13. Solutions PDF - Chemistry Chapter

13: SOLUTIONS  CHAPTER OVERVIEW 13: SOLUTIONS Solutions play a very important role in many biological, laboratory, and industrial applications of chemistry. Of particular importance are solutions involving substances dissolved in water, or aqueous solutions. Solutions represent equilibrium systems...

13: SOLUTIONS  CHAPTER OVERVIEW 13: SOLUTIONS Solutions play a very important role in many biological, laboratory, and industrial applications of chemistry. Of particular importance are solutions involving substances dissolved in water, or aqueous solutions. Solutions represent equilibrium systems, and the lessons learned in the last chapter will be of particular importance again. Quantitative measurements of solutions are another key component of this chapter. Solutions can involve all physical states—gases dissolved in gases (the air around us), solids dissolved in solids (metal alloys), and liquids dissolved in solids (amalgams—liquid mercury dissolved in another metal such as silver, tin or copper). This chapter is almost exclusively concerned with aqueous solutions, substances dissolved in water. 13.1: Tragedy in Cameroon 13.2: Solutions - Homogeneous Mixtures 13.3: Solutions of Solids Dissolved in Water- How to Make Rock Candy 13.4: Solutions of Gases in Water 13.5: Specifying Solution Concentration- Mass Percent 13.6: Specifying Solution Concentration- Molarity 13.7: Solution Dilution 13.8: Solution Stoichiometry 13.9: Freezing Point Depression and Boiling Point Elevation 13.10: Osmosis 13: Solutions is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 1 13.1: TRAGEDY IN CAMEROON Lake Nyos is a deep crater lake in the Northwest region of Cameroon, high on the flank of an inactive volcano in the Oku volcanic plain along the Cameroon line of volcanic activity. A volcanic dam impounds the lake waters. A pocket of magma lies beneath the lake and leaks carbon dioxide (CO ) into the water, changing it into carbonic acid. Nyos is one of only three known exploding lakes to be saturated with 2 carbon dioxide in this way. In 1986, more than 1700 people in Cameroon were killed when a cloud of gas, almost certainly carbon dioxide, bubbled from Lake Nyos (Figure 13.1.1 ), a deep lake in a volcanic crater. It is believed that the lake underwent a turnover due to gradual heating from below the lake, and the warmer, less-dense water saturated with carbon dioxide reached the surface. Consequently, tremendous quantities of dissolved CO2 were released, and the colorless gas, which is denser than air, flowed down the valley below the lake and suffocated humans and animals living in the valley. Figure 13.1.1: Two photos are shown. The first is an aerial view of a lake surrounded by green hills. The second shows a large body of water with a fountain sending liquid up into the air several yards or meters above the surface of the water. (a) It is believed that the 1986 disaster, that killed more than 1700 people near Lake Nyos in Cameroon, resulted when a large volume of carbon dioxide gas was released from the lake. (b) A CO2 vent has since been installed to help outgas the lake in a slow, controlled fashion and prevent a similar catastrophe from happening in the future. (Credit a: modification of work by Jack Lockwood; credit b: modification of work by Bill Evans.) Following the Lake Nyos tragedy, scientists investigated other African lakes to see if a similar phenomenon could happen elsewhere. Lake Kivu in Democratic Republic of Congo, 2,000 times larger than Lake Nyos, was also found to be supersaturated, and geologists found evidence for out-gassing events around the lake about every one thousand years. CONTRIBUTIONS & ATTRIBUTIONS Wikipedia Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected]). 13.1: Tragedy in Cameroon is shared under a CC BY-SA license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 13.1.1 https://chem.libretexts.org/@go/page/47545 13.2: SOLUTIONS - HOMOGENEOUS MIXTURES  LEARNING OBJECTIVES Learn terminology involving solutions. Explain the significance of the statement "like dissolves like." Explain why certain substances dissolve in other substances. The major component of a solution is called the solvent. The minor component of a solution is called the solute. By major and minor we mean whichever component has the greater or lesser presence by mass or by moles. Sometimes this becomes confusing, especially with substances with very different molar masses. However, here we will confine the discussion to solutions for which the major component and the minor component are obvious. Figure 13.2.1: Making a saline water solution by dissolving table salt (NaCl) in water. The salt is the solute and the water the solvent. (CC- BY-SA 3.0; Chris 73). Solutions exist for every possible phase of the solute and the solvent. Salt water, for example, is a solution of solid NaCl in liquid water, while air is a solution of a gaseous solute (O ) in a gaseous solvent (N ). In all cases, however, the overall phase of the solution is the same 2 2 phase as the solvent. Table 13.2.1 lists some common types of solutions, with examples of each. Table 13.2.1: Types of Solutions Solvent Phase Solute Phase Example gas gas air liquid gas carbonated beverages liquid liquid ethanol (C2H5OH) in H2O (alcoholic beverages) liquid solid salt water solid gas H2 gas absorbed by Pd metal solid liquid Hg(ℓ) in dental fillings solid solid steel alloys  EXAMPLE 13.2.1: SUGAR AND WATER A solution is made by dissolving 1.00 g of sucrose (C 12 H22 O11 ) in 100.0 g of liquid water. Identify the solvent and solute in the resulting solution. Solution Either by mass or by moles, the obvious minor component is sucrose, so it is the solute. Water—the majority component—is the solvent. The fact that the resulting solution is the same phase as water also suggests that water is the solvent.  EXERCISE 13.2.1 A solution is made by dissolving 3.33 g of HCl(g) in 40.0 g of liquid methyl alcohol (CH 3 OH ). Identify the solvent and solute in the resulting solution. Answer solute: HCl(g) solvent: CH 3 OH 13.2.1 https://chem.libretexts.org/@go/page/47547 LIKE DISSOLVES LIKE A simple way to predict which compounds will dissolve in other compounds is the phrase "like dissolves like". What this means is that polar compounds dissolve polar compounds, nonpolar compounds dissolve nonpolar compounds, but polar and nonpolar do not dissolve in each other. Even some nonpolar substances dissolve in water but only to a limited degree. Have you ever wondered why fish are able to breathe? Oxygen gas, a nonpolar molecule, does dissolve in water—it is this oxygen that the fish take in through their gills. The reason we can enjoy carbonated sodas is also due to a nonpolar compound that dissolves in water. Pepsi-cola and all the other sodas have carbon dioxide gas, CO , a nonpolar compound, dissolved in a sugar-water solution. In this case, to keep as much gas in solution as possible, the sodas are kept 2 under pressure. This general trend of "like dissolves like" is summarized in the following table: Table 13.2.2: Summary of Solubilities Solute (Polarity of Compound) Solvent (Polarity of Compound) Dominant Intermolecular Force Is Solution Formed? Polar Polar Dipole-Dipole Force and/or Hydrogen Bond yes Non-polar Non-polar Dispersion Force yes Polar Non-polar no Non-polar Polar no Ionic Polar Ion-Dipole yes Ionic Non-polar no Note that every time charged particles (ionic compounds or polar substances) are mixed, a solution is formed. When particles with no charges (nonpolar compounds) are mixed, they will form a solution. However, if substances with charges are mixed with other substances without charges, a solution does not form. When an ionic compound is considered "insoluble", it doesn't necessarily mean the compound is completely untouched by water. All ionic compounds dissolve to some extent. An insoluble compound just doesn't dissolve in any noticeable or appreciable amount. What is it that makes a solute soluble in some solvents but not others? The answer is intermolecular interactions. The intermolecular interactions include London dispersion forces, dipole-dipole interactions, and hydrogen bonding (as described in Chapter 10). From experimental studies, it has been determined that if molecules of a solute experience the same intermolecular forces that the solvent does, the solute will likely dissolve in that solvent. So, NaCl—a very polar substance because it is composed of ions—dissolves in water, which is very polar, but not in oil, which is generally nonpolar. Nonpolar wax dissolves in nonpolar hexane, but not in polar water. Figure 13.2.2: Water (clear liquid) and oil (yellow) do not form liquid solutions. (CC BY-SA 1.0 Generic; Victor Blacus)  EXAMPLE 13.2.2: POLAR AND NONPOLAR SOLVENTS Would I be more soluble in CCl or H 2 4 2 O ? Explain your answer. Solution I2 is nonpolar. Of the two solvents, CCl is nonpolar and H 4 2 O is polar, so I would be expected to be more soluble in CCl. 2 4 13.2.2 https://chem.libretexts.org/@go/page/47547   EXERCISE 13.2.2 Would C 3 H OH 7 be more soluble in CCl or H 4 2 O ? Explain your answer. Answer H O 2 , because both experience hydrogen bonding.  