Cells Part 2: Osmosis
LAB 2B:SOLUTIONS, OSMOSIS, OSMOLARITY & TONICITYREADING in HUMAN PHYSIOLOGY, 7th Edition, Silverthorn.Ions: p. 33Solutions and solubility: pp. 40, 42-43, 133Osmolarity: pp. 123-129Objectives:Gain an understanding of solutions, solubility and concentrations, and their relevance to physiologyPerform calculations for % solution, molarity, osmolarity, milliosmolarity, and dilutions.Prepare NaCl solutions based on your calculations. Demonstrate the effects of hypertonic, isotonic and hypotonic solutions on RBCs.Our bodies are full of solutions. Our cells are filled with a solution known as the intracellular fluid (ICF), and our cells are surrounded by solutions known as the extracellular fluid (ECF). The ECF of our blood is known as the plasma, while the ECF of our tissues is known as interstitial fluid. So what is a solution? 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 ions and molecules such as salts (e.g. NaCl), sugars (e.g. glucose), proteins, and gases. Ex. 1: Solubility and a Solubility Demo: Mineral Oil, Water and DyeIn this exercise, you will make observations as your instructor perform a demonstration on solubility using mineral oil, water and a colored dye.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 (water-loving), 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 cell 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 cell 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.Now you instructor will do a demonstration involving mineral oil and water to illustrate the solubility differences in nonpolar and polar solutions. 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?Answer the questions on p. 6.Ex. 2: Concentrations, Dilutions, Calculations & Preparing Solutions In this exercise, you will perform calculations and describe the preparation of the solutions that you will use in exercise 3.Ex. 2A. Calculations for the preparation of a10% and 0.9% NaCl solutionConcentration:The concentration of a solution relates the amount of solute to the amount of solvent in a solution (concentration = solute/solvent). Concentrations can be expressed in a variety of ways, and we will encounter many of these forms of concentration in physiology. 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: Weight per volume e.g. mg/dl and mg/100 ml are common weight per volume units used in physiologyPercentage (% ) solutionsPercentage 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:% = (grams solute/volume of solvent in milliliters) X 100EXAMPLE 1: How would you prepare 500 ml of a 4% NaCl solution?4% = (X g/500 ml) x 100 X = 20 g NaClEXAMPLE 2: Another way to think about it and perform this calculation is:4% = 4g/100mL, since want 500mL 4% = 4g/100mL x 500mL= 20gAdd 20 g NaCl to a container, then fill up to the 500 ml level with water.Molarity (M) and molality (m)The equations for molarity and molality are as follows.Molarity (M) = moles of solute/L of solutionMolality (m) = moles solute/ kg of solventAs you can see, both molar (M) and molal (m) solutions express the amount of solute in terms of moles. A mole is just a specific number of atoms or molecules. One mole of any substance contains 6.02 x 1023 molecules. The weight in grams of one mole of a substance equals the molecular weight (or atomic weight or molar mass) of that substance. Atomic weights (molar mass) for the elements are given in the periodic table of the elements. Molecular weights are computed by adding the individual atomic weights of the atoms that make up a molecule.Dilutions:Since we often need to dilute concentrated solutions in the laboratory, another important calculation related to solutions are dilution calculations. To make a dilute solution, the following formula applies: C1V1 = C2V2, where C1 = concentration of solution 1, V1 = volume of solution 1 that you need to dilute with water, C2 = concentration of solution 2, and V2 = volume of solution 2. When making a number of dilute solutions, it is sometimes tedious and inaccurate to repeatedly weigh 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 0.01 M NaCl. It is easiest to prepare a stock (or parent) solution of 1.0 M NaCl first and then dilute it to make a 0.01M NaCl solution. (1.0 M)(V1) = (0.01M)(100 ml) V1 = 1 mlAdd 1 ml of 1.0 M NaCl in a container and then add 99ml of