Solubility Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-profit organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. Copyright © 2013 CK-12 Foundation, www.ck12.org The names “CK-12” and “CK12” and associated logos and the terms “FlexBook®” and “FlexBook Platform®” (collectively “CK-12 Marks”) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: November 24, 2013 www.ck12.org C ONCEPT Concept 1. Solubility 1 Solubility Lesson Objectives • Define the term solubility and describe factors affecting the solubility of a particular solution. • Predict whether a substance will dissolve or dissociate in water. Describe what it means for a solution to be saturated, supersaturated, or unsaturated. • Define and give examples of both miscible and immiscible mixtures. • Explain the statement "like dissolves like" at the molecular level, and give specific examples of this concept. • Describe how pressure and temperature affect the solubility of a liquid or solid in solution. • Describe how pressure and temperature affect the solubility of gases in solutions, and use Henry’s Law to predict the solubility of a gas in a solution given the necessary variables. Vocabulary • solubility: The degree to which a solute dissolves in a solvent. • saturated: The point at which no more solute is able to dissolve. • unsaturated: A solution in which more solute could be dissolved, solute concentration is less than predicted by solubility properties. • supersaturated: When the amount of solute dissolved exceeds the solubility. Occurs when a solution is saturated and the temperature slowly drops. • miscible: Molecules mix well with one another, and form a homogeneous mixture. • immiscible: Molecules don’t mix well together, and form a heterogeneous mixture exhibiting a noticeable bilayer. • van’t Hoff factor: Describes the number of moles of particles that dissociate from solid. • Henry’s law: Mathematically describes the relationship between the vapor pressure of the solution and the solute concentration. Check Your Understanding 1. What type of intermolecular forces will exist between molecules of the following substances? a. b. c. d. e. f. H2 O CO2 CH4 N2 CO NH3 1 www.ck12.org Introduction Have you ever wondered why it is easier to stir sugar into hot tea rather than into ice tea? Or, why no matter how much you shake a jar of oil and vinegar, it always seems to separate? These observations can be explained by a property of solutions known as solubility. In this section, you’ll learn how molecular structure and binding forces contribute to the solubility properties of various solutions and mixtures. Solubility Solubility is the degree to which a given solute dissolves in a particular solvent. It depends on various factors, including temperature and pressure. A common way to express the solubility of a given solute-solvent pair is to state the maximum amount of solute that can be dissolved by 100 grams of the solvent. The temperature dependence of the solubilities for various compounds in water is shown in Figure 1.1. FIGURE 1.1 Solubility Curve for Ionic Solids 2 www.ck12.org Concept 1. Solubility The aqueous solubility of a given substance is determined experimentally by dissolving increasing amounts into a known mass of water at a specific temperature until no more solid dissolves. A solution that cannot hold any more of a given solute is said to be saturated. For example, in Figure 1.1 we see that the solubility of sodium nitrate (NaNO3 ) is approximately 90 g per 100 g of H2 O at 20°C. At this temperature, a 100 gram sample of H2 O in which 90 grams of sodium nitrate is dissolved would be saturated, and a solution in which less sodium nitrate is dissolved would be unsaturated. Note that this ratio also holds for samples in which the solvent is present in different amounts; 50 grams of water would hold 45 grams of sodium nitrate at 20°C, and 300 grams of water would hold 270 g of NaNO3 . Notice that even relative solubilities of various compounds are temperature dependent. For example, at 20°C, KCl has a higher solubility than NaCl, but at 50°C, this relationship is reversed. Solutions can also become supersaturated, where the amount of solute dissolved exceeds its solubility. Supersaturation most commonly occurs when a saturated solution is slowly cooled. They occur frequently in geological and meteorological processes. Supersaturated systems are unstable, and eventually, the solute will precipitate until a saturated solution is regenerated. We can quantify supersaturation by looking at solubility curves. If the ratio of solute to solvent is above the saturation