EXAMPLE 13.2.3 Water is considered a polar solvent. Which substances should dissolve in water? a. methanol (CH OH) 3 b. sodium sulfate (\ce{Na2SO4}\)) c. octane (C H ) 8 18 Solution Because water is polar, substances that are polar or ionic will dissolve in it. a. Because of the OH group in methanol, we expect its molecules to be polar. Thus, we expect it to be soluble in water. As both water and methanol are liquids, the word miscible can be used in place of soluble. b. Sodium sulfate is an ionic compound, so we expect it to be soluble in water. c. Like other hydrocarbons, octane is nonpolar, so we expect that it would not be soluble in water.  EXERCISE 13.2.3: TOLUENE Toluene (C 6 H CH 5 3 ) is widely used in industry as a nonpolar solvent. Which substances should dissolve in toluene? a. water (H2O) b. sodium sulfate (Na2SO4) c. octane (C8H18) Answer Octane (C 8 H 18 ) will dissolve. It is also non-polar. SUMMARY Solutions are composed of a solvent (major component) and a solute (minor component). “Like dissolves like” is a useful rule for deciding if a solute will be soluble in a solvent. 13.2: Solutions - Homogeneous Mixtures is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 13.2.3 https://chem.libretexts.org/@go/page/47547 13.3: Solutions of Solids Dissolved in Water- How to Make Rock Candy  Learning Objectives Define electrolytes and non electrolytes Explain why solutions form. Discuss the idea of water as the "universal solvent". Explain how water molecules attract ionic solids when they dissolve in water. We have learned that solutions can be formed in a variety of combinations using solids, liquids, and gases. We also know that solutions have constant composition, and that this composition can be varied up to a point to maintain the homogeneous nature of the solution. But how exactly do solutions form? Why is it that oil and water will not form a solution, and yet vinegar and water will? Why could we dissolve table salt in water, but not in vegetable oil? The reasons why solutions will form will be explored in this section, along with a discussion of why water is used most frequently to dissolve substances of various types. Solubility and Saturation Table salt (NaCl) readily dissolves in water. In most cases, only a certain maximum amount of solute can be dissolved in a given amount of solvent. This maximum amount is specified as the solubility of the solute. It is usually expressed in terms of the amount of solute that can dissolve in 100 g of the solvent at a given temperature. Table 13.3.1 lists the solubilities of some simple ionic compounds. These solubilities vary widely. NaCl can dissolve up to 31.6 g per 100 g of H2O, while AgCl can dissolve only 0.00019 g per 100 g of H2O. Table 13.3.1 : Solubilities of Some Ionic Compounds Solute Solubility (g per 100 g of H2O at 25°C) AgCl 0.00019 CaCO3 0.0006 KBr 70.7 NaCl 36.1 NaNO3 94.6 When the maximum amount of solute has been dissolved in a given amount of solvent, we say that the solution is saturated with solute. When less than the maximum amount of solute is dissolved in a given amount of solute, the solution is unsaturated. These terms are also qualitative terms because each solute has its own solubility. A solution of 0.00019 g of AgCl per 100 g of H2O may be saturated, but with so little solute dissolved, it is also rather dilute. A solution of 36.1 g of NaCl in 100 g of H2O is also saturated, but rather concentrated. In some circumstances, it is possible to dissolve more than the maximum amount of a solute in a solution. Usually, this happens by heating the solvent, dissolving more solute than would normally dissolve at regular temperatures, and letting the solution cool down slowly and carefully. Such solutions are called supersaturated solutions and are not stable; given an opportunity (such as dropping a crystal of solute in the solution), the excess solute will precipitate from the solution.The figure below illustrates the above process and shows the distinction between unsaturated and saturated. 13.3.1 https://chem.libretexts.org/@go/page/47549  Figure 13.3.1 : When 30.0 g of NaCl is added to 100 mL, it all dissolves, forming an unsaturated solution. When 40.0 g is added, 36.0 g dissolves and 4.0 g remains undissolved, forming a saturated solution. How can you tell if a solution is saturated or unsaturated? If more solute is added and it does not dissolve, then the original solution was saturated. If the added solute dissolves, then the original solution was unsaturated. A solution that has been allowed to reach equilibrium, but which has extra undissolved solute at the bottom of the container, must be saturated. Rock Candy Recipe - Crystallization of S… S… Electrolyte Solutions: Dissolved Ionic Solids When some substances are dissolved in water, they undergo either a physical or a chemical change that yields ions in solution. These substances constitute an important class of compounds called electrolytes. Substances that do not yield ions when dissolved are called nonelectrolytes. If the physical or chemical process that generates the ions is essentially 100% efficient (all of the dissolved compound yields ions), then the substance is known as a strong electrolyte (good conductor). If only a relatively small fraction of the dissolved substance undergoes the ion-producing process, the substance is a weak electrolyte (does not conduct electricity as well). Substances may be identified as strong, weak, or nonelectrolytes by measuring the electrical conductance of an aqueous solution containing the substance. To conduct electricity, a substance must contain freely mobile, charged species. Most familiar is the conduction of electricity through metallic wires, in which case the mobile, charged entities are electrons. Solutions may also conduct electricity if they contain dissolved ions, with conductivity increasing as ion concentration increases. Applying a voltage to electrodes immersed in a solution permits assessment of the relative concentration of dissolved ions, either quantitatively, by measuring the electrical current flow, or qualitatively, by observing the brightness of a light bulb included in the circuit (Figure 13.3.1). 13.3.2 https://chem.libretexts.org/@go/page/47549  Figure 13.3.1 : Solutions of nonelectrolytes, such as ethanol, do not contain dissolved ions and cannot conduct electricity. Solutions of electrolytes contain ions that permit the passage of electricity. The conductivity of an electrolyte solution is related to the strength of the electrolyte. This diagram shows three separate beakers. Each has a wire plugged into a wall outlet. In each case, the wire leads from the wall to the beaker and is split resulting in two ends. One end leads to a light bulb and continues on to a rectangle labeled with a plus sign. The other end leads to a rectangle labeled with a minus sign. The rectangles are in a solution. In the first beaker, labeled “Ethanol No Conductivity,” four pairs of linked small green spheres suspended in the solution between the rectangles. In the second beaker, labeled “K C l Strong Conductivity,” six individual green spheres, three labeled plus and three labeled minus are suspended in the solution. Each of the six spheres has an arrow extending from it pointing to the rectangle labeled with the opposite sign. In the third beaker, labeled “Acetic acid solution Weak conductivity,” two pairs of joined green spheres and two individual spheres, one labeled plus and one labeled minus are shown suspended between the two rectangles. The plus labeled sphere has an arrow pointing to the rectangle labeled minus and the minus labeled sphere has an arrow pointing to the rectangle labeled plus. Water and other polar molecules are attracted to ions, as shown in Figure 13.3.2. The electrostatic attraction between an ion and a molecule with a dipole is called an ion-dipole attraction. These attractions play an important role in the dissolution of ionic compounds in water. Figure 13.3.2 : As potassium chloride (KCl) dissolves in water, the ions are hydrated. The polar water molecules are attracted by the charges on the K+ and Cl− ions. Water molecules in front of and behind the ions are not shown. The diagram shows eight purple spheres labeled K superscript plus and eight green spheres labeled C l superscript minus mixed and touching near the center of the diagram. Outside of this cluster of spheres are seventeen clusters of three spheres, which include one red and two white spheres. A red sphere in one of these clusters is labeled O. A white sphere is labeled H. Two of the green C l superscript minus spheres are surrounded by three of the red and white clusters, with the red spheres closer to the green spheres than the white spheres. One of the K superscript plus purple spheres is surrounded by four of the red and white clusters. The white spheres of these clusters are closest to the purple spheres. When ionic compounds dissolve in water, the ions in the solid separate and disperse uniformly throughout the solution because water molecules surround and solvate the ions, reducing the strong electrostatic forces between them. This process represents a physical change known as dissociation. Under most conditions, ionic compounds will dissociate nearly completely when dissolved, and so they are classified as strong electrolytes.  