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BIOLOGY 207 LABORATORY 2
SOLUTIONS, DIFFUSION AND OSMOSIS
STUDENT LEARNING OBJECTIVES
- Define various solutions in terms of solute and solvent.
- Learn to calculate solution concentrations based on osmolality and percent solutions.
- Understand solubility in the context of polar, nonpolar and ionic solutes and solvents, and understand how solubility affects membrane transport and other aspects of physiology
- Describe diffusion, osmosis, osmotic pressure and tonicity from the perspective of cells.
- Examine cells exposed to various hypotonic, hypertonic and isotonic solutions. Understand the changes in cells after exposure to these solutions and how this relates to basic human physiology. GENERAL STATEMENT REGARDING SUBJECT MATTER Solutions The concepts of solute, solvent and solution are very important in physiology. To increase your familiarity with these concepts, the first part of this laboratory deals with different ways of expressing concentrations of solutions and serial dilutions. In this lab, you will be required to do fundamental calculations regarding concentrations. These calculations will allow you to become familiar the concept of solutions and concentration. It is imperative that you understand the various concepts in this section, as you will be expected to use them throughout the semester. Diffusion, Osmosis, Tonicity Solubility, diffusion, osmosis, and the property of tonicity are important in human physiology. For instance, you have several hormones that contribute to maintaining a nearly constant osmotic condition in your bodily fluids. A rapid change in the osmotic properties of your bodily fluids can be harmful, especially in very young and very old persons since the homeostatic mechanisms required to maintain osmotic properties may be sluggish or not fully developed. Numerous proteins embedded in the plasma membranes of your cells allow for the diffusion of ions and other molecules down their concentration gradients, and your cells use these gradients to move materials into and out of them. In contrast, the movement of nonpolar molecules and gases into and out of your cells relies on simple diffusion which doesn’t involve proteins. Finally, tonicity is a concept that describes the movement of water into or out of cells as dictated by osmotic pressure and solute concentration. We will explore the effect of isotonic, hypotonic and hypertonic solutions on cells. Again, these concepts will be with you all semester so it is important that you become familiar with them. READING in HUMAN PHYSIOLOGY , 14th^ Edition, Stuart Ira Fox. molarity and molality: pp. 138-139 osmolality and tonicity: pp. 139- diffusion, osmosis and osmotic pressure: pp. 133-138 chemical bonds and hydrophobicity and hydrophilicity: pp. 26- selective permeability of membranes: p. 52-54, 132-133, much of chapter 6 LAB 2 CONVERSIONS & FORMULAS YOU MUST MEMORIZE & KNOW HOW TO USE FOR LAB QUIZ 1 AND LAB EXAM 1: For water, 1 kg = 1 L % solution = (grams solute/ml solvent) x 100 C 1 V 1 =C 2 V 2 1000 ml = 1 L Osmolality = molality x # particles in water 10 dl = 1 L Milliosmolality = osmolality x 1000 Ask you lab instructor if they will require that you memorize any additional conversions, values or formulas.
Ex. 2.1: Solutions and Calculating Concentrations (background for calculations on p. 2.10)
A solution is a homogenous mixture of two or more substances. The substance present in the greatest amount is called the solvent , while the substances present in the smaller amounts are called the solutes. Another way of describing the relationship of solute and solutes is that solutes are the molecules that are dissolved in the solvent. For instance, in a salt water solution, the salt is the solute and the solvent is the water. In physiology, the solutions that we deal with are usually liquid solutions, in which water is the solvent, and the solutes can be a variety of molecules such as salts (e.g. NaCl), sugars (e.g. glucose), proteins, and gases. Our bodies are full of solutions. Our cells are filled with a solution known as the cytoplasm. In the blood, our cells are bathed in a solution known as the plasma, and in our tissues are cells are surrounded by a solution known as the interstitial fluid. Concentration of solutions : The concentration of a solution relates the amount of solute to the amount of solvent in a solution (concentration = solute/solvent). In physiology you will frequently encounter concentrations. For instance, fasting blood glucose levels must be maintained at 75-100 mg/dl, normal saline solutions found in IV fluids are at 0.9% NaCl, pH is related to the molar hydrogen ion concentration in a solution, and blood plasma has a set point of around 300 mOsm. In the tonicity activity today we’ll see that the concentration of solutes inside and outside the cell is critical to cell shape, function and survival. Concentration always relates the amount of solute to solvent in a solution, but the amounts can be expressed as mass, volumes, or even in terms of the number of atoms, molecules or particles. Mass is the amount of matter. Kilograms (kg), grams (g) and milligrams (mg) are all examples of units of mass. Volume is the amount of space that matter takes up. Liters (L), deciliters (dl), and milliliters (ml) are all examples of units of measurements of volume. Following are some common ways that concentration can be expressed: a) Weight per volume e.g. mg/dl and mg/100 ml are common weight per volume units used in physiology b) Percentage (% ) solutions Percentage solutions are a subtype of a weight per volume concentration. Percentage means "parts in 100." For example, percentage is the number of grams of solute dissolved in 100 ml (or a dl)* of solution. It can be expressed as follows: percentage = (grams solute/volume of solvent in milliliters) X 100 EXAMPLE 1: a 9% solution of sodium chloride (NaCl): 9% NaCl = (9 grams of NaCl/ 100 ml of solution) X 100 EXAMPLE 2: a 0.09% solution of glucose is the average human blood concentration of glucose. 0.09% = (90 mg of glucose/ 100 ml of blood) X 100 Note: 90 mg (milligrams) is the same as 0.090 grams. A milligram is 1/1000 of a gram so to get milligrams, you multiply the number of grams by 1000. Thus, 0.09 grams in milligrams is 0.09 X 1000 or 90 milligrams (mg).