curve at the given temperature, the solution is supersaturated. If it is on the curve, the solution is saturated, and if it is below the curve, the solution is unsaturated. Solubility can be described for any solute-solvent pairing, but because water is such a fundamentally important solvent, we are mainly focusing on aqueous solutions. Example 16.1 You dissolve 40 g of KCl in 100 g of water at 40°C. You then cool the solution to 20°C, during which you notice solid KCl precipitating. How many grams of KCl would you expect to precipitate? Answer: Consult Figure 1.1 to find the solubility of KCl at 20°C (approximately 32 g KCl/100 g H2 O). Therefore, we would expect approximately 8 g KCl to precipitate out (40 g – 32 g = 8 g). Factors Affecting Solubility There are three main factors that control solubility. 1. Identities of the solute and solvent 2. Temperature 3. Pressure (for gases only) Solute and Solvent Ultimately, the ability of a solute to dissolve in a particular solvent will be dictated by the relative favorability of solute-solvent interactions compared to solute-solute and solvent-solvent interaction. In particular, the polarity of these two substances has a major effect on whether a significant amount of solute is able to dissolve. Polar solutes are typically quite soluble in polar solvents (e.g., ethanol in water), and nonpolar solutes generally dissolve well in nonpolar solvents (e.g., grease in gasoline). Conversely, polar solutes will have low solubilities in nonpolar solvents (e.g., NaCl in CCl4 ), and solubilities will be low for nonpolar solutes in polar solvents (e.g., oil in vinegar). Temperature As you can see in Figure 1.1, solid and liquid solutes generally become more soluble as the temperature increases. This is true for solvents other than water as well. This effect varies quite a bit by substance. For example, the solubility of KNO3 has a very strong temperature dependence (its solubility curve has a large slope), whereas the solubility of NaCl is minimally influenced by temperature (its solubility curve is nearly flat). For gaseous solutes, solubility decreases at higher temperatures. We will look more at this effect later in the lesson. Pressure 3 www.ck12.org Higher pressures increase the solubility of gases. You are probably familiar with this concept as it relates to carbonated beverages. Before opening the container, the inside is pressurized, so a large amount of CO2 is dissolved in the liquid. After opening, the pressure decreases (to the ambient pressure), so the solubility of CO2 drops, causing it to bubble out of solution. Because they are not compressible like gases, solid and liquid solutes do not have noticeable changes in solubility at different external pressures. A Review of Intermolecular Forces Our understanding of the behavior of solutes and solvents can be largely explained at the molecular level using our model of intermolecular forces. Some substances will mix freely while others barely mix at all. This is due to the interactions between particles of the solvent and solute. Recall that nonpolar molecular substances are held together in the solid and liquid phases by relatively weak London dispersion forces, in which induced dipoles line up into a favorable arrangement. An example of this is the interactions found between molecules of iodine (I2 ). In contrast, polar molecules are held together by stronger dipole-dipole interactions. Additionally, molecules that contain N-H, O-H, or F-H bonds exhibit a special dipole-dipole interaction called hydrogen bonding, which is unusually strong even for a polar interaction. Ammonia (NH3 ) and water are examples of small molecules that exhibit hydrogen bonding. The cations and anions in an ionic compound are held together by very strong ionic bonds, but ion-dipole interactions are nearly as strong. Ion-dipole interactions would be found, for example, when an ionic substance like NaCl is dissolved in water. Each ion is attracted to the appropriate end of the dipole on surrounding molecules of water. FIGURE 1.2 Decision tree for types of intermolecular interactions Liquid Solutes When combining two liquids, we can generally predict whether they will mix to form a homogeneous solution or not by looking at the relative polarity of each substance. We will consider three scenarios: the combination of two polar liquids, the combination of one polar and one nonpolar liquid, and the mixing of two nonpolar liquids. Polar-Polar Interactions Polar-polar interactions occur when two or more polar liquids are mixed. An example of this is when methanol mixes with water. Both of these