Example 13.3.1: Identifying Ionic Compounds Which compound(s) will dissolve in solution to separate into ions? a. LiF 13.3.3 https://chem.libretexts.org/@go/page/47549  b. P2 F 5 c. C2 H OH 5 Solution LiF will separate into ions when dissolved in solution, because it is an ionic compound. P F and C H OH are both covalent 2 5 2 5 and will stay as molecules in a solution.  Exercise 13.3.1 Which compounds will dissolve in solution to separate into ions? a. C6H12O11, glucose b. CCl4 c. CaCl2 d. AgNO3 Answer c&d How Temperature Influences Solubility The solubility of a substance is the amount of that substance that is required to form a saturated solution in a given amount of solvent at a specified temperature. Solubility is often measured as the grams of solute per 100 g of solvent. The solubility of sodium chloride in water is 36.0 g per 100 g water at 20 C. The temperature must be specified because solubility varies with o temperature. For gases, the pressure must also be specified. Solubility is specific for a particular solvent. We will consider solubility of material in water as solvent. The solubility of the majority of solid substances increases as the temperature increases. However, the effect is difficult to predict and varies widely from one solute to another. The temperature dependence of solubility can be visualized with the help of a solubility curve, a graph of the solubility vs. temperature (Figure 13.3.4). Figure 13.3.4 : Solubility curves for several compounds. Notice how the temperature dependence of NaCl is fairly flat, meaning that an increase in temperature has relatively little effect on the solubility of NaCl. The curve for KNO , on the other hand, is very steep and so an increase in temperature dramatically 3 increases the solubility of KNO. 3 13.3.4 https://chem.libretexts.org/@go/page/47549 Several substances—HCl, NH , and SO —have solubility that decreases as temperature increases. They are all gases at standard 3 2 pressure. When a solvent with a gas dissolved in it is heated, the kinetic energy of both the solvent and solute increase. As the kinetic energy of the gaseous solute increases, its molecules have a greater tendency to escape the attraction of the solvent molecules and return to the gas phase. Therefore, the solubility of a gas decreases as the temperature increases. Solubility curves can be used to determine if a given solution is saturated or unsaturated. Suppose that 80 g of KNO is added to 3 100 g of water at 30 C. According to the solubility curve, approximately 48 g of KNO will dissolve at 30 C. This means that the o o 3 solution will be saturated since 48 g is less than 80 g. We can also determine that there will be 80 − 48 = 32 g of undissolved KNO 3 remaining at the bottom of the container. Now suppose that this saturated solution is heated to 60 C. According to the o curve, the solubility of KNO at 60 C is about 107 g. Now the solution is unsaturated since it contains only the original 80 g of 3 o dissolved solute. Now suppose the solution is cooled all the way down to 0 C. The solubility at 0 C is about 14 g, meaning that o o 80 − 14 = 66 g of the KNO will re-crystallize. 3 Summary Solubility is the specific amount of solute that can dissolve in a given amount of solvent. Saturated and unsaturated solutions are defined. Ionic compounds dissolve in polar solvents, especially water. This occurs when the positive cation from the ionic solid is attracted to the negative end of the water molecule (oxygen) and the negative anion of the ionic solid is attracted to the positive end of the water molecule (hydrogen). Water is considered the universal solvent since it can dissolve both ionic and polar solutes, as well as some nonpolar solutes (in very limited amounts). The solubility of a solid in water increases with an increase in temperature. Vocabulary Miscible - Liquids that have the ability to dissolve in each other. Immiscible - Liquids that do not have the ability to dissolve in each other. Electrostatic attraction - The attraction of oppositely charged particles. 13.3: Solutions of Solids Dissolved in Water- How to Make Rock Candy is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 13.3.5 https://chem.libretexts.org/@go/page/47549 13.4: SOLUTIONS OF GASES IN WATER  LEARNING OBJECTIVES Explain how temperature and pressure affect the solubility of gases. In an earlier module of this chapter, the effect of intermolecular attractive forces on solution formation was discussed. The chemical structures of the solute and solvent dictate the types of forces possible and, consequently, are important factors in determining solubility. For example, under similar conditions, the water solubility of oxygen is approximately three times greater than that of helium, but 100 times less than the solubility of chloromethane, CHCl3. Considering the role of the solvent’s chemical structure, note that the solubility of oxygen in the liquid hydrocarbon hexane, C6H14, is approximately 20 times greater than it is in water. Other factors also affect the solubility of a given substance in a given solvent. Temperature is one such factor, with gas solubility typically decreasing as temperature increases (Figure 13.4.1 ). This is one of the major impacts resulting from the thermal pollution of natural bodies of water. Figure 13.4.1: The solubilities of these gases in water decrease as the temperature increases. All solubilities were measured with a constant pressure of 101.3 kPa (1 atm) of gas above the solutions. When the temperature of a river, lake, or stream is raised abnormally high, usually due to the discharge of hot water from some industrial process, the solubility of oxygen in the water is decreased. Decreased levels of dissolved oxygen may have serious consequences for the health of the water’s ecosystems and, in severe cases, can result in large-scale fish kills (Figure 13.4.2 ). Figure 13.4.2: (a) The small bubbles of air in this glass of chilled water formed when the water warmed to room temperature and the solubility of its dissolved air decreased. (b) The decreased solubility of oxygen in natural waters subjected to thermal pollution can result in large-scale fish kills. (Credit a: modification of work by Liz West; credit b: modification of work by U.S. Fish and Wildlife Service.) 13.4.1 https://chem.libretexts.org/@go/page/47551 The solubility of a gaseous solute is also affected by the partial pressure of solute in the gas to which the solution is exposed. Gas solubility increases as the pressure of the gas increases. Carbonated beverages provide a nice illustration of this relationship. The carbonation process involves exposing the beverage to a relatively high pressure of carbon dioxide gas and then sealing the beverage container, thus saturating the beverage with CO2 at this pressure. When the beverage container is opened, a familiar hiss is heard as the carbon dioxide gas pressure is released, and some of the dissolved carbon dioxide is typically seen leaving solution in the form of small bubbles (Figure 13.4.3 ). At this point, the beverage is supersaturated with carbon dioxide and, with time, the dissolved carbon dioxide concentration will decrease to its equilibrium value and the beverage will become “flat.” Figure 13.4.3: Opening the bottle of carbonated beverage reduces the pressure of the gaseous carbon dioxide above the beverage. The solubility of CO2 is thus lowered, and some dissolved carbon dioxide may be seen leaving the solution as small gas bubbles. A dark brown liquid is shown in a clear, colorless container. A thick layer of beige bubbles appear at the surface of the liquid. In the liquid, thirteen small clusters of single black spheres with two red spheres attached to the left and right are shown. Red spheres represent oxygen atoms and black represent carbon atoms. Seven white arrows point upward in the container from these clusters to the bubble layer at the top of the liquid. (Credit: modification of work by Derrick Coetzee.)  "FIZZ" The dissolution in a liquid, also known as fizz, usually involves carbon dioxide under high pressure. When the pressure is reduced, the carbon dioxide is released from the solution as small bubbles, which causes the solution to become effervescent, or fizzy. A common example is the dissolving of carbon dioxide in water, resulting in carbonated water. Carbon dioxide is weakly soluble in water, therefore it separates into a gas when the pressure is released. This process is generally represented by the following reaction, where a pressurized dilute solution of carbonic acid in water releases gaseous carbon dioxide at decompression: H2 CO3 (aq) → H2 O(l) + CO2 (g) In simple terms, it is the result of the chemical reaction occurring in the liquid which produces a gaseous product. For many gaseous solutes, the relation between solubility, C , and partial pressure, P , is a proportional one: g g Cg = k Pg where k is a proportionality constant that depends on the identities of the gaseous solute and solvent, and on the solution temperature. This is a mathematical statement of Henry’s law: The quantity of an ideal gas that dissolves in a definite volume of liquid is directly proportional to the pressure of the gas.  EXAMPLE 13.4.1: APPLICATION OF HENRY’S LAW At 20 °C, the concentration of dissolved oxygen in water exposed to gaseous oxygen at a partial pressure of 101.3 kPa (760 torr) is 1.38 × 10−3 mol L−1. Use Henry’s law to determine the solubility of oxygen when its partial pressure is 20.7 kPa (155 torr), the approximate pressure of oxygen in earth’s atmosphere. Solution According to Henry’s law, for an ideal solution the solubility, C , of a gas (1.38 × 10−3 mol L−1, in this case) is directly proportional to g the pressure, P , of the undissolved gas above the solution (101.3 kPa, or 760 torr, in this case). Because we know both C and P , we g g g can rearrange this expression to solve for k. 13.4.2 https://chem.libretexts.org/@go/page/47551  Cg = kPg Cg k = Pg −3 −1 1.38 × 10 mol L = 101.3 kPa −5 −1 −1 = 1.36 × 10 mol L kPa −6 −1 −1 (1.82 × 10 mol L torr ) Now we can use k to find the solubility at the lower pressure. Cg = k Pg −5 −1 −1 1.36 × 10 mol L kPa × 20.7 kPa −6 −1 −1 (or 1.82 × 10 mol L torr × 155 torr) −4 −1 = 2.82 × 10 mol L Note that various units may be used to express the quantities involved in these sorts of computations. Any combination of units that yield to the constraints of dimensional analysis are acceptable.  EXERCISE 13.4.1 A 100.0 mL sample of water at 0 °C to an atmosphere containing a gaseous solute at 20.26 kPa (152 torr) resulted in the dissolution of 1.45 × 10−3 g of the solute. Use Henry’s law to determine the solubility of this gaseous solute when its pressure is 101.3 kPa (760 torr). Answer 7.25 × 10−3 g  CASE STUDY: DECOMPRESSION SICKNESS (“THE BENDS”) Decompression sickness (DCS), or “the bends,” is an effect of the increased pressure of the air inhaled by scuba divers when swimming underwater at considerable depths. In addition to the pressure exerted by the atmosphere, divers are subjected to additional pressure due to the water above them, experiencing an increase of approximately 1 atm for each 10 m of depth. Therefore, the air inhaled by a diver while submerged contains gases at the corresponding higher ambient pressure, and the concentrations of the gases dissolved in the diver’s blood are proportionally higher per Henry’s law. As the diver ascends to the surface of the water, the ambient pressure decreases and the dissolved gases becomes less soluble. If the ascent is too rapid, the gases escaping from the diver’s blood may form bubbles that can cause a variety of symptoms ranging from rashes and joint pain to paralysis and death. To avoid DCS, divers must ascend from depths at relatively slow speeds (10 or 20 m/min) or otherwise make several decompression stops, pausing for several minutes at given depths during the ascent. When these preventative measures are unsuccessful, divers with DCS are often provided hyperbaric oxygen therapy in pressurized vessels called decompression (or recompression) chambers (Figure 13.4.4 ). Figure 13.4.4: (a) US Navy divers undergo training in a recompression chamber. (b) Divers receive hyperbaric oxygen therapy. Two photos are shown. The first shows two people seated in a steel chamber on benches that run length of the chamber on each side. The chamber has a couple of small circular windows and an open hatch-type door. One of the two people is giving a thumbs up gesture. The second image provides a view through a small, circular window. Inside the two people can be seen with masks over their mouths and noses. The people appear to be reading. 13.4.3 https://chem.libretexts.org/@go/page/47551 Deviations from Henry’s law are observed when a chemical reaction takes place between the gaseous solute and the solvent. Thus, for example, the solubility of ammonia in water does not increase as rapidly with increasing pressure as predicted by the law because ammonia, being a base, reacts to some extent with water to form ammonium ions and hydroxide ions. This reaction diagram shows three H atoms bonded to an N atom above, below, and two the left of the N. A single pair of dots is present on the right side of the N. This is followed by a plus, then two H atoms bonded to an O atom to the left and below the O. Two pairs of dots are present on the O, one above and the other to the right of the O. A double arrow, with a top arrow pointing right and a bottom arrow pointing left follows. To the right of the double arrow, four H atoms are shown bonded to a central N atom. These 5 atoms are enclosed in brackets with a superscript plus outside. A plus follows, then an O atom linked by a bond to an H atom on its right. The O atom has pairs of dots above, to the left, and below the atom. The linked O and H are enclosed in brackets with superscript minus outside. Gases can form supersaturated solutions. If a solution of a gas in a liquid is prepared either at low temperature or under pressure (or both), then as the solution warms or as the gas pressure is reduced, the solution may become supersaturated. CONTRIBUTIONS & ATTRIBUTIONS Wikipedia Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected]). 13.4: Solutions of Gases in Water is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 13.4.4 https://chem.libretexts.org/@go/page/47551 13.5: Specifying Solution Concentration- Mass Percent  Learning Objectives Express the amount of solute in a solution in various concentration units. To define a solution precisely, we need to state its concentration: how much solute is dissolved in a certain amount of solvent. Words such as dilute or concentrated are used to describe solutions that have a little or a lot of dissolved solute, respectively, but these are relative terms with meanings that depend on various factors. Introduction Concentration is the measure of how much of a given substance is mixed with another substance. Solutions are said to be either dilute or concentrated. When we say that vinegar is 5% acetic acid in water, we are giving the concentration. If we said the mixture was 10% acetic acid, this would be more concentrated than the vinegar solution. Figure 13.5.1 : The solution on the left is more concentrated than the solution on the right because there is a greater ratio of solute (red balls) to solvent (blue balls) particles. The solution particles are closer together. The solution on the right is more dilute (less concentrated). (CC-SA-BY-3.0 Tracy Poulsen). A concentrated solution is one in which there is a large amount of solute in a given amount of solvent. A dilute solution is one in which there is a small amount of solute in a given amount of solvent. A dilute solution is a concentrated solution that has been, in essence, watered down. Think of the frozen juice containers you buy in the grocery store. To make juice, you have to mix the frozen juice concentrate from inside these containers with three or four times the container size full of water. Therefore, you are diluting the concentrated juice. In terms of solute and solvent, the concentrated solution has a lot of solute versus the dilute solution that would have a smaller amount of solute. The terms "concentrated" and "dilute" provide qualitative methods of describing concentration. Although qualitative observations