Chapter 8 Grade 11 Chemistry PDF

Solutions and Their Concentrations Y our environment is made up of many important solutions, or homogeneous mixtures. The air you breathe and the liquids you drink are Chapter Preview...

Solutions and Their Concentrations Y our environment is made up of many important solutions, or homogeneous mixtures. The air you breathe and the liquids you drink are Chapter Preview 8.1 Types of Solutions solutions. So are many of the metallic objects that you use every day. The quality of a solution, such as tap water, depends on the substances that 8.2 Factors That Affect Rate of Dissolving are dissolved in it. “Clean” water may contain small amounts of dissolved and Solubility substances, such as iron and chlorine. “Dirty” water may have dangerous chemicals dissolved in it. 8.3 The Concentration of Solutions The difference between clean water and undrinkable water often depends on concentration: the amount of a dissolved substance in a 8.4 Preparing Solutions particular quantity of a solution. For example, tap water contains a low concentration of fluoride to help keep your teeth healthy. Water with a high concentration of fluoride, however, could be harmful to your health. Water is a good solvent for many substances. You may have noticed, however, that grease-stained clothing cannot be cleaned by water alone. Grease is one substance that does not dissolve in water. Why doesn’t it dissolve? In this chapter, you will find out why. You will learn how solutions form. You will explore factors that affect a substance’s ability to dissolve. You will find out more about the concentration of solutions, Concepts and Skills and you will have a chance to prepare your own solutions as well. Yo u W i l l N e e d Before you begin this chapter, review the following concepts As water runs and skills: through soil and classifying mixtures rocks, it dissolves (Chapter 1, section 1.3) minerals such as predicting molecular iron, calcium, and polarity (Chapter 3, magnesium. What section 3.3) makes water such distinguishing between a good solvent? intermolecular and intramolecular forces (Chapter 3, section 3.2) describing the shape and bonding of the water molecule (Chapter 3, section 3.3) calculating molar mass (Chapter 5, section 5.3) calculating molar amounts (Chapter 5, section 5.3) Chapter 8 Solutions and Their Concentrations MHR 283  8.1 Types of Solutions Section Preview/ As stated in the chapter opener, a solution is a homogeneous mixture. It Specific Expectations is uniform throughout. If you analyze any two samples of a solution, you In this section, you will will find that they contain the same substances in the same relative amounts. The simplest solutions contain two substances. Most common explain solution formation, referring to polar and non- solutions contain many substances. polar solvents A solvent is any substance that has other substances dissolved in it. identify examples of solid, In a solution, the substance that is present in the largest amount (whether liquid, and gas solutions by volume, mass, or number of moles) is usually referred to as the solvent. from everyday life The other substances that are present in the solution are called the solutes. communicate your under- Pure substances (such as pure water, H2O) have fixed composition. standing of the following You cannot change the ratio of hydrogen, H, to oxygen, O, in water terms: solution, solvent, without producing an entirely new substance. Solutions, on the other solutes, variable composi- hand, have variable composition. This means that different ratios of tion, aqueous solution, solvent to solute are possible. For example, you can make a weak or a miscible, immiscible, alloys, strong solution of sugar and water, depending on how much sugar you solubility, saturated solution, add. Figure 8.1 shows a strong solution of tea and water on the left, and a unsaturated solution weak solution of tea and water on the right. The ratio of solvent to solute in the strong solution is different from the ratio of solvent to solute in the weak solution. Figure 8.1 How can a solution have variable composition yet be uniform throughout? COURSE CHALLENGE When a solute dissolves in a solvent, no chemical reaction occurs. How can you find out what Therefore, the solute and solvent can be separated using physical solutes are dissolved in a properties, such as boiling point or melting point. For example, water sample of water? What physi- and ethanol have different boiling points. Using this property, a solution cal properties might be useful? of water and ethanol can be separated by the process of distillation. Refer You will need answers to these questions when you do your back to Chapter 1, section 1.2. What physical properties, besides boiling Chemistry Course Challenge. point, can be used to separate the components of solutions and other mixtures? 