Fluids, solutes and the cell

Goals:

1. Calculate the molarity, osmolality or osmolarity of a solution

2. Explain and use the dissociation constant to calculate osmolality

3. Explain the processes of diffusion and osmosis when cells are placed in solutions of different concentrations.

4. Predict the ion and water movements across a semi-permeable membrane when given information about the selectivity of the membrane and the concentrations of solutes on either side.

5. Visualize the effects of osmolarity on artificial membrane and living cells.

Tutorial:

Particles in solution are always in constant random motion. As a consequence the particles will tend to move from regions of high concentration to regions of lower concentration, along a concentration gradient. This process is known as diffusion. Both particles in a solution and water molecules move randomly in the solution. When the movement of particles is blocked by a semi-permeable membrane, they can not move to the other side of the membrane.  Water molecules, smaller than the particles are able to move and will do so in response to their own concentration gradient.  This process is known as osmosis.  Living organisms are subject to the same physical and chemical laws, with the cell membrane being the divider between the two regions containing potentially different concentration solutions – the intracellular and extracellular compartments. Clearly, if water moves across the cell membrane, either into or out of the cell, the volume of the cell will change. If water leaves the cell, it shrinks (a process known as crenation). If water enters the cell, it enlarges. If too much water enters the cell, it will burst (in red blood cells, this is known as hemolysis).

In order to understand osmosis and its effect on cell volume it is important to be clear about the terms and descriptions of the properties of solutions that influence these fluid movements. Solutions we encounter in everyday life have their concentrations expressed in a variety of ways. Some of the most common are:

Molarity                                   Example: 1 M sucrose

Amount/Volume                       Example: 2 gm/L

Percent solution                        Example: 0.9% NaCl

A Mole is 6.02 x 1023 atoms of a compound. The number "6.02 x 1023" is the Avogadro's number (NA).  The molecular weight of a compound is the weight of 1 mole or the weight of 6.02 x 1023 atoms of this compound.  Ex:  the molecular weight of Carbon-12 = 12 = NA x (mass of carbon-12). For further reading on this subject: http://gemini.tntech.edu/~tfurtsch/scihist/avogadro.htm

http://encyclopedia.laborlawtalk.com/Atomic_mass_unit

Molarity is defined as the number of moles per liter of solution (mol/L or M). The weight of a mole is equal to the atomic mass or molecular weight of the substance, expressed in grams (gram molecular weight). Molecular weight is obtained by adding the atomic weights of each atom in the molecule. What is the molecular weight of glucose, C6H12O6? (Atomic weights: carbon = 12, hydrogen = 1, oxygen = 16) __________________________

Remember not to confuse molarity with molality which is the number of moles of solute in 1 Kg of solvent (not total solution). The density of water is 1.00 g/ml, thus 1 Kg of water occupies a volume of 1 liter. Therefore, to make a 1 molar (1M) solution of NaCl you would take 1 mole of NaCl and add sufficient to make a final total volume of 1 liter. To make a 1 molal (1m) solution, you would simply add 1 liter of water to 1 mole of NaCl (total volume thus greater than 1 liter). How you would make up 1 liter of 2M NaCl solution:

Osmolarity is defined as the concentration of a solution expressed in terms of osmotically active particles, or osmoles (osmol/L or OsM).  Osmotically active particles are any solutes that contribute to the concentration gradient. These can be either intact, uncharged molecules or charged ions.  Physiological solutions are very dilute and are usually expressed in milliosmoles/L (mOsM). To convert between molarity and osmolarity, use the following equation:

Molarity (mol/L) X # of particles/molecule = osmolarity (osmol/L)

How would you make up 500 ml of a 100 mOsm NaCl solution?