are necessary and have their place in every part of science, including chemistry, we have seen throughout our study of science that there is a definite need for quantitative measurements in science. This is particularly true in solution chemistry. In this section, we will explore some quantitative methods of expressing solution concentration. Mass Percent There are several ways of expressing the concentration of a solution by using a percentage. The mass/mass percent (% m/m) is defined as the mass of a solute divided by the mass of a solution times 100: mass of solute % m/m = × 100% mass of solution mass of solution = mass of solute + mass solvent If you can measure the masses of the solute and the solution, determining the mass/mass percent is easy. Each mass must be expressed in the same units to determine the proper concentration. Suppose that a solution was prepared by dissolving 25.0 g of sugar into 100.0 g of water. The mass of the solution is mass of solution = 25.0g sugar + 100.0g water = 125.0 g The percent by mass would be calculated by: 25.0 g sugar Percent by mass = × 100% = 20.0% sugar 125.0 g solution 13.5.1 https://chem.libretexts.org/@go/page/47553   Example 13.5.1 A saline solution with a mass of 355 g has 36.5 g of NaCl dissolved in it. What is the mass/mass percent concentration of the solution? Solution We can substitute the quantities given in the equation for mass/mass percent: 36.5 g % m/m = × 100% = 10.3% 355 g  Exercise 13.5.1 A dextrose (also called D-glucose, C6H12O6) solution with a mass of 2.00 × 102 g has 15.8 g of dextrose dissolved in it. What is the mass/mass percent concentration of the solution? Answer 7.90 % Using Mass Percent in Calculations Sometimes you may want to make up a particular mass of solution of a given percent by mass and need to calculate what mass of the solute to use. Using mass percent as a conversion can be useful in this type of problem. The mass percent can be expressed as a g solute 100 gsolution conversion factor in the form or 100 gsolution g solute For example, if you need to make 3000.0 g of a 5.00% solution of sodium chloride, the mass of solute needs to be determined. Solution Given: 3000.0 g NaCl solution 5.00% NaCl solution Find: mass of solute = ? g NaCl Other known quantities: 5.00 g NaCl is to 100 g solution The appropriate conversion factor (based on the given mass percent) can be used follows: To solve for the mass of NaCl, the given mass of solution is multiplied by the conversion factor. 5.00 g N aC l gN aC l = 3, 000.0 g N aC l solution × = 150.0g N aC l 100 g N aC l solution You would need to weigh out 150 g of NaCl and add it to 2850 g of water. Notice that it was necessary to subtract the mass of the NaCl (150 g) from the mass of solution (3000 g) to calculate the mass of the water that would need to be added.  Exercise 13.5.1 What is the amount (in g) of hydrogen peroxide (H2O2) needed to make a 6.00 kg, 3.00 % (by mass) H2O2 solution? Answer 13.5.2 https://chem.libretexts.org/@go/page/47553  180 g H2O2 Contributors and Attributions 13.5: Specifying Solution Concentration- Mass Percent is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 13.5.3 https://chem.libretexts.org/@go/page/47553 13.6: Specifying Solution Concentration- Molarity  Learning Objectives Use molarity to determine quantities in chemical reactions. Use molarity as a conversion factor in calculations. Another way of expressing concentration is to give the number of moles of solute per unit volume of solution. Of all the quantitative measures of concentration, molarity is the one used most frequently by chemists. Molarity is defined as the number of moles of solute per liter of solution. number of moles of solute molarity = (13.6.1) number of liters of solution The symbol for molarity is M or moles/liter. Chemists also use square brackets to indicate a reference to the molarity of a substance. For example, the expression [ Ag ] refers to the molarity of the silver ion in solution. Solution concentrations expressed in molarity are the easiest + to perform calculations with, but the most difficult to make in the lab. Such concentration units are useful for discussing chemical reactions in which a solute is a product or a reactant. Molar mass can then be used as a conversion factor to convert amounts in moles to amounts in grams. It is important to remember that “mol” in this expression refers to moles of solute and that “L” refers to liters of solution. For example, if you have 1.5 mol of NaCl dissolved in 0.500 L of solution, its molarity is 1.5 mol NaCl = 3.0 M NaCl 0.500 L solution Sometimes (aq) is added when the solvent is water, as in “3.0 M NaCl (aq).” This is read as "a 3.00 molar sodium chloride solution," meaning that there are 3.00 moles of NaOH dissolved per one liter of solution. Be sure to note that molarity is calculated as the total volume of the entire solution, not just volume of solvent! The solute contributes to total volume. If the quantity of the solute is given in mass units, you must convert mass units to mole units before using the definition of molarity to calculate concentration. For example, what is the molar concentration of a solution of 22.4 g of HCl dissolved in 1.56 L? Step 1: First, convert the mass of solute to moles using the molar mass of HCl (36.5 g/mol): 1 mol H C l 22.4 gH C l × = 0.614 mol H C l 36.5 gH C l Step 2: Now we can use the definition of molarity to determine a concentration: 0.614 mol H C l M = = 0.394 M H C l 1.56L solution Before a molarity concentration can be calculated, the amount of the solute must be expressed in moles, and the volume of the solution must be expressed in liters, as demonstrated in the following example.  Example 13.6.1 A solution is prepared by dissolving 42.23 g of NH 4 Cl into enough water to make 500.0 mL of solution. Calculate its molarity. Solution Solutions to Example 13.6.1 Steps for Problem Solving Given: Identify the "given" information and what the problem is asking you to Mass = 42.23 g NH Cl 4 "find." Volume solution = 500.0 mL = 0.5000 L Find: Molarity = ? M List other known quantities. Molar mass NH 4 Cl = 53.50 g/mol 13.6.1 https://chem.libretexts.org/@go/page/47555  Steps for Problem Solving 1. The mass of the ammonium chloride is first converted to moles. Plan the problem. 2. Then the molarity is calculated by dividing by liters. Note the given volume has been converted to liters. mol N H4 Cl M = L solution Now substitute the known quantities into the equation and solve. 1 mol NH Cl 4 42.23 g NH Cl × = 0.7893 mol NH Cl 4 4 Cancel units and calculate. 53.50 g NH Cl 4 0.7893 mol NH Cl 4 = 1.579 M 0.5000 L solution The molarity is 1.579 M, meaning that a liter of the solution would contain Think about your result. 1.579 mol NH Cl. Four significant figures are appropriate. 4  Exercise 13.6.1A What is the molarity of a solution made when 66.2 g of C6H12O6 are dissolved to make 235 mL of solution? Answer 1.57 M C6H12O6  Exercise 13.6.1B What is the concentration, in mol/L, where 137 g of NaCl has been dissolved in enough water to make 500 mL of solution? Answer 4.69 M NaCl Using Molarity in Calculations Concentration can be a conversion factor between the amount of solute and the amount of solution or solvent (depending on the definition of the concentration unit). As such, concentrations can be useful in a variety of stoichiometry problems. In many cases, it is best to use the original definition of the concentration unit; it is that definition that provides the conversion factor. A simple example of using a concentration unit as a conversion factor is one in which we use the definition of the concentration unit and rearrange; we can do the calculation again as a unit conversion, rather than as a definition.  Example 13.6.2: Determining Moles of Solute, Given the Concentration and Volume of a Solution For example, suppose we ask how many moles of solute are present in 0.108 L of a 0.887 M NaCl solution. Because 0.887 M means 0.887 mol/L, we can use this second expression for the concentration as a conversion factor: Solution 0.887 mol N aC l 0.108 L solution × = 0.0958 mol N aC l 1L solution 13.6.2 https://chem.libretexts.org/@go/page/47555  If we used the definition approach, we get the same answer, but now we are using conversion factor skills. Like any other conversion factor that relates two different types of units, the reciprocal of the concentration can be also used as a conversion factor.  