284 MHR Unit 3 Solutions and Solubility Figure 8.2 Can you identify the components of some of these solutions? A solution can be a gas, a liquid, or a solid. Figure 8.2 shows some examples of solutions. Various combinations of solute and solvent states are possible. For example, a gas can be dissolved in a liquid, or a solid can be dissolved in another solid. Solid, liquid, and gaseous solutions are all around you. Steel is a solid solution of carbon in iron. Juice is a liquid solution of sugar and flavouring dissolved in water. Air is an example of a gaseous solution. The four main components of dry air are nitrogen (78%), oxygen (21%), argon (0.9%), and carbon dioxide (0.03%). Table 8.1 lists some other common solutions. Table 8.1 Types of Solutions Original state of solute Solvent Examples gas gas air; natural gas; oxygen-acetylene mixture used in welding gas liquid carbonated drinks; water in rivers and lakes containing oxygen gas solid hydrogen in platinum liquid gas water vapour in air; gasoline-air mixture liquid liquid alcohol in water; antifreeze in water liquid solid amalgams, such as mercury in silver solid gas mothballs in air Take another look at the four solid liquid sugar in water; table salt in water; components of dry air. Which amalgams component would you call the solid solid alloys, such as the copper-nickel alloy solvent? Which components used to make coins are the solutes? Chapter 8 Solutions and Their Concentrations MHR 285  Electronic Learning Partner You are probably most familiar with liquid solutions, especially aqueous solutions. An aqueous solution is a solution in which water is Your Chemistry 11 Electronic the solvent. Because aqueous solutions are so important, you will focus Learning Partner can help on them in the next two sections of this chapter and again in Chapter 9. you learn more about the Some liquids, such as water and ethanol, dissolve readily in each properties of water. other in any proportion. That is, any amount of water dissolves in any amount of ethanol. Similarly, any amount of ethanol dissolves in any amount of water. Liquids such as these are said to be miscible with each other. Miscible liquids can be combined in any proportions. Thus, either ethanol or water can be considered to be the solvent. Liquids that do not readily dissolve in each other, such as oil and water, are said to CHEM be immiscible. FA C T As you know from Chapter 4, solid solutions of metals are called An alloy that is made of a alloys. Adding even small quantities of another element to a metal metal dissolved in mercury is changes the properties of the metal. Technological advances throughout called an amalgam. A tradition- history have been linked closely to the discovery of new alloys. For al dental amalgam, used to fill example, bronze is an alloy of copper and tin. Bronze contains only cavities in teeth, contains 50% about 10% tin, but it is much stronger than copper and more resistant to mercury. Due to concern over corrosion. Also, bronze can be melted in an ordinary fire so that castings the use of mercury, which is toxic, dentists now use other can be made, as shown in Figure 8.3. materials, such as ceramic materials, to fill dental cavities. Solubility and Saturation The ability of a solvent to dissolve a solute depends on the forces of attraction between the particles. There is always some attraction between solvent and solute particles, so some solute always dissolves. The solubility of a solute is the amount of solute that dissolves in a given quantity of solvent, at a certain temperature. For example, the solubility of sodium chloride in water at 20˚C is 36 g per 100 mL of water. A saturated solution is formed when no more solute will dissolve in a solution, and excess solute is present. For example, 100 mL of a saturated solution of table salt (sodium chloride, NaCl) in water at 20˚C contains 36 g of sodium chloride. The solution is saturated with respect to sodium chloride. If more sodium chloride is added to the solution, it will not dissolve. The solution may still be able to dissolve other solutes, however. An unsaturated solution is a solution that is not yet saturated. Therefore, it can dissolve more solute. For example, a solution that contains 20 g of sodium chloride dissolved in 100 mL of water at 20˚C is unsat- urated. This solution has the potential to dissolve another 16 g of salt, as Figure 8.4 demonstrates. Figure 8.3 The introduction of the alloy bronze around 3000 BCE led to the production of better-quality tools and weapons. 