protein.0461

It's g/mol- that's the unit of molar mass.0472

Well, we already have the mass of the protein; now, let me go ahead and go to blue here.0475

We have the mass, that is 0.001g.0480

Now, all we need is to find the moles of the protein, so now we need to find the denominator.0486

OK.0491

Let's go ahead and just circle a couple of things.0493

We have 1mL of solution, we are at 1.22 Torr.0495

Those are the other numbers that are given to us, so we need to find moles of protein.0501

OK, so here's what we are going to do.0504

We are going to use R, so let me go ahead and write moles of protein, this is what we are looking for.0508

We are going to use our equation: π = iMRT.0514

The osmotic pressure equals the van 't Hoff factor times the molarity times the gas constant times the absolute temperature; and now what we're going to do is we are going to manipulate this equation.0519

We are going to rearrange it and solve for moles of protein given other information, and again, what you have in the equation, you have 1,2,3,4,5 parameters.0530

Well, if you're given any four of them, you can find the fifth.0539

It just depends on what the particular problem is asking for.0542

OK.0546

Let me write this out as i x mol/L, so I'm going to actually write out this molarity thing as its full unit: i x mol/L x R x T.0547

We want that.0567

We want the moles of the protein, so I'm going to rearrange this equation and solve for mole; and when I do, I get the following.0569

When I move everything over, I get the osmotic pressure times the volume divided by iRT, that is equal to the number of moles.0576

So, that's it.0588

I have the osmotic pressure, that's 1.22 Torr.0589

I have the liters, that's 1mL of solution.0592

I have i, the van 't Hoff factor, in this particular case it is 1, because it's not an ionic compound, a protein is a covalent compound so it's just 1.0595

I have R, and I have T, temperature is 25°C, R is the gas constant, so I can solve for mole.0606

The only thing I have to be careful of, though, have to watch the units.0613

Remember the gas constant R is equal to 0.08206 Latm/molK.0617

Liter is fine; I can change the 1mL to liter.0626

Kelvin is fine, 25°C is going to be 298K.0629

The pressure atmosphere- so the unit when you use this equation, it has to be atmospheres but notice the pressure was given to us in Torricelli, so we have to do that conversion first, so let's go ahead and do that conversion.0633

We have 1.22 Torr x 1 atm happens to be 760 Torr, and the Torr cancels Torr, and we're left with is 0.001605 atm.0647

There we go.0664

Now, we just plug everything in, and I'm going to go ahead and do it up here.0665

OK, so let me go ahead and use red.0670

So, our osmotic pressure is 0.001605 - OK, let me go ahead and do the units - so you see, that is going to be atmosphere.0674

Volume, liter, that's going to be 1mL, so it's a 0.001L, and down here, the van 't Hoff factor is 1, and this is 0.08206, that's Latm/molK, and of course, the temperature is 25°C which is 298K.0686

Now, let's see, K cancels K.0710

L on top cancels L on the bottom.0715

Atm cancels atm.0717

What we are left with is 1/mol in the denominator.0720

Well, 1/mol on the denominator is mole in the numerator, so the units work out just fine.0723

When we do this, we end up with the following: the number is 6.56 x 10^-8mol.0728

There we go.0738

We found the number of moles of the protein.0739

Now, we can solve for our molar mass.0742

Molar mass of our protein, we said we had 0.001g of it, and we now know based on the osmotic pressure measurement, we have 6.56 x 10^-8mol; and when we do this division, we get 1.5 x 10⁴ g/mol.0746

This is a very large protein.0773

There you go.0776

That's it.0778

Osmotic pressure is used in the laboratory to actually measure the molar mass of a protein.0779

OK.0787

Now, let's do a couple of other definitions here.0790

Let me go back to blue.0794

We are going to define something called osmolarity.0797

Now, it is not necessary to actually define this thing, but it is something that comes up.0801

We have been talking in terms of osmotic pressure and like we said: when you separate two solutions by a semi-permeable membrane that have two different osmotic pressures - in other words, one has more particles in it floating around than the other - solvent is going to flow from the solution of lower osmotic pressure to the solution of higher osmotic pressure, in other words, the one that has more particles floating around it in order to dilute it, in order to make the concentrations equal.0807

Well, osmolarity is i x M.0831

You had the equation osmotic pressure = iMRT, if we just take this part iM, we call that the osmolarity.0838

OK.0846

Let's go ahead and do an example here, and I think it'll make sense.0848

The unit of osmolarity - as you'll see in a minute - the unit of osmolarity is particles per liter, and you'll see how that works out in just a second when we do the example.0852

So again, we can speak of two solutions that have different osmolarity, or we can speak of two solutions that have different osmotic pressure.0870

It's actually the same thing.0877

The only difference is you're multiplying osmolarity by this constant RT, or by the two constants; but when you multiply them together, it's going to be some constants, some number.0878

That's it.0888

Osmolarity and osmotic pressure are just two different ways of looking at the same thing- a certain number of particles per liter.0889

That's all that is happening here, so make sure that you have the general idea.0895

It's just which has the greater number of particles floating around.0899

You can call it osmolarity, that's one number; or you can use osmotic pressure, it's a different number- it doesn't matter.0903