Example 13.6.3: Determining Volume of a Solution, Given the Concentration and Moles of Solute Using concentration as a conversion factor, how many liters of 2.35 M CuSO4 are needed to obtain 4.88 mol of CuSO4? Solution This is a one-step conversion, but the concentration must be written as the reciprocal for the units to work out: 1 L solution 4.88 mol C uSO4 × = 2.08 L of solution 2.35 mol C uSO4 In a laboratory situation, a chemist must frequently prepare a given volume of solutions of a known molarity. The task is to calculate the mass of the solute that is necessary. The molarity equation can be rearranged to solve for moles, which can then be converted to grams. The following example illustrates this.  Example 13.6.4 A chemist needs to prepare 3.00 L of a 0.250 M solution of potassium permanganate (KMnO ). What mass of KMnO does she need to 4 4 make the solution? Solution Solutions to Example 13.6.4 Steps for Problem Solving Given: Identify the "given" information and what the problem is asking you to Molarity = 0.250 M "find." Volume = 3.00 L Find: Mass KMnO =? g 4 Molar mass KMnO = 158.04 g/mol List other known quantities. 4 0.250 mol KMnO4 to 1 L of KMnO4 solution Plan the problem. Now substitute the known quantities into the equation and solve. mol KMnO = 0.250 M KMnO × 3.00 L = 0.750 mol KM 4 4 Cancel units and calculate. 0.250 mol KMnO 4 158.04 g KMnO 4 3.00 L solution × × = 119 g KMnO 1 L solution 1 mol KMnO 4 When 119 g of potassium permanganate is dissolved into water to make Think about your result. 3.00 L of solution, the molarity is 0.250 M.  Exercise 13.6.4A Using concentration as a conversion factor, how many liters of 0.0444 M CH2O are needed to obtain 0.0773 mol of CH2O? Answer 13.6.3 https://chem.libretexts.org/@go/page/47555  1.74 L  Exercise 13.6.4B Answer the problems below using concentration as a conversion factor. a. What mass of solute is present in 1.08 L of 0.0578 M H2SO4? b. What volume of 1.50 M HCl solution contains 10.0 g of hydrogen chloride? Answer a 6.12 g Answer b 183 mL or 0.183L  How to Indicate Concentration Square brackets are often used to represent concentration, e.g., [NaOH] = 0.50 M. Use the capital letter M for molarity, not a lower case m (this is a different concentration unit called molality). How To Prepare Solutions Watch as the Flinn Scientific Tech Staff demonstrates "How To Prepare Solutions." It is important to note that there are many different ways you can set up and solve your chemistry equations. Some students prefer to answer multi-step calculations in one long step, while others prefer to work out each step individually. Neither method is necessarily better or worse than the other method—whichever makes the most sense to you is the one that you should use. In this text, we will typically use unit analysis (also called dimension analysis or factor analysis). Contributors and Attributions Peggy Lawson (Oxbow Prairie Heights School). Funded by Saskatchewan Educational Technology Consortium. 13.6: Specifying Solution Concentration- Molarity is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 13.6.4 https://chem.libretexts.org/@go/page/47555 13.7: SOLUTION DILUTION  LEARNING OBJECTIVES Explain how concentrations can be changed in the lab. Understand how stock solutions are used in the laboratory. We are often concerned with how much solute is dissolved in a given amount of solution. We will begin our discussion of solution concentration with two related and relative terms: dilute and concentrated. A dilute solution is one in which there is a relatively small amount of solute dissolved in the solution. A concentrated solution contains a relatively large amount of solute. These two terms do not provide any quantitative information (actual numbers), but they are often useful in comparing solutions in a more general sense. These terms also do not tell us whether or not the solution is saturated or unsaturated, or whether the solution is "strong" or "weak". These last two terms will have special meanings when we discuss acids and bases, so be careful not to confuse them. STOCK SOLUTIONS It is often necessary to have a solution with a concentration that is very precisely known. Solutions containing a precise mass of solute in a precise volume of solution are called stock (or standard) solutions. To prepare a standard solution, a piece of lab equipment called a volumetric flask should be used. These flasks range in size from 10 mL to 2000 mL and are carefully calibrated to a single volume. On the narrow stem is a calibration mark. The precise mass of solute is dissolved in a bit of the solvent, and this is added to the flask. Then, enough solvent is added to the flask until the level reaches the calibration mark. Often, it is convenient to prepare a series of solutions of known concentrations by first preparing a single stock solution, as described in the previous section. Aliquots (carefully measured volumes) of the stock solution can then be diluted to any desired volume. In other cases, it may be inconvenient to weigh a small mass of sample accurately enough to prepare a small volume of a dilute solution. Each of these situations requires that a solution be diluted to obtain the desired concentration. DILUTIONS OF STOCK (OR STANDARD) SOLUTIONS Imagine we have a salt water solution with a certain concentration. That means we have a certain amount of salt (a certain mass or a certain number of moles) dissolved in a certain volume of solution. Next, we will dilute this solution. This is done by adding more water, not more salt: → Before Dilution and After Dilution The molarity of solution 1 is moles 1 M1 = liter 1 and the molarity of solution 2 is moles 2 M2 = liter 2 rearrange the equations to find moles: moles 1 = M1 liter 1 and moles 2 = M2 liter 2 What stayed the same and what changed between the two solutions? By adding more water, we changed the volume of the solution. Doing so also changed its concentration. However, the number of moles of solute did not change. So, moles1 = moles2 13.7.1 https://chem.libretexts.org/@go/page/47557 Therefore M1 V1 = M2 V2 (13.7.1) where M1 and M are the concentrations of the original and diluted solutions 2 V1 and V are the volumes of the two solutions 2 Preparing dilutions is a common activity in the chemistry lab and elsewhere. Once you understand the above relationship, the calculations are simple. Suppose that you have 100. mL of a 2.0 M solution of HCl. You dilute the solution by adding enough water to make the solution volume 500. mL. The new molarity can easily be calculated by using the above equation and solving for M. 2 M1 × V1 2.0 M × 100. mL M2 = = = 0.40 M HCl V2 500. mL The solution has been diluted by one-fifth since the new volume is five times as great as the original volume. Consequently, the molarity is one-fifth of its original value. Another common dilution problem involves calculating what amount of a highly concentrated solution is required to make a desired quantity of solution of lesser concentration. The highly concentrated solution is typically referred to as the stock solution.  EXAMPLE 13.7.1: DILUTING NITRIC ACID Nitric acid (HNO ) is a powerful and corrosive acid. When ordered from a chemical supply company, its molarity is 16 M. How much 3 of the stock solution of nitric acid needs to be used to make 8.00 L of a 0.50 M solution? Solution Solutions to Example13.7.1 Steps for Problem Solving Identify the "given" Given: information M1, Stock HNO = 16 M and what 3 V = 8.00 L 2 the M = 0.50 M2 problem is Find: Volume stock HNO (V1 ) =? L asking you 3 to "find." List other known none quantities. First, rearrange the equation algebraically to solve for V.1 Plan the M2 × V2 problem. V1 = M1 Now substitute the known quantities into the equation and solve. Calculate and cancel 0.50 M × 8.00 L V1 = = 0.25 L (13.7.2) units. 16 M Think 0.25 L (250 mL) of the stock HNO needs to be diluted with water to a final volume of 8.00 L. The dilution is by a factor of 32 to go from 16 M to about your 3 0.5 M. result.  