286 MHR Unit 3 Solutions and Solubility  A B Figure 8.4 At 20˚C, the solubility of table salt in water C is 36 g/100 mL. A 20 g of NaCl dissolve to form an unsaturated solution. B 36 g of NaCl dissolve to form a saturated solution. C 40 g of NaCl are added to 100 mL of water. 36 g dissolve to form a saturated solution. 4 g of undissolved solute are left. Suppose that a solute is described as soluble in a particular solvent. This generally means that its solubility is greater than 1 g per 100 mL of solvent. If a solute is described as insoluble, its solubility is less than 0.1 g per 100 mL of solvent. Substances with solubility between these limits are Imagine that you are given a called sparingly soluble, or slightly soluble. Solubility is a relative term, filtered solution of sodium however. Even substances such as oil and water dissolve in each other to chloride. How can you decide some extent, although in very tiny amounts. whether the solution is The general terms that are used to describe solubility for solids and saturated or unsaturated? liquids do not apply to gases in the same way. For example, oxygen is described as soluble in water. Oxygen from the air dissolves in the water of lakes and rivers. The solubility of oxygen in fresh water at 20˚C is only 9 mg/L, or 0.0009 g/100 mL. This small amount of oxygen is enough to ensure the survival of aquatic plants and animals. A solid solute with the same solubility, however, would be described as insoluble in water. Identifying Suitable Solvents Water is a good solvent for many compounds, but it is a poor solvent for others. If you have grease on your hands after adjusting a bicycle chain, Electronic Learning Partner you cannot use water to dissolve the grease and clean your hands. You need to use a detergent, such as soap, to help dissolve the grease in the Go to your Chemistry 11 water. You can also use another solvent to dissolve the grease. How can Electronic Learning Partner you find a suitable solvent? How can you predict whether a solvent will to find out more about the properties of two solvents: dissolve a particular solute? Try the Thought Lab on the next page to find water and benzene. out for yourself. Chapter 8 Solutions and Their Concentrations MHR 287 ThoughtLab Matching Solutes and Solvents kerosene kerosene water A B water Although there is a solvent for every solute, In photo B, the same experiment was repeated not all mixtures produce a solution. Table salt with crystals of cobalt(II) chloride. This time, the dissolves in water but not in kerosene. Oil crystal dissolved in the water but not in the dissolves in kerosene but not in water. What kerosene. properties must a solvent and a solute share in order to produce a solution? Analysis In an investigation, the bottom of a Petri dish 1. Water is a polar molecule. Therefore, it acts was covered with water, as shown in photo A. An as a polar solvent. equal amount of kerosene was added to a second (a) Think about the compounds you classified Petri dish. When a crystal of iodine was added in the Procedure. Which compounds are to the water, it did not dissolve. When a second soluble in water? crystal of iodine was added to the kerosene, (b) Assume that the interaction between solutes however, it did dissolve. and solvents that you examine here applies to a wide variety of substances. Make a Procedure general statement about the type of solute Classify each compound as ionic (containing that dissolves in polar solvents. ions), polar (containing polar molecules), or 2. Kerosene is non-polar. It acts as a non-polar non-polar (containing non-polar molecules). solvent. (a) iodine, I2 (a) Which of the compounds you classified in (b) cobalt(II) chloride, CoCl2 the Procedure is soluble in kerosene? (c) sucrose, C12H22O11 (Hint: Sucrose contains (b) Make a general statement about the type of 8 O–H bonds.) solute that dissolves in non-polar solvents. Section Wrap-up In this section, you learned the meanings of several important terms, such as solvent, solute, saturated solution, unsaturated solution, aqueous solution, and solubility. You need to know these terms in order to understand the material in the rest of the chapter. In section 8.2, you will examine the factors that affect the rate at which a solute dissolves in a solvent. You will also learn about factors that affect solubility. 288 MHR Unit 3 Solutions and Solubility  Section Review 1 K/U Name the two basic components of a solution. 2 K/U Give examples of each type of solution. (a) solid solution (b) liquid solution (c) gaseous solution (at room temperature) 3 K/U Explain the term “homogeneous mixture.” 4 C How do the properties of a homogeneous mixture differ from the properties of a heterogeneous mixture, or mechanical mixture? Use diagrams to explain. 5 K/U Give examples of each type of mixture. (a) homogeneous mixture (b) mechanical mixture (heterogeneous mixture) 6 K/UDistinguish between the following terms: soluble, miscible, and immiscible. 