Your answer above probably relies on the assumption that every molecule of NaCl can dissociate completely into separate Na+ and Cl- ions.  In reality, not every molecule will dissociate in solution. For example, one mole of NaCl in solution actually gives a 1.8 OsM solution, not a 2 OsM solution. This is because not every NaCL molecule dissociates into two ions. The way to calculate the actual osmolarity of a solution is to use the dissociation constant of the molecule rather than the number of particles per molecule according to the following:

Molarity (mol/L) X dissociation constant = osmolarity (osmol/L)

Knowing this, recalculate how you would make up 500 ml of a 100 mOsM NaCl solution.

What is the osmolarity of two liters of 1 M NaCl mixed with one liter of 3 M glucose?  It is possible to compare the osmolarities of any two solutions simply by comparing their concentrations in osmoles/liter.

To compare osmolarities, make sure the two solutions being compared have the same units, then compare numbers. Follow these rules:

A.     Find the total concentration of the resulting solution.

B.     Find the concentration of NaCL in moles.

C.     Find the concentration of NaCl in osmoles.

D.     Find the concentration of glucose in moles and osmoles.

Osmosis and Cells

-     if solution A has more particles/unit volume than solution B, we say that solution A is hyperosmotic to B

-     this also means that B has fewer particles/unit volume than A, so we can say that B is hyposmotic to A

-     If A and B have the same osmotic concentrations, we say they are isosmotic

Fill in the blanks in the following table with the correct term:

 Solution A = 1 OsM glucose Solution B = 2 OsM glucose Soln. C   = 1M NaCl  =          ___OsM A is _____________to B A is _____________to C B is _______________to A B is _______________to C C is _______________to A C is _______________to B

Laboratory Experiments (for today)

1-  Diffusion, Osmosis and the dialysis bags

You will observe the movement of water across a semipermeable membrane (dialysis bags).

a- Prepare the following solutions:

1- solution 1: 10% M glucose

2- solution 2: 10% M albumin

3- solution 3: 10% NaCl

4- solution 4: 10% glucose + 10% albumin + 10% NaCl (final concentration)

Explain how you would prepare each solution.  Calculate the concentration in % w/v and osmolarity of each solution.

b- Wet four dialysis bags (15 cm long) so they become pliable and do not break. Tie one end of a dialysis bag with a knot.  Fill the bags with solutions 1, 2, 3, and 4.  Tie a Pasteur pipette with the thin end broken off to the other end of the bag.  The solution in the bag should reach the base of the Pasteur pipette.  Take care not to spill the solution and make sure the bag is not leaky.

Let the dialysis bag sit for 2 hours in water.  Every half an hour, measure the high of the liquid in the Pasteur pipette.

d- At the end of 2 hours, test for the presence of glucose, NaCl, starch in the beaker and albumin in the dialysis bag.

- Use glucose sticks to measure the amount of glucose in a solution.

- Use sodium nitrate in to measure the amount of NaCl.  If you have a refractometer, you can use it.

- Use a simple BCA assay to see if any bluish color appears.  A urine stick is adequate also.

e- Make a chart, with the dialysis bag volume on the Y axis, the time on the X axis.   Analyze your results in light of the solutions present in the beaker and the dialysis bag.

Questions:

1. Did any diffusion, osmosis, or both occur? Explain. Which chemical diffused out?  Which ones played a part with osmosis? Support your conclusions with data obtained from this experiment.

2. Assume that instead of placing a dialysis bag in tap water that it was placed in a solution containing 5 M glucose.  The dialysis bag contains solution 4. What will happen to the following:

a. Glucose

b. Albumin

c. Water

d. Will the dialysis bag gain or lose weight?

2- Diffusion. Osmosis and your red blood cells

a- Prick the tip of your finger and deposit a small drop of blood on 5 glass slides.  Quickly, mix the blood with the following 5 solutions:

1) tap water, 2) 0.3% NaCl, 3) 0.9% NaCl, 4) 3% NaCl, 5) 1% glucose, 6) 5% glucose,

7) 10% glucose and 8) 0.3% NaCl + 5% glucose

Note: You will mix your own solutions.

b) Observe and draw the behavior of the red blood cells.

c) Explain your results in light of diffusion and osmosis.  Which solutions are hypotonic, isotonic and hypertonic to the RBCs?