When we discuss osmotic pressure and tonicity, we’ll see that determining the osmolality of a solution is useful for predicting the movement of water across a semi-permeable membrane, and thus the shape, function, and survival of the cell. For this activity, you will complete the calculations and questions on p. 2.9.
Ex. 2.2: Serial Dilutions (background for calculations on p. 2.10)
To save space, we often store small quantities of very concentrated solutions and then dilute the concentrated solutions with water. We do this in everyday life. For instance, if you buy liquid laundry detergent, your wash machine dilutes the concentrated soap solution with water. We also often need to dilute concentrated solutions in the laboratory. To make a dilute solution, the following formula applies: M 1 V 1 = M 2 V 2 , where M 1 = concentration of solution 1, V 1 = volume of solution 1 that you need to dilute with water, M 2 = concentration of solution 2, and V 2 = volume of solution 2. When making a number of dilute solutions, it is sometimes tedious and inaccurate to repeatedly weight out very small quantities of a compound. The serial dilution method enables us to rapidly and accurately prepare several more dilute solutions after making one concentrated solution. EXAMPLE: For example, you may wish to obtain 100 ml of 0.001 m NaCl. It is easiest to prepare a so-called parent solution of 1.0 m NaCl first and then dilute it to make a 0.1 m NaCl solution, and use the 0.1 m NaCl solution to make the 0.01 m NaCl solution, and then finally use the 0.01 m NaCl solution to make a 0.001 m NaCl solution. a. To prepare a 0.1 m NaCl solution from a 1.0 m NaCl stock: (1.0 m)(V 1 ) = (0.1 m)(100 ml) à V 1 = 10 ml Place 10 ml of 1.0 m NaCl in a container and then add 90 ml of water for a final volume of 100 ml. b. To prepare a 0.01 m NaCl solution from a 0.1 m NaCl stock: (0.1 m)(V 1 ) = (0.01 m)(100 ml) à V 1 = 10 ml Place 10 ml of 0.1 m NaCl in a container and then add 90 ml of water for a final volume of 100 ml. c. To prepare a 0.001 m NaCl solution from a 0.01 m NaCl stock: (0.01m)(V 1 ) = (0.001 m)(100 ml) à V 1 = 10 ml Place 10 ml of 0.01 m NaCl in a container and then add 90 ml of water for a final volume of 100 ml. Now perform the serial dilution calculations and fill out the diagram on p. 2.10. Once your group has checked your work with your instructor, follow the procedures below. PROCEDURES FOR SERIAL DILUTION ACTIVITY:
- Label 3 test tubes 0.2 M, 0.04 M, and 0.008 M.
- Fill a 250-ml beaker with tap water and bring it back to your table to be used for the serial dilutions.
- The concentrated 1 M dye stock can be found on the countertop. Using the pipette labeled “1 M stock” to measure out the correct volume of 1 M concentrated dye stock solution into the appropriate test tube. Return to your table and continue with the serial dilution steps that you diagramed on p. 2..