water for a final volume of 100 ml.Now perform the calculations and answer the questions on p.7.Ex. 2B: Preparation of 0%, 10% and 0.9% NaCl SolutionsOnce you have verified your calculations and preparation descriptions with your instructor, you will prepare the solutions from Ex. 2A, and then you will use them for Ex. 3A.Procedure: Label three 50-mL beakers as distilled water, 10% NaCl and 0.9% NaCl. Prepare the solutions as follows: Distilled water (0% NaCl): Add 10 ml distilled water to the beaker. Distilled water is available in the back of the lab.10% NaCl: Prepare the 10 ml of solution according to your approved calculations on p.70.9% NaCl: Prepare 10 ml of this solution as a dilution according to your approved calculations on p. 7.You will use these solutions for Ex. 3.Exercise 3: Osmosis & TonicityOsmosis is the diffusion of a solvent, such as water, through a semi-permeable or selectively permeable membrane that prevents the movement of certain solutes. The solutes that are unable to move are termed osmotically active particles. The plasma membranes of our cells are selectively permeable. In general, polar molecules and ions, have a very limited ability to cross the plasma membrane of our cells, and thus, they often serve as osmotically active molecules. Although water is polar, it can easily cross the plasma membrane due to the abundance of water channels known as aquaporins. The movement of water is toward the solution with a higher concentration of osmotically active solute particles. Tonicity refers to how a solution would affect cell volume. In this part of the lab we will explore the effects of solute concentration on red blood cells with a hypertonic solution (a solution that draws water out of the cell resulting in crenation (wrinkling)), hypotonic solution (a solution that causes the cell to gain water resulting in swelling or even lysis (bursting)) and isotonic solution (a solution that does not change the shape of the cell since water moves in and out of the cell at equal rates). Ex. 3A: Demonstrating the effects of hypertonic, isotonic and hypotonic solutions on RBCs.Procedures:Label 3 test tubes as follows: distilled water, 10% NaCl, and 0.9% NaCl. Then transfer 1 ml of each solution that you prepared for Ex. 2B to the appropriate tube.Add one drop of sheep’s blood to each tube and gently swirl. Allow the tubes to sit about a minute.Hold each tube up and examine the turbidity (cloudiness). The contents of all tubes will appear pinkish red, but two should be cloudy, and one should be translucent. Record the appearance of each.Label three slides, and then add a drop of each solution to the appropriate slide. Cover with a cover slip and observe the cell morphology of the RBCs under the microscope. NOTE: You are not interested in the number of cells (that just has to do with the size of the drop of blood that you used) or the “movement” of the cells (that just has to do with the air causing the solution that contains the cells to move). Instead, you should note the shape of the cells (round, wrinkled/crenated, or absent (lysed)). Record observations on the worksheet p. 8.Clean up: Once you have recorded your observations, please thoroughly clean and dry your test tubes and beakers. Make sure that you remove all markings. Slides should be discarded in the sharps plete the table and answer the questions on p. 8.Ex. 3B: OsmolarityThe tonicity of a solution depends on the osmolarity and the ability of the solute particles to cross the membrane. Osmolarity (Osm) is related to molarity, but while molarity considers the moles per L of solute, osmolarity also takes into account the number of particles that a molecule would become if dissolved in water. Molecules held together by ionic bonds dissociate when placed in water as water forms spheres of hydration around the ions. Thus, a molecule such as NaCl, which is held together by ionic bonds will dissociate into separate Na+ and Cl- ions in water. However, molecules held together by covalent bonds stay intact when dissolved in water. For instance, if you dissolve the sugar sucrose (C12H22O11) in water, the sucrose molecules will not break apart into separate C, H and O atoms, but instead will remain as C12H22O11 molecules. While osmolarity alone is not sufficient to predict the tonicity of a