are small polar molecules containing O-H bonds, which means that they can 4 www.ck12.org Concept 1. Solubility both participate in hydrogen bonding. Figure 1.3 shows molecules of methanol and water connected by hydrogen bonds. Because of these strong interactions, the two substances mix freely to form a homogeneous mixture; they are miscible. One way to remember this interaction is the phrase "like dissolves like." In this case, a polar solvent dissolves polar solutes. FIGURE 1.3 Hydrogen bonding between methanol and water Nonpolar-Polar Interactions Toluene (C6 H5 CH3 is an organic compound that is often used as a solvent in paint thinners. Toluene is a nonpolar compound. When mixed with water, the two substances will separate into two layers rather than forming a homogeneous solution; these two liquids are immiscible. Toluene is a nonpolar chain that cannot form hydrogen bonds with water. Dissolving this chain in water would break up the strong hydrogen bonds between water molecules and replace them with weaker dispersion forces. This is generally not energetically favorable, so the liquids tend to separate themselves to maximize the number of strong attractive interactions. Example 16.2 Can you think of two other liquids that are immiscible and form a heterogeneous mixture? Answer: Another familiar example is the mixing of vinegar and olive oil. Olive oil is a nonpolar substance, while vinegar (which is mostly water and acetic acid) is polar. The result is a heterogeneous mixture that exhibits a bilayer. Nonpolar-Nonpolar Interactions Nonpolar-nonpolar interactions occur when two nonpolar liquids are mixed. An example of this is the interaction between toluene and octane (see Figures above and 1.5). The interactions between a molecule of toluene and a molecule of octane are relatively weak, but so are the toluene-toluene and octane-octane interactions. Because no strong intermolecular forces (like those between water molecules) need to be broken for mixing to occur, no strong interactions need to be formed in order for mixing to be a favorable process. Toluene and octane will form a homogeneous mixture. The phrase "like dissolves like" applies to these mixtures as well. In this case, nonpolar dissolves nonpolar. Example 16.3 Can you think of another example of a nonpolar-nonpolar interaction between two different liquids that form a homogeneous mixture? Answer: Another example of a nonpolar-nonpolar interaction between two different liquids would be the mixing of motor oil and gasoline. Both of these substances are nonpolar, so they are miscible and form a homogeneous mixture when combined. 5 www.ck12.org FIGURE 1.4 FIGURE 1.5 6 www.ck12.org Concept 1. Solubility Solid Solutes Solutions can also be comprised of a solid solute and a liquid solvent. These interactions are governed by the same three interactions we discussed earlier: polar-polar, nonpolar-polar, and nonpolar-nonpolar. The table below describes these interactions. TABLE 1.1: 16.2 Solid (solute) polar nonpolar polar nonpolar Liquid (solvent) polar polar nonpolar nonpolar Example NaCl + H2 O I2 + H2 O NaCl + Toluene I2 + Toluene Result Homogeneous solution Heterogeneous mixture Heterogeneous mixture Homogeneous mixture Example 16.4 Using the data in the above table, could you replace the solute or the solvent in the heterogeneous mixtures with another material to make them homogeneous? Answer: There are several combinations that could be described. For instance, in the polar-nonpolar mixture between NaCl and toluene, the NaCl could be replaced with a nonpolar solid, like I2 , or the toluene could be replaced with a polar substance, like water. Ionic Solids in Water When placed in water, ionic solids dissolve to varying degrees. Some ionic solids have a high solubility in water (e.g., NaCl), while others barely dissolve at all (e.g., AgCl). Still others are moderately soluble (e.g., Ag2 CO3 ). The solubility rules we studied in the chapter on Chemical Reactions provide guidelines for predicting the relative solubility of a given ionic compound in water. In this chapter, we will focus primarily on water-soluble ionic solids. When a soluble ionic solid is added to water, it interacts with water molecules and dissociates into isolated ions that diffuse out into the solution. These charged particles become solvated by surrounding water molecules (Figure 1.6). Although the strong ionic bonds in the solid are broken up, they are replaced by numerous favorable interactions between the charged ions and the partial charges on the appropriate ends of the polar water molecules. Notice that for each