- Once you’re done, rinse the test tubes and pipettes and save them for the next class. Then answer the remaining questions on p. 2.. SOLUBILITY Whether or not a solute molecule is soluble in a solvent will determine whether they will mix. The principle “ like dissolves like ” applies to solubility. For instance, polar molecules, such as sugar molecules, dissolve well in polar water, whereas oils and other nonpolar molecules do not mix with polar water. In addition, ionic molecules, such as salts, are held together by ionic bonds rather than polar covalent bonds, ionic molecules dissolve well in polar water due to the attraction between the charged ions and the partial charges on the atoms within polar covalent bonds of water. Thus, ions and polar molecules are hydrophilic (waterloving), while lipids like oils are hydrophobic (water-fearing). The solubility of a particular solute molecule in its solvent is critical for the transport of substances through the blood and across the plasma membranes of your cells. Water-soluble molecules (ions and polar molecules) are able to travel freely through the blood, whereas nonpolar molecules that are not soluble in the watery solvent of the plasma membrane must be bound to special carrier proteins when they travel through the blood. Solubility also affects the movement of molecules in and out of the cell. The plasma membrane , which surrounds the cell, consists of proteins embedded in a phospholipid bilayer. The lipid portion of the phospholipid bilayer serves as a barrier to many types of substances. Only lipid-soluble (nonpolar, hydrophobic) molecules can “dissolve” in the lipid layer of the plasma membrane and directly cross the membrane by simple diffusion. In contrast, polar molecules and ions cannot dissolve in the lipid layer, and therefore are unable to cross. Cells still need to move polar molecules like glucose and ions like Na+ in and out of the cell, and in order to do so they rely on protein channels, transporters and carriers, among other processes to transport these types of substances.
Ex. 2.3 : Solubility Demo: Mineral Oil and Water (background for p. 2.11)
This activity will demonstrate the concept of the solubility differences in nonpolar and polar solutions. This activity will be done as a demonstration by the lab instructor. Your responsibility will be to understand what happened and to answer the questions regarding this demonstration. Your lab instructor will mix about 2 milliliter mineral oil and about 2 milliliters of water in a test tube. What should happen when your lab instructor mixes the two substances? Why? Since the mineral oil and water are both clear, colorless liquids, we can’t easily distinguish between the two substances by sight alone. To help you determine which layer is water and which layer is the nonpolar liquid, a colored salt (ionic dye) will be added to the test tube containing the nonpolar liquid and water. If the colored salt is ionic, which in which layer will it dissolve? Why? Which layer is which? Turn to p.2. to answer the questions.
- Obtain two shell-less eggs from the 10% NaCl solution, and GENTLY rinse them under running water.
- GENTLY pat the eggs dry.
- Place a plastic 250-ml beaker on the scale, and zero the scale by pressing “TARE.” Once the scale is zeroed, place the one egg in the beaker and record the mass in the table on p. 2.12.
- Remove the beaker containing the first egg and cover the egg with salt (or sugar).
- Place a second plastic 250-ml beaker on the scale, and zero the scale by pressing “TARE.” Once the scale is zeroed, place the second egg in the beaker and record the mass in the table on p. 2..
- Remove the beaker containing the second egg from the scale and cover the egg with distilled water (white tap) from the sink.
- Let the eggs stand for one hour.
- After 1 hour, gently rinse and pat dry each egg (make sure you don’t mix them up!) and record the new mass of each egg.
- Return the eggs to the 10% NaCl solution, so they can be reused by the next class.
Ex. 2.4B : Thistle Tubes and Osmosis Demo (background for p. 2.13)
In this exercise you will explore the concepts of both osmosis and osmotic pressure. In addition, you will see how differences in concentration gradients can affect the rate of the movement of molecules. Two thistle tubes are filled with two different concentrations of sugar solutions. The two solutions differ in the concentration of sugar (the solute) and thus water (the solvent). At the bottom of each thistle tube is a selectively permeable membrane that allows water, but not sugar to cross. The thistle tubes will be lowered into beakers of pure water, and you will measure changes in the height of the liquid over an hour. You should see that the liquid level goes higher faster in one tube. By observing the change in the height of the liquid in each thistle tube, and by applying what you know about osmosis and osmotic pressure, you will determine which tube has the higher sugar concentration. Turn to p.2.13 to answer the questions. TONICITY The tonicity of a solution refers to that solution’s effect on the osmotic movement of water when it is separated from another solution by a semi-permeable membrane. Since water moves towards areas of higher concentrations of osmotically active particles, determining the osmolality of a solution is useful for predicting the movement of water across a semi-permeable membrane, and thus the tonicity of a solution. Recall that osmolality takes into account the number of particles that a molecule becomes when dissolved in
water, and thus, indicates the concentration of osmotically active particles in a solution (see p. 2.3 for more explanation of osmolality. If a solution has the same osmolality as the solution inside the cell, it is termed isotonic , and there will be no net of movement of water in or out of cell. Thus, the cell will retain its shape, and function well. If a solution has a lower osomolality than the solution inside the cell, it is termed hypotonic , and there will be a net movement of water towards the inside of the cell where there is a higher concentration of osmotically active solute particles. Thus, the cells will expand, and possibly lyse (burst). Expansion of cells, and thus the tissues made up by the cells, is dangers since when tissues swell, they can begin to press up against bones and other structures, impairing organ function (e.g. imagine the effects of the brain swelling and pressing up against the skull). In addition, a lysed cell is a dead cell. Finally, if the solution has a greater osmolality than the solution inside the cell, it is termed hypertonic , and there will be a net movement of water out of the cell where there is a higher concentration of osmotically active solute particles. Thus, the cell will crenate (shrink), and not function properly.