solution (see pp. 127-132 of your textbook), determining the osmolarity of a solution and knowing whether or not the solute within a solution is osmotically active (incapable of penetrating the cell membrane) can be useful for predicting the movement of water across a semi-permeable membrane, and thus the shape, function, and survival of the cell. The normal osmolarity of our cells (ICF) and body fluids (ECF) is about 0.3 OsM = 300 mOsM. Normally changes in osmolarity occur slowly and the cell is able to maintain homeostasis, but if the fluid surrounding a cell changes too quickly or falls far outside normal ranges, it can be dangerous and even life-threatening. If a solution has a higher concentration of osmotically active, non-penetrating particles as compared to the cell, then you would predict that the solution will be hypertonic. If a solution has a lower concentration of osmotically active, non-penetrating particles as compared to the cell, you would predict that the solution will be hypotonic. Following are the formulas for osmolarity (Osm) and milliosmolarity (mOsm):Osmolarity (OsM) = molarity (mol/L) x # of active particles/moleculeMilliosmolarity (mOsM) = osmolarity x 1000EXAMPLE 1: What is the osmolarity of 0.1 M sucrose solution, assuming that sucrose is held together by covalent bonds?0.1 M x 1 particle = 0.1 OsMEXAMPLE 2: What is the osmolarity of a 0.1 M NaCl solution, assuming that NaCl is held together by ionic bonds?M x 2 particles = 0.2 OsM EXAMPLE 3: What is the milliosmolarity of a 0.2 OsM solultion?0.2 OsM x 1000 mOsM/1 OsM = 200 mOsMNote on question #9a: Using concepts from chemistry it is possible to convert a % solution to a molar solution (M).? Such calculations are discussed in more detail in the textbook but will not be addressed here. Instead, we have provided you with the M concentration of each of the % solutions you prepared earlier. You will determine the milliosmolarity (mOsM) of those solutions using the molarity (M) values provided.Perform the calculations and answer the questions on p.9. LAB DAY & TIME: Monday 7:00- 10:10 AM NAME: Frankie Guevara GROUP MEMBER NAMES: Christy Balderrama, Lisa Reyes, Alexandrea RuizLAB 2B:SOLUTIONS, OSMOSIS, OSMOLARITY & TONICITYWORKSHEET (to be turned in INDVIDUALLY)Ex. 1. Solubility and a Solubility Demo: Mineral Oil and Water (from p. 2)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!There were two layers present. Two layers were present because the chemical properties and solubility differs from the water and oil. Water is polar and oil is nonpolar because of this the oil cannot dissolve with water creating two layers. Like dissolves like. a) In which layer (top or bottom?) did the ionic dye dissolve? BottomBased on this, which layer is water? Water layer is the bottomEXPLAIN why the dye dissolved in the layer that it did.The dye dissolved in the layer that was polar because since dye is also considered polar and water is polar, like dissolves like. a) What types of molecules pass relatively easily across the lipid part of the membranes of your cells? Small and nonpolar What types of molecules do not pass freely across the lipid part of the membranes of your cells? Large and polarHOW do molecules that CANNOT easily cross the lipid part of the cell membrane enter or exit the cell? Molecules that cannot easily cross the lipid part of the cell membrane enter or exit the cell through protein channels that allow substances to diffuse across the membrane. Would you predict that the dye used in this demonstration could cross the lipid part of the cell membrane? Why or why not?I would predict that the dye used in this demonstration could cross the lipid part of the cell membrane because the cell membrane has polar heads and nonpolar tails. Since the dye is polar it can cross and eventually dissolve. Ex. 2A: Calculations for the preparation of 10% and 0.9% NaCl solutions (from pp. 2-3)Perform the calculations and describe the procedure that you will use to prepare 10 mL of a 10% NaCl solution. a) Calculate the # of grams of NaCl needed. Show your work! 1 g NaCl10g equals 100 ml 1 g equals 10 mlb) Briefly describe the procedure that you will use to prepare 10mL of the 10% NaCl solution.To prepare 10 ml of the 10% Nacl solution I will gather 10 ml of water and 0.1g of NaCl solution (or 0.1 ml of NaCl solution) and mix it together. Now perform the calculations and describe the procedure that you will use to prepare 10 ml of the 0.9% NaCl solution using the 10% NaCl solution as your concentrated stock solution. Note: You will be performing a dilution in this case.a) Calculate the volume of NaCl the 10% NaCl solution that you will need to prepare 10 mL of the 0.9% NaCl solution. Show your work!C?V? = C?V?10(V?) = 0.9(10)V?