unit of NaCl that dissolves, two particles are freed into solution, the Na+ cation and the Cl− anion. This means that if one mole of NaCl is dissolved, 2 moles of solute particles are found in the homogeneous solution (one mole of each ion). This dissociation is quantified by something called the van’t Hoff factor. The van’t Hoff factor (i) describes the number of moles of solute particles that are found in a solution when one mole of a substance is completely dissolved. The van’t Hoff factor for NaCl would be expressed as i = 2. Example 16.5 If one mole of magnesium fluoride (MgF2 ) is added to water and fully dissociates, how many moles of particles will be formed? Answer: We can describe the dissociation of magnesium fluoride as follows: MgF2 → Mg2+ + 2 F− Each unit of magnesium fluoride contains three ions (one Mg2+ ion and two F− ions). Using the van’t Hoff factor to describe this dissociation, we would say that i = 3, because three moles of ions are produced from the dissociation 7 www.ck12.org FIGURE 1.6 of one mole of the ionic solid. Gas Solutes Gases are also capable of dissolving in liquids. There are many examples of this in our everyday lives. For example, carbonated beverages contain dissolved carbon dioxide. We notice this when bubbles come out of solution when the beverage is opened. Another example is when oxygen from the air we breathe dissolves in our blood, where it is transported throughout the body. Fish and other aquatic organisms use gills to capture dissolved oxygen from their environments. Because determining the mass of a gaseous sample is generally less convenient than determining how many moles are present, solubilities for gases are often expressed as concentrations instead of as the mass that can be dissolved in a specified amount of solvent. A solution in which one mole of a gas is dissolved in one liter of solution has a concentration of 1 molar (1 M). Because the solubility of most gases is much less than that, the molar solubility is often given in millimolar (mM). A one millimolar solution contains 1/1000 mol of solute per liter of solution. Other methods of expressing concentration, such as parts per million (ppm) or parts per billion (ppb), will be discussed in the following lesson. Example 16.6 What does it mean if the molar solubility of a gas is 2.0 mM? Answer: Each liter of solution can hold a maximum of 2.0 millimoles of that particular gas at the indicated temperature and pressure. 8 www.ck12.org Concept 1. Solubility Temperature Effects As with all substances, the solubility of gases is temperature dependent. However, in contrast to the situation for most solids and liquids, higher temperatures will decrease the solubility of a gaseous solute. Figure 1.7 shows this relationship with several common gases. FIGURE 1.7 Solubility-Temperature Curves for Various Common Gases This inverse relationship between solubility and temperature can be understood by looking at a molecular model. Recall that higher temperatures are associated with faster particles. Gas particles are held in solution by attractive interactions with the solvent molecules. If the particles are moving slowly, these attractive forces will pull back any particles that try to escape the surface of the solution and re-enter the gas phase. However, if the gas particles are moving fast enough, these interactions will not be sufficiently strong to prevent this process from occurring. As a result, more particles are able to escape, and the amount of dissolved solute is less than it would be at a lower temperature. Pressure Effects At a constant temperature, the amount a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in the area immediately adjacent to the solution. This principle is called Henry’s Law and is illustrated in Figure 1.8. Mathematically, Henry’s Law is expressed as follows: ρ = kH c where ρ is the partial pressure of the gas, c is its molar solubility at the given temperature and pressure, and kH is a constant that depends on the temperature and the identities of both the solute and solvent. Some kH values for various gases dissolved in water at 298 K are presented below. TABLE 1.2: 16.3 Gas He O2 N2 H2 CO2 NH3 Constant (Pa*mol−1 *L) 282.7 x 106 74.68 x 106 155. x 106 121.2 x 106 2.937 x 106 5.69 x 106 Constant (atm*mol−1 *L) 2865 756.7 1600. 