Ex. 2.4D : Red Blood Cells and Tonicity (background for pp. 2.14-2.15)
As we have learned, the plasma membrane of the cell will allow water to pass fairly rapidly, but not NaCl or glucose. In this exercise, you will use non-human red blood cells to determine the osmotic properties of three different NaCl solutions. You will determine which solution is HYPERTONIC, which is ISOTONIC, and which is HYPOTONIC by placing red blood cells in the solutions and then observing the shape of the cells under the microscope. PROCEDURES for RED BLOOD CELLS AND TONICITY:
- Label 3 test tubes A, B and C.
- Use the correctly labeled pipette (A, B or C) to add 2 ml of the appropriate sodium chloride (NaCl) solution (A, B or C) to the appropriate test tube.
- Add one drop of blood to each of the 3 test tubes containing the NaCl solution.
- Gently mix the solution in each tube by holding the test tube with one hand and gently tapping the side of the test tube as demonstrated by your lab instructor.
- Label 3 slides A, B and C, and then add a drop of each suspension to the appropriately labeled slide.
- Place a coverslip over the drop of blood as demonstrated by your lab instructor.
- Turn on the microscope and increase the light intensity until you can see that the light under the stage is on.
- Lower the stage as far as it will go using the large knob (course focus) on the side of the microscope. Then secure the slide on the stage of the microscope, while being careful to not let the slide touch the objective lens above the stage.
- Use the knobs (x-y knobs) below the stage to center the slide over the light source below.
- With the shortest objective lens (4X scanning power) in place, use the large knob (course focus) on the side of the microscope to focus bring the cells into focus. At this power, the cells will look like many tiny, gray dots.
- Once the cells are in focus, shift the 10X objective lens into place. Now use the small knob (fine focus) to bring the cells into focus. [DO NOT USE THE LARGE KNOB (COURSE FOCUS) at with the 10X lens in place because you could damage the lens or slide]. At this power the cells will appear as slightly larger gray dots.
- Then shift the 40X objective into place, and use the small knob (fine focus) to bring the cells into focus.. [DO NOT USE THE LARGE KNOB (COURSE FOCUS) at with the 40X lens in place because you could damage the lens or slide]. At this power, the cells will still be tiny, but you should
0.2m(1) = 0.2 0sm - 0.2 Osm(1000) = 200 mOsm b) What is the milliosmolality of a 0.2 m CaCl 2 solution, assuming that the atoms in CaCl 2 are held together by ionic bonds? Show your work! c) If the 0.2 m glucose solution and 0.2 m CaCl 2 were on two different sides of a semi-permeable membrane that was only permeable to water, towards which solution would the water move? (see background on p. 2.6) Water would move towards the 0.2 m CaCl2 solution, because it has a higher solute concentration and therefore has a lower solvent concentration. This allows water to move towards it.
Ex. 2.2: Serial Dilutions (from pp. 2.4)
Imagine that you have a 1 M concentrated dye solution, but you need 5 ml of 0.008 M dye solution. However, going directly from a 1 M to a 0.008 M solution would require measuring volumes that would be too small to accurately measure. Thus, you need to perform a series of dilutions. M= Molarity; V= Volume ; M1 x V1+ what you start with; M2 x V2= what you want to end up with
4. Beginning with the 1 M concentrated dye solution, calculate the series of dilutions needed to make 5 ml each of first a 0.2 M , then a 0.04 M and finally a 0.008M solution. Make sure they’re serial dilutions! a) Show your work for each of the three serial dilution calculations. 0.2 M: (1M)V1=(0.2M)(5ml) --> V1= 1ml —> 1ml stock, 4 ml water 0.04 M: (0.2M)V1=(0.04M)(5ml) --> V1= 1ml —> 1ml stock, 4 ml water 0.008 M: (0.04M)V1=(0.008M)(5ml) --> V1= 1ml —> 1ml stock, 4 ml water
b) Label the diagram below to indicate how much of which concentration you would dilute with what quantity of water to prepare the solution! Add 4 ml of water to each tube Check your work with your instructor, and then prepare your solutions as described on p.2.4.
**5. a) What happened to the color of the solution as you prepared each dilution?
7.** The color of the solution would get lighter as we prepared each dilution. b) Explain this observed color change of the solutions you prepared in terms of the ratio of the relative number of dye particles to water for each dilution. ANSWER: the ratio decreases.
Ex. 2.3 : Mineral Oil and Water Solubility Demo (from pp. 2.5)
6. How many layers formed when your instructor mixed the mineral oil and water? Explain why this happened in terms of the chemical properties of the substances and solubility! Mineral oil and water when mixed, form 2 layers. They are not soluble in each other. Water is a polar solven. 7. a) After your lab instructor added the dye (an ionic compound), which layer (top or bottom) became colored?_______________ bottom part.
0.2 M 0.04 M 0.008 M
Ex. 2.4A Shell-less
Eggs and Osmosis (from
p. 2.6-2.7) (^) Selectively permeable membrane of shell- less egg
10. a) What happened to the once dry salt (or sugar) surrounding Egg #1 after the hour incubation? The salt stayed outside the egg because the salt is not able to cross the semipermeable membrane of the egg. Only small molecules such as water ions can cross through it b) Based on the appearance of the salt (or sugar) after incubation, which material moved through the egg’s membrane: salt (or sugar) **or water?
- Did Egg #1 gain or lose mass?_____lost______________. Based on your understanding of osmosis,** **explain why this makes sense.
- Did Egg #2 gain or lose mass?____gained_______________ Why did the mass of Egg #2 change?** EXPLAIN in terms of the i) egg’s membrane permeability to the salt (or sugar), water, and proteins (i.e. what allows to cross or not easily cross the membrane, assuming that the egg’s membrane is similar to a cell’s plasma membrane) ii) relative concentration of osmotically active solutes inside versus outside the egg, and iii) the direction of the movement of the molecule to which the membrane was permeable. Water & proteins yolk Initial mass (g) Final mass (g) Egg #1 –egg in salt (or sugar)
71.78g 50.24g
Egg #2 –egg in water
70.26g 71.04g
Ex. 2.4B Thistle Tubes and Osmosis Demo (from pp. 2.6-2.7)
13. Record the height of the liquid in each of the thistle tubes at the time intervals indicated below. Time (minutes) Height of purple liquid (mm) Height of green liquid (mm)
0 6.5 cm 6.5 cm
30 10.5 cm 7.5 cm
45 12.5 cm 8 cm
60 14.5 cm 8.5 cm
14. Which tube (purple or green) has the highest solute (sugar) concentration? Explain how you made this conclusion in terms of the height of the liquid in each tube after 60 minutes, osmotic pressure of each solution and the movement of water. Green. 15. APPLICATION QUESTION : Edema is tissue swelling that occurs when a high level of fluid accumulates in the tissues. Although there are many causes of edema, one way that edema can occur is due increased protein concentration in the interstitial fluid of the tissues during inflammation or thyroid disease. Why would this situation result in edema? Filtration of the fluid into the tissue spaces from the blood capillaries will occur continuously.
ii) 0.9 g/dl NaCl = 0.15 m = ______________Osm =_______________mOsm = ii) 0.9 g/dl NaCl = 0.15 m= mOsm Osm iii) 0.15 g/dl NaCl= 0.026 m = _______________Osm =______________mOsm = ii 0.15 g/dl NaCl= 0.026m
19. The inside of a red blood cell is about 300 mOsm. a) Label the milliosmolality (mOsm) of interior of the red blood cell, and of each of the NaCl solutions surrounding the cells (see question 18). b) Draw arrows to indicate the direction of water movement for each test tube. c) Below each illustration, label whether the each NaCl solution is isotonic, hypertonic or hypotonic. Red blood cell in 9 g/dl NaCl: Red blood cell in 0.9 g/dl NaCl Red blood cell in 0.15 g/dl NaCl ___________________ _____________________ ____________________ NaCl solution Red blood cell 20. APPLICATION QUESTION: Your body has multiple mechanisms in place to maintain the osmotic properties of your blood plasma and extracellular fluid. Why is it so critical that the osmotic properties be maintained? In other words, what would happen to your cells , tissues AND overall well-being if: When the body becomes dehydrated and blood plasma and extracellular fluids become hypertonic, it leads to expulsion. a) you became dehydrated and your blood plasma and extracellular fluids became HYPERTONIC****? The dehydration (fluid loss) causes direct increase in the serum osmolality triggers the kidney to retain the water and increases the thirst.Therefore,the extracellular fluid and plasma becomes hypertonic.