7 K/U Distinguish between an alloy and an amalgam. Give one example of each. 8 K/U What type of solute dissolves in a polar solvent, such as water? Give an example. 9 I Potassium bromate, KBrO3 , is sometimes added to bread dough to make it easier to work with. Suppose that you are given an aqueous solution of potassium bromate. How can you determine if the solution is saturated or unsaturated? 10 K/U Two different clear, colourless liquids were gently heated in an evaporating dish. Liquid A left no residue, while liquid B left a white residue. Which liquid was a solution, and which was a pure substance? Explain your answer. 11 I You are given three liquids. One is a pure substance, and the second is a solution of two miscible liquids. The third is a solution composed of a solid solute dissolved in a liquid solvent. Describe the procedure you would follow to distinguish between the three solutions. 12 MC In 1989, the oil tanker Exxon Valdez struck a reef in Prince William Sound, Alaska. The accident released 40 million litres of crude oil. The oil eventually covered 26 000 km2 of water. (a) Explain why very little of the spilt oil dissolved in the water. (b) The density of crude oil varies. Assuming a value of 0.86 g/mL, estimate the average thickness of the oil slick that resulted from the Exxon Valdez disaster. (c) How do you think most of the oil from a tanker accident is dispersed over time? Why would this have been a slow process in Prince William Sound? 13 MC Food colouring is often added to foods such as candies, ice cream, and icing. Are food colouring dyes more likely to be polar or non-polar molecules? Explain your answer. Chapter 8 Solutions and Their Concentrations MHR 289  8.2 Factors That Affect Rate of Dissolving and Solubility Section Preview/ As you learned in section 8.1, the solubility of a solute is the amount of Specific Expectations solute that dissolves in a given volume of solvent at a certain temperature. In this section, you will Solubility is determined by the intermolecular attractions between solvent and solute particles. You will learn more about solubility and the factors explain some important properties of water that affect it later in this section. First, however, you will look at an important property of a solution: the rate of dissolving, or how quickly a explain solution formation in terms of intermolecular solute dissolves in a solvent. forces between polar, ionic, The rate of dissolving depends on several factors, including tempera- and non-polar substances ture, agitation, and particle size. You have probably used these factors describe the relationship yourself when making solutions like the fruit juice shown in Figure 8.5. between solubility and temperature for solids, liquids, and gases communicate your under- standing of the following terms: rate of dissolving, dipole, dipole-dipole attraction, hydrogen bond- ing, ion-dipole attractions, hydrated, electrolyte, non-electrolytes Figure 8.5 Fruit juice is soluble in water. The concentrated juice in this photograph, however, will take a long time to dissolve. Why? Factors That Affect the Rate of Dissolving You may have observed that a solute, such as sugar, dissolves faster in hot water than in cold water. In fact, for most solid solutes, the rate of dissolving is greater at higher temperatures. At higher temperatures, the solvent molecules have greater kinetic energy. Thus, they collide with the undissolved solid molecules more frequently. This increases their rate of dissolving. Suppose that you are dissolving a spoonful of sugar in a cup of hot coffee. How can you make the sugar dissolve even faster? You can stir the coffee. Agitating a mixture by stirring or by shaking the container increases the rate of dissolving. Agitation brings fresh solvent into contact with undissolved solid. Finally, you may have noticed that a large lump of solid sugar dissolves more slowly than an equal mass of powdered sugar. Decreasing the size of the particles increases the rate of dissolving. When you break Figure 8.6 Chemists often grind up a large mass into many smaller masses, you increase the surface area solids into powders using a mortar and pestle. This increases the rate that is in contact with the solvent. This allows the solid to dissolve faster. of dissolving. Figure 8.6 shows one way to increase the rate of dissolving. 290 MHR Unit 3 Solutions and Solubility Solubility and Particle Attractions Math LINK By now, you are probably very familiar with the process of dissolving. You already know what it looks like when a solid dissolves in a liquid. Calculate the surface area Why, however, does something dissolve? What is happening at the of a cube with the dimensions molecular level? 