1228 29.76 56.9 9 www.ck12.org FIGURE 1.8 Illustration of Pressure effects on gas solubility If we solve the Henry’s Law equation for the molar solubility, we get the following useful relationship: c= ρ kH Example 16.7 What are the molar solubilities of oxygen gas and nitrogen gas in pure water at 298 K and 1 atm of pressure? Which gas is more soluble under these conditions? c O2 = 1 atm 756.7 (atm∗mol−1 ∗L) =0.00132 mol/L =1.32 mM c N2 = 1 atm 1600. (atm∗mol−1 ∗L) =0.000625 mol/L =0.625 mM Oxygen gas is more soluble under these conditions. A saturated solution of O2 would have a concentration of 1.32 mM, whereas a saturated solution of N2 would have a concentration of 0.625 mM. Lesson Summary • Solubility describes the degree to which a solute will dissolve in a particular solvent. • Water is a common solvent for dissolving various solids, liquids, and gases. • The solubilities of solids and liquids are commonly expressed as grams of solute that can be dissolved by 100 g of water at a specified temperature. • Solutions can be unsaturated, saturated, or supersaturated, depending on the relationship between the solubility of a substance and the amount that is actually dissolved. 10 www.ck12.org Concept 1. Solubility • The solubility of gases, liquids, and solids are affected by changes in temperature. • Solutes and solvents that have similar polarities (i.e., both are polar, or both are nonpolar) tend to mix well, creating homogeneous solutions. Solutes and solvents with very different polarities (i.e., one is polar, and the other is nonpolar) often do not mix well, resulting in the formation of heterogeneous mixtures. • The solubilities of gases are often described as concentrations (i.e., mM or M) rather than the mass that can be dissolved by a given mass of solvent. • The solubility of a gaseous solute is inversely related to the temperature of the solvent and directly proportional to the partial pressure of the gas in the surrounding atmosphere. • Henry’s law describes the mathematical relationship between the concentration of a gaseous solute and its partial pressure in the gas above the solution. Review Questions 1. Draw the mixture that would be formed when oil and water are combined. 2. Give an example of molecular solid that dissolves in water due to polar-polar interactions. 3. Ammonia (NH3 ) dissolves well in water. Explain how this interaction might occur and the type(s) of intermolecular forces that would be involved. 4. Using the solubility-temperature curve in Figure 1.1, describe conditions under which a solution of potassium chloride would be unsaturated, saturated, and supersaturated at 20°C. 5. A solution is formed by dissolving 10. grams of potassium chlorate in 100. g of water at 30°C. If the solution were heated to 40°C, how many more grams of solute could be dissolved? 6. A solution that is saturated with both methane and oxygen gas at 1 atm and 20°C is then heated to 40°C. What will happen to the dissolved gases as the temperature increases? Referring back to Figure 1.7, how much of each gas would leave the solution? How much would remain? 7. Which of the gases in Table 1.2 would have the highest solubility in water at 298K? Which would have the lowest solubility? 8. Urea (CH4 N2 O) is a molecular solid that has a relatively high solubility in water. How would you account for this fact? The molecular structure for urea is shown below. FIGURE 1.9 9. Which of the following substances would dissolve better in water: iodine crystals (I2 ) or liquid methanol (CH3 OH)? How would you categorize the resulting solute-solvent interactions? 11 www.ck12.org Further Reading / Supplemental Links • Solubility-pressure curves for a variety of gases: http://www.engineeringtoolbox.com/gases-solubility-water -d_1148.html • Supersaturated solution demonstration: http://www.youtube.com/watch?v=D1PDE5OawuI Points to Consider • When you open a can of soda or sparkling water, you can usually see some gas escape or bubble out of the can. Based on the relationship between pressure and gas solubility, why might this occur? • Why do you suppose it is better to wash your dishes with warm water than with cold water? Explain these effects in terms of solubility. References 1. 2. 3. 4. 5. 6. 7. 8. 9. 12 CK-12 Foundation - Christopher Auyeung. . CC-BY-NC-SA 3.0 CK-12 Foundation - Jodi So. . CC-BY-NC-SA 3.0 Jü. http://commons.wikimedia.org/wiki/File:Methanol_Hydrogen_Bridge_V.2.svg. CC 0 public domain NEUROtiker. http://commons.wikimedia.org/wiki/File:Toluol.svg. Public Domain Bangin. https://commons.wikimedia.org/wiki/File:Octane.svg. Public Domain Taxman. http://commons.wikimedia.org/wiki/File:Na%2BH2O.svg. Public Domain CK-12 Foundation - Jodi So and Steven Lai. . CC-BY-NC-SA 3.0 CK-12 Foundation - Jodi So. . CC-BY-NC-SA 3.0 NEUROtiker. http://commons.wikimedia.org/wiki/File:Harnstoff.svg. Public Domain
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