EXERCISE 13.7.1 A 0.885 M solution of KBr with an initial volume of 76.5 mL has more water added until its concentration is 0.500 M. What is the new volume of the solution? Answer 135.4 mL 13.7.2 https://chem.libretexts.org/@go/page/47557  Note that the calculated volume will have the same dimensions as the input volume, and dimensional analysis tells us that in this case we don't need to convert to liters, since L cancels when we divide M (mol/L) by M (mol/L). DILUTING AND MIXING SOLUTIONS Diluting and Mixing Solutions How to Dilute a Sol… Sol… How to Dilute a Solution by CarolinaBiological This page titled 13.7: Solution Dilution is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Ed Vitz, John W. Moore, Justin Shorb, Xavier Prat-Resina, Tim Wendorff, & Adam Hahn. 13.7.3 https://chem.libretexts.org/@go/page/47557 13.8: SOLUTION STOICHIOMETRY  LEARNING OBJECTIVES Determine amounts of reactants or products in aqueous solutions. As we learned previously, double replacement reactions involve the reaction between ionic compounds in solution and, in the course of the reaction, the ions in the two reacting compounds are “switched” (they replace each other). Because these reactions occur in aqueous solution, we can use the concept of molarity to directly calculate the number of moles of reactants or products that will be formed, and therefore their amounts (i.e. volume of solutions or mass of precipitates). As an example, lead (II) nitrate and sodium chloride react to form sodium nitrate and the insoluble compound, lead (II) chloride. Pb(NO ) (aq) + 2 NaCl(aq) → PbCl (s) + 2 NaNO (aq) (13.8.1) 3 2 2 3 In the reaction shown above, if we mixed 0.123 L of a 1.00 M solution of NaCl with 1.50 M solution of Pb(NO ) 3 2 , we could calculate the volume of Pb(NO ) solution needed to completely precipitate the Pb ions. 2+ 3 2 The molar concentration can also be expressed as the following: 1.00 mol NaCl 1.00 M NaCl = 1 L NaCl solution and 1.50 mol Pb(NO ) 3 2 1.50 M Pb(NO ) = 3 2 1 L Pb(NO ) solution 3 2 First, we must examine the reaction stoichiometry in the balanced reaction (Equation 13.8.1 ). In this reaction, one mole of Pb(NO 3 ) 2 reacts with two moles of NaCl to give one mole of PbCl precipitate. Thus, the concept map utilizing the stoichiometric ratios is: 2 so the volume of lead (II) nitrate that reacted is calculated as: 1.00 mol NaCl 1 mol Pb(NO ) 1 L Pb(NO ) solution 3 2 3 2 0.123 L NaCl solution × × × = 0.041 Pb(NO ) 3 2 1 L NaCl solution 2 mol NaCl 1.5 mol Pb(NO ) 3 2 L solution This volume makes intuitive sense for two reasons: (1) the number of moles of Pb(NO ) required is half of the number of moles of NaCl , based off of the 3 2 stoichiometry in the balanced reaction (Equation 13.8.1 ); (2) the concentration of Pb(NO ) solution is 50% greater than the NaCl solution, so less volume is 3 2 needed.  EXAMPLE 13.8.1 What volume (in L) of 0.500 M sodium sulfate will react with 275 mL of 0.250 M barium chloride to completely precipitate all Ba in the solution? 2+ Solution 13.8.1 https://chem.libretexts.org/@go/page/52193  Solutions to Example 13.8.1 Steps for Problem Example 13.8.1 Solving Identify the "given" Given: 275 mL BaCl2 information 0.250 molBaC l2 0.250 M BaCl or and what 2 1 L BaC l2 solution the 0.500 M Na SO or 0.500 molN a2 S O4 2 4 problem is 1 L N a2 S O4 solution asking you Find: Volume Na 2 SO 4 solution. to "find." Set up and balance the Na SO (aq) + BaCl (aq) ⟶ BaSO (s) + 2 2 4 2 – NaCl(aq) 4 chemical An insoluble product is formed after the reaction. equation. List other 1 mol of Na2SO4 to 1 mol BaCl2 known 1000 mL = 1 L quantities. Prepare a concept map and use the proper conversion factor. Cancel 1 L 0.250 mol BaC l2 1 mol N a2 S O4 1 L N a2 S O4 solution 275 mL BaC l2 solution × × × × units and 1000 mL 1 L BaC l2 solution 1 mol BaC l2 0.500 molN a2 S O4 calculate. = 0.1375 L sodium sulfate Think The lesser amount (almost half) of sodium sulfate is to be expected as it is more concentrated than barium chloride. about your Also, the units are correct. result.  EXERCISE 13.8.1 What volume of 0.250 M lithium hydroxide will completely react with 0.500 L of 0.250 M of sulfuric acid solution? Answer 0.250 L LiOH solution This page titled 13.8: Solution Stoichiometry is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew, Henry Agnew, Paul Young, & Paul Young (ChemistryOnline.com). 13.8.2 https://chem.libretexts.org/@go/page/52193 13.9: Freezing Point Depression and Boiling Point Elevation  Learning Objectives Explain what the term "colligative" means, and list the colligative properties. Indicate what happens to the boiling point and the freezing point of a solvent when a solute is added to it. Calculate boiling point elevations and freezing point depressions for a solution. People who live in colder climates have seen trucks put salt on the roads when snow or ice is forecast. Why is this done? As a result of the information you explore in this section, you will understand why these events occur. You will also learn to calculate exactly how much of an effect a specific solute can have on the boiling point or freezing point of a solution. The example given in the introduction is an example of a colligative property. Colligative properties are properties that differ based on the concentration of solute in a solvent, but not on the type of solute. What this means for the example above is that people in colder climates do not necessarily need salt to get the same effect on the roads—any solute will work. However, the higher the concentration of solute, the more these properties will change. Boiling Point Elevation Water boils at 100 C at 1 atm of pressure, but a solution of saltwater does not. When table salt is added to water, the resulting o solution has a higher boiling point than the water did by itself. The ions form an attraction with the solvent particles that prevents the water molecules from going into the gas phase. Therefore, the saltwater solution will not boil at 100 C. In order for the o saltwater solution to boil, the temperature must be raised about 100 C. This is true for any solute added to a solvent; the boiling o point will be higher than the boiling point of the pure solvent (without the solute). In other words, when anything is dissolved in water, the solution will boil at a higher temperature than pure water would. The boiling point elevation due to the presence of a solute is also a colligative property. That is, the amount of change in the boiling point is related to the number of particles of solute in a solution and is not related to the chemical composition of the solute. A 0.20 m solution of table salt and a 0.20 m solution of hydrochloric acid would have the same effect on the boiling point. Freezing Point Depression The effect of adding a solute to a solvent has the opposite effect on the freezing point of a solution as it does on the boiling point. A solution will have a lower freezing point than a pure solvent. The freezing point is the temperature at which the liquid changes to a solid. At a given temperature, if a substance is added to a solvent (such as water), the solute-solvent interactions prevent the solvent from going into the solid phase. The solute-solvent interactions require the temperature to decrease further in order to solidify the solution. A common example is found when salt is used on icy roadways. Salt is put on roads so that the water on the roads will not freeze at the normal 0 C but at a lower temperature, as low as −9 C. The de-icing of planes is another common example of o o freezing point depression in action. A number of solutions are used, but commonly a solution such as ethylene glycol, or a less toxic monopropylene glycol, is used to de-ice an aircraft. The aircrafts are sprayed with the solution when the temperature is predicted to drop below the freezing point. The freezing point depression is the difference in the freezing points of the solution from the pure solvent. This is true for any solute added to a solvent; the freezing point of the solution will be lower than the freezing point of the pure solvent (without the solute). Thus, when anything is dissolved in water, the solution will freeze at a lower temperature than pure water would. The freezing point depression due to the presence of a solute is also a colligative property. That is, the amount of change in the freezing point is related to the number of particles of solute in a solution and is not related to the chemical composition of the solute. A 0.20 m solution of table salt and a 0.20 m solution of hydrochloric acid would have the same effect on the freezing point. 13.9.1 https://chem.libretexts.org/@go/page/47559  Figure 13.9.1 : Comparison of boiling and freezing points of a pure liquid (right side) with a solution (left side). Comparing the Freezing and Boiling Point of Solutions Recall that covalent and ionic compounds do not dissolve in the same way. Ionic compounds break up into cations and anions when they dissolve. Covalent compounds typically do not break up. For example a sugar/water solution stays as sugar and water, with the sugar molecules staying as molecules. Remember that colligative properties are due to the number of solute particles in the solution. Adding 10 molecules of sugar to a solvent will produce 10 solute particles in the solution. When the solute is ionic, such as NaCl however, adding 10 formulas of solute to the solution will produce 20 ions (solute particles) in the solution. Therefore, adding enough NaCl solute to a solvent to produce a 0.20 m solution will have twice the effect of adding enough sugar to a solvent to produce a 0.20 m solution. Colligative properties depend on the number of solute particles in the solution. "i" is the number of particles that the solute will dissociate into upon mixing with the solvent. For example, sodium chloride, NaCl, will dissociate into two ions so for NaCl, i = 2 ; for lithium nitrate, LiNO , i = 2 ; and for calcium chloride, CaCl , i = 3. 3 2 For covalent compounds, i is always equal to 1. By knowing the molality of a solution and the number of particles a compound will dissolve to form, it is possible to predict which solution in a group will have the lowest freezing point. To compare the boiling or freezing points of solutions, follow these general steps: 1. Label each solute as ionic or covalent. 2. If the solute is ionic, determine the number of ions in the formula. Be careful to look for polyatomic ions. 3. Multiply the original molality (m) of the solution by the number of particles formed when the solution dissolves. This will give you the total concentration of particles dissolved. 4. Compare these values. The higher total concentration will result in a higher boiling point and a lower freezing point.  Example 13.9.1 Rank the following solutions in water in order of increasing (lowest to highest) freezing point: 0.1 m NaCl 0.1 m C H O 6 12 6 0.1 m CaI 2 Solution To compare freezing points, we need to know the total concentration of all particles when the solute has been dissolved. 0.1 m NaCl : This compound is ionic (metal with nonmetal), and will dissolve into 2 parts. The total final concentration is: (0.1 m) (2) = 0.2 m 0.1 m C H O : This compound is covalent (nonmetal with nonmetal), and will stay as 1 part. The total final 6 12 6 concentration is: (0.1 m) (1) = 0.1 m 0.1 m CaI : This compound is ionic (metal with nonmetal), and will dissolve into 3 parts. The total final concentration is: 2 (0.1 m) (3) = 0.3 m Remember, the greater the concentration of particles, the lower the freezing point will be. 0.1 m CaI will have the lowest 2 freezing point, followed by 0.1 m NaCl, and the highest of the three solutions will be 0.1 m C H O , but all three of them 6 12 6 will have a lower freezing point than pure water. 13.9.2 https://chem.libretexts.org/@go/page/47559 The boiling point of a solution is higher than the boiling point of a pure solvent, and the freezing point of a solution is lower than the freezing point of a pure solvent. However, the amount to which the boiling point increases or the freezing point decreases depends on the amount of solute that is added to the solvent. A mathematical equation is used to calculate the boiling point elevation or the freezing point depression. The boiling point elevation is the amount that the boiling point temperature increases compared to the original solvent. For example, the boiling point of pure water at 1.0 atm is 100 C while the boiling point of a 2% saltwater solution is about 102 C. o o Therefore, the boiling point elevation would be 2 C. The freezing point depression is the amount that the freezing temperature o decreases. Both the boiling point elevation and the freezing point depression are related to the molality of the solution. Looking at the formula for the boiling point elevation and freezing point depression, we see similarities between the two. The equation used to calculate the increase in the boiling point is: ΔTb = kb ⋅ m ⋅ i (13.9.1) Where: ΔTb = the amount the boiling point increases. kb = the boiling point elevation constant which depends on the solvent (for water, this number is 0.515 C/m). o m = the molality of the solution. i = the number of particles formed when that compound dissolves (for covalent compounds, this number is always 1). The following equation is used to calculate the decrease in the freezing point: ΔTf = kf ⋅ m ⋅ i (13.9.2) Where: ΔTf = the amount the freezing temperature decreases. kf = the freezing point depression constant which depends on the solvent (for water, this number is 1.86 C/m). o m = the molality of the solution. i = the number of particles formed when that compound dissolves (for covalent compounds, this number is always 1).  Example 13.9.2: Adding Antifreeze to Protein Engines Antifreeze is used in automobile radiators to keep the coolant from freezing. In geographical areas where winter temperatures go below the freezing point of water, using pure water as the coolant could allow the water to freeze. Since water expands when it freezes, freezing coolant could crack engine blocks, radiators, and coolant lines. The main component in antifreeze is ethylene glycol, C H (OH). What is the concentration of ethylene glycol in a solution of water, in molality, if the freezing 2 4 2 point dropped by 2.64 C? The freezing point constant, k , for water is 1.86 C/m. o f o Solution Use the equation for freezing point depression of solution (Equation 13.9.2): ΔTf = kf ⋅ m ⋅ i Substituting in the appropriate values we get: o o 2.64 C = (1.86 C/m) (m) (1) Solve for m by dividing both sides by 1.86 o. C/m m = 1.42  Example 13.9.3: Adding Salt to Elevate Boiling Temperature A solution of 10.0 g of sodium chloride is added to 100.0 g of water in an attempt to elevate the boiling point. What is the boiling point of the solution? k for water is 0.52 C/m. b o Solution 13.9.3 https://chem.libretexts.org/@go/page/47559  Use the equation for boiling point elevation of solution (Equation 13.9.1): ΔTb = kb ⋅ m ⋅ i We need to be able to substitute each variable into this equation. o kb = 0.52 C/m m: We must solve for this using stoichiometry. Given: 10.0 g NaCland 100.0 g H 2 O Find: mol NaCl/kg H 2. Ratios: O molar mass of NaCl, 1000 g = 1 kg 10.0 g NaCl 1000 g H O 1 mol NaCl 2 ⋅ ⋅ = 1.71 m 1 kg H O 100.0 g H O 58.45 g NaCl 2 2 For NaCl, i = 2 Substitute these values into the equation ΔT b = kb ⋅ m ⋅ i. We get: o C o ΔTb = (0.52 ) (1.71 m ) (2) = 1.78 C m Water normally boils at 100 o , but our calculation shows that the boiling point increased by 1.78 C o. Our new boiling point is C 101.78 C. o Note: Since sea water contains roughly 28.0 g of NaCl per liter, this saltwater solution is approximately four times more concentrated than sea water (all for a 2° C rise of boiling temperature). Summary Colligative properties are properties that are due only to the number of particles in solution, and are not related to the chemical properties of the solute. Boiling points of solutions are higher than the boiling points of the pure solvents. Freezing points of solutions are lower than the freezing points of the pure solvents. Ionic compounds split into ions when they dissolve, forming more particles. Covalent compounds stay as complete molecules when they dissolve. Vocabulary Colligative property - A property that is due only to the number of particles in solution, and not the type of the solute. Boiling point elevation - The amount that the boiling point of a solution increases from the boiling point of the pure solvent. Freezing point depression - The amount that the freezing point of a solution decreases from the freezing point of the pure solvent. 13.9: Freezing Point Depression and Boiling Point Elevation is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 13.9.4 https://chem.libretexts.org/@go/page/47559 13.10: Osmosis  Learning Objectives Explain the following laws within the Ideal Gas Law Before we introduce the final colligative property, we need to present a new concept. A semipermeable membrane is a thin membrane that will pass certain small molecules, but not others. A thin sheet of cellophane, for example, acts as a semipermeable membrane. Consider the system in Figure 13.10.1. Figure 13.10.1: Osmosis. (a) Two solutions of differing concentrations are placed on either side of a semipermeable membrane. (b) When osmosis occurs, solvent molecules selectively pass through the membrane from the dilute solution to the concentrated solution, diluting it until the two concentrations are the same. The pressure exerted by the different height of the solution on the right is called the osmotic pressure. (CC BY-SA-NC 3.0; anonymous) a. A semipermeable membrane separates two solutions having the different concentrations marked. Curiously, this situation is not stable; there is a tendency for water molecules to move from the dilute side (on the left) to the concentrated side (on the right) until the concentrations are equalized, as in Figure 13.10.1b. b. This tendency is called osmosis. In osmosis, the solute remains in its original side of the system; only solvent molecules move through the semipermeable membrane. In the end, the two sides of the system will have different volumes. Because a column of liquid exerts a pressure, there is a pressure difference (Π) on the two sides of the system that is proportional to the height of the taller column. This pressure difference