5.0 cm × 5.0 cm × 5.0 cm. The reasons why a solute may or may not dissolve in a solvent are Now imagine cutting this cube to form smaller cubes related to the forces of attraction between the solute and solvent particles. with the dimensions These forces include the attractions between two solute particles, the 1.0 cm × 1.0 cm × 1.0 cm. attractions between two solvent particles, and the attractions between a How many smaller cubes could solute particle and a solvent particle. When the forces of attraction you make? Calculate their total between different particles in a mixture are stronger than the forces of surface area. attraction between like particles in the mixture, a solution forms. The strength of each attraction influences the solubility, or the amount of a solute that dissolves in a solvent. To make this easier to understand, consider the following three steps in the process of dissolving a solid in a liquid. The Process of Dissolving at the Molecular Level Step 1 The forces between the particles in the solid must be broken. This step always requires energy. In an ionic solid, the forces that are holding the ions together must be broken. In a molecu- lar solid, the forces between the molecules must be broken. Step 2 Some of the intermolecular forces between the particles in the liquid must be broken. This step also requires energy. Step 3 There is an attraction between the particles of the solid and the particles of the liquid. This step always gives off energy. The solid is more likely to dissolve in the liquid if the energy change in step 3 is greater than the sum of the energy changes in steps 1 and 2. (You will learn more about energy and dissolving in Unit 5.) Polar and Non-Polar Substances In the Thought Lab in section 8.1, you observed that solid iodine is insoluble in water. Only a weak attraction exists between the non-polar iodine molecules and the polar water molecules. On the other hand, the intermolecular forces between the water molecules are very strong. As a result, the water molecules remain attracted to each other rather than attracting the iodine molecules. You also observed that iodine is soluble in kerosene. Both iodine and kerosene are non-polar substances. The attraction that iodine and Remember that the rate at kerosene molecules have for each other is greater than the attraction which a solute dissolves is between the iodine molecules in the solid and the attraction between the different from the solubility of the solute. In your notebook, kerosene molecules in the liquid. explain briefly and clearly the The Concept Organizer shown on the next page summarizes the difference between rate of behaviour of polar and non-polar substances in solutions. You will learn dissolving and solubility. more about polar and non-polar substances later in this section. Chapter 8 Solutions and Their Concentrations MHR 291  Concept Organizer Polar and Non-Polar Compounds Polar compounds dissolve Polar solvents e.g. sucrose, C12H22O11 in e.g. water, H2O() do not dissolve in Non-Polar compounds dissolve Non-Polar solvents e.g. iodine, l2(s) in e.g. benzene, C6H6() Solubility and Intermolecular Forces H You have learned that solubility depends on the forces between particles. x Thus, polar substances dissolve in polar solvents, and non-polar substances dissolve in non-polar solvents. What are these forces that act 104.5˚ between particles? O In Chapter 3, section 3.3, you learned that a water molecule is polar. x H It has a relatively large negative charge on the oxygen atom, and positive charges on both hydrogen atoms. Molecules such as water, which have charges separated into positive and negative regions, are said to have a permanent dipole. A dipole consists of two opposite charges that are sepa- Figure 8.7 The bent shape and rated by a short distance. Figure 8.7 shows the dipole of a water molecule. polar bonds of a water molecule give it a permanent dipole. Dipole-Dipole Attractions The attraction between the opposite charges on two different polar mole- cules is called a dipole-dipole attraction. Dipole-dipole attractions are intermolecular. This means that they act between molecules. Usually they are only about 1% as strong as an ionic or covalent bond. In water, there is a special dipole-dipole attraction called hydrogen bonding. It occurs between the oxygen atom on one molecule and the hydrogen atoms on a nearby molecule. Hydrogen bonding is much stronger than an ordinary Electronic Learning Partner dipole-dipole attraction. It is much weaker, however, than the covalent bond between the oxygen and hydrogen atoms in a water molecule. Go to the Chemistry 11 Figure 8.8 illustrates hydrogen bonding between water molecules. Electronic Learning Partner to find out how hydrogen bonding leads to water’s amazing surface tension. Figure 8.8 Hydrogen bonding between water molecules is shown as dotted lines. The H atoms on each molecule are attracted to O atoms on other water molecules. 