Potential test questions

1.      Because the osmotic pressures of a solution depend on solute concentration, osmotic water movements are interrelated with solute behavior. Make a diagram to show the direction of water and solutes.  Decide and then explain why you concluded the following statements are true or false.

a)      If glucose concentrations in solution A and B separated by a semi permeable membrane are the same, the net movement of water will be zero.

b)      If glucose concentration in solution A is half that of solution B, twice as much water will leave solution B compared to that from solution A.

c)      The osmotic pressure of a 2M solution of urea is the same as that of a 1M solution of potassium chloride.

d)      To make a 5 milliosmolar solution of sodium sulfate, 50g of the salt are placed in a liter of water.

2.   Calculate the osmolality for each of these six solutions 0.2, 0.5, 0.6, 0.8, 0.9, 1.1% NaCl.  Which concentration is closest to that of plasma (290 mOsm/kg)?

3.   Assume the dissociation constant for NaCl is 1.7.

a)      Show how you would prepare 750 ml of a 0.9% NaCl solution?

b)      Would the above solution be hyper-, iso-, or hypo-osmotic to a 55mM KCl solution?

4.   Describe and explain what would happen to a cell containing 35mM NaCl, 150 mM glucose, and 100 mM protein if placed into:

a)   distilled water? (describe the volume change if the cell were permeable to water only).

b)   a solution of 35mM NaCl, IF the cell were permeable to water only.

c)      a solution of 35mM NaCl, IF the cell were permeable to salt and glucose as well as water.

d)      A solution of 200mM glucose, IF the cell were permeable to water only.

5. Imagine you have 2 chambers connected by an opening - one chamber contains 100ml of a 10mM NaCl solution (A), and the other (B) contains 100 ml of a 100mM NaCl solution? If the opening permits both water and NaCl to cross:

a)  describe and explain which way the water and the Na+ and Cl- ions would flow?

b)  describe and explain what the volumes and concentrations of the two compartments would be at equilibrium?

c)  Describe and explain how the result would differ if the opening were permeable to water only?

6. Red blood cell membranes are completely impermeable to sodium, potassium and chloride ions; however they are permeable to both water and urea.  The concentration inside a red blood cell is 300mOsm and consists predominantly of sodium and potassium chloride.

a)   With the aid of diagrams, decide what would happen to red blood cells when placed in the following solutions.

i.       Fresh water.

ii.       Salt solution at a concentration of 150mOsm.

iii.       Salt solution at a concentration of 300mOsm.

iv.      Salt solution at a concentration of 400mOsm.

v.       A urea solution at a concentration of 300mOsm.

b)   Which solutions were isosmotic to the red blood cells?

c)      Which solutions were isotonic to the red blood cells?

7. Suppose a salt and a glucose solution are separated by a membrane that is permeable to water but not to the solutes. The NaCl solution has a concentration of 1.95 g per 250 ml (molecular weight = 58.5). The glucose solution has a concentration of 9.0 g per 250 ml (molecular weight= 180).  Calculate the molarity, millimolality, and milliosmolality of both solutions. State whether or not osmosis will occur and, if it will, in which direction. Explain your answer.

8. When the body needs to conserve water, the kidneys excrete a hypertonic urine. What do the terms isotonic and hypertonic mean? Since the fluid that is to become urine begins as an isotonic solution, what must happen to change it to a hypertonic urine?

9. The receptors for thirst are located in a part of the brain called the hypothalamus. These receptors are osmoreceptors—they are stimulated by an increase in blood osmolality. How would dehydration lead to a sense of thirst? What effect would drinking seawater have on his sense of thirst? Explain.

10. Before the invention of refrigerators, pioneers preserved meat by salting it. Explain how meat can be preserved by this procedure. (Hint: Think about what salting the meat would do to decomposer organisms, such as bacteria and fungi.)