= 0.9(10)10 = 0.9 ml + 9.1 ml of water = 10 ml b) How much distilled water was needed for 10mL of 0.9% solution? Show work.For the 10 ml of 0.9% solution there needs to be 9.1 ml of water.c) Briefly describe or diagram how you will prepare 10 mL of the 0.9% NaCl solutionFor the solution I will gather 0.9 ml of the solution and 9.1 ml of water and mix together to create a 10 ml of 0.9% NaCl solution.Verify your calculations and procedures for the preparation of the 10% and 0.9% NaCl solutions with your instructor!!! Then prepare the solutions as described in Ex. 2B on p.3. Ex. 3A: Demonstrating the effects of hypertonic, isotonic and hy?potonic solutions on RBCs. (from p. 4)In the table below, describe the appearance of each solution once the RBCs have been added, describe and/or draw the appearance of RBCs in each of the 3 solutions as viewed under the microscope, and draw the direction of water movement (e.g. arrows showing water entering or leaving the cell) as inferred based on the appearance of the cells, and indicate tonicity.SolutionTurbidity (cloudy or translucent?)Appearance of RBCs andDirection of Water Movement as Inferred Based on AppearanceTonicity (Hypertonic/ Hypotonic/ Isotonic)Distilled waterTranslucentCells eruptedHypotonic0.9% NaClSlightly cloudyNormal meaning no change to red blood cells Isotonic10%NaClCloudyBunched together, bumpy lookingHypertonicWhich tube contained a translucent suspension? Why wasn’t it cloudy like the solutions in the other tubes? The distilled water contained a translucent suspension because the cells erupted when the red blood cells were added to the solution creating a hypotonic environment. Since the cells erupted the solution did not appear like the others which were cloudy. Ex. 1D: Determining milliosmolarity from a % solution. (from p.4-5)a) Calculate the milliosmolarity (mOsM) for each of the solutions you prepared for the tonicity experiment. The molarity (M) has been provided (see note p. 5). Show your work!Distilled water 0% NaCl = 0.0 M = 0 mOsM0.0 M x 2 particles = 0 OsM x 1000 = 00.9% NaCl = 0.15 M = 300 mOsM0.15 M x 2 particles = 0.3 OsM x 1000 = 30010% NaCl = 1.5 M = 3000 mOsM1.5 M x 2 particles = 3 OsM x 1000 = 3000b) What is the normal milliosmolarity found in blood and most cells of the body? (see p. 5) The normal osmolarity of our cells (ICF) and body fluids (ECF) is about 0.3 OsM = 300 mOsMc) If NaCl is a non-penetrating solution, do the results of the tonicity experiment make sense? Explain for each solution.For each solution, the results of tonicity do make sense since the environment for the distilled water was less concentrated than inside the cell where water moved in causing the cells to erupt because there was too much. For the 0.9% NaCl the environment and the cell were both equal to the point of creating an isotonic environment causing the cell to appear slightly cloudy. For the 10% NaCl the environment was more concentrated making the cell shrivel up because water must leave the cell. Intravenous (IV) fluids are given to patients to replace lost solutes and/or fluids.a) If a person is dehydrated, which type of solution (Hypotonic/Isotonic/Hypertonic) would replace intracellular fluid lost from cells? Explain!A hypotonic solution would replace intracellular fluid loss because since the environment would be less concentrated than the inside of the cell water would then move from the environment into the cells allowing for rehydration of intracellular fluid lost from the cells. b) Which type of solution would replace extracellular fluid caused by blood loss? Explain! Extra cellular fluid would be replaced by a hypertonic solution because since the cells environment would be more concentrated the water from the replacement cells would leave the cells into the extracellular fluid environment. ................
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