292 MHR Unit 3 Solutions and Solubility Ion-Dipole Attractions Ionic crystals consist of repeating patterns of oppositely charged ions, as shown in Figure 8.9. What happens when an ionic compound comes in contact with water? The negative end of the dipole on some water mole- cules attracts the cations on the surface of the ionic crystal. At the same time, the positive end of the water dipole attracts the anions. These attrac- tions are known as ion-dipole attractions: attractive forces between an ion and a polar molecule. If ion-dipole attractions can replace the ionic bonds between the cations and anions in an ionic compound, the compound will dissolve. Generally an ionic compound will dissolve in a polar solvent. For example, table salt (sodium chloride, NaCl) is an ionic Cl– compound. It dissolves well in water, which is a polar solvent. When ions are present in an aqueous solution, each ion is hydrated. Na+ This means that it is surrounded by water molecules. Hydrated ions can move through a solution and conduct electricity. A solute that forms an Figure 8.9 Ionic crystals have aqueous solution with the ability to conduct electricity is called an very ordered structures. electrolyte. Figure 8.10 shows hydrated sodium chloride ions, which are electrolytes. hydrated water molecules ions Na+ ion Cl– ion Figure 8.10 Ion-dipole attrac- tions help to explain why sodium chloride dissolves in water. An Exception: Insoluble Ionic Compounds Although most ionic compounds are soluble in water, some are not very soluble at all. The attraction between ions is difficult to break. As a result, compounds with very strong ionic bonds, such as silver chloride, tend to be less soluble in water than compounds with weak ionic bonds, such as sodium chloride. Predicting Solubility You can predict the solubility of a binary compound, such as mercury(II) sulfide, HgS, by comparing the electronegativity of each element in the compound. If there is a large difference in the two electronegativities, the bond between the elements is polar or even ionic. This type of compound probably dissolves in water. If there is only a small difference in the two electronegativities, the bond is not polar or ionic. This type of compound probably does not dissolve in water. For example, the electronegativity of mercury is 1.9. The electronegativity of sulfur is 2.5. The difference in these two electronegativities is small, only 0.6. Therefore, you can predict that mercury(II) sulfide is insoluble in water. In Chapter 9, you will learn another way to predict the solubility of ionic compounds in water. Chapter 8 Solutions and Their Concentrations MHR 293  The Solubility of Covalent Compounds Look back at the Concept Many covalent compounds do not have negative and positive charges to Organizer on page 292. Where attract water molecules. Thus they are not soluble in water. There are do ionic compounds belong in some exceptions, however. Methanol (a component of windshield washer this diagram? fluid), ethanol (the “alcohol” in alcoholic beverages), and sugars (such as sucrose) are examples of covalent compounds that are extremely soluble in water. These compounds dissolve because their molecules contain polar bonds, which are able to form hydrogen bonds with water. For example, sucrose molecules have a number of sites that can form a hydrogen bond with water to replace the attraction between the sucrose molecules. (See Figure 8.11.) The sucrose molecules separate and Electronic Learning Partner become hydrated, just like dissolved ions. The molecules remain neutral, however. As a result, sucrose and other soluble covalent compounds do The Chemistry 11 Electronic Learning Partner contains a not conduct electricity when dissolved in water. They are non-electrolytes. video clip describing how water dissolves ionic and H some covalent compounds. H H O This will be useful if you are having difficulty visualizing H O O H particle attractions. O H H CH2 O CH2 C O O H H H H H H H O C C C H C H O H H O Figure 8.11 A sucrose mole- CH2 O H O C C O C C cule contains several O–H atom O H connections. The O–H bond H H H O H O H H is highly polar, with the H atom having the positive charge. H H O H The negative charges on water H O molecules form hydrogen bonds H O with a sucrose molecule, as H shown by the dotted lines. Insoluble Covalent Compounds The covalent compounds that are found in oil and grease are insoluble in water. They have no ions or highly polar bonds, so they cannot form hydrogen bonds with water molecules. Non-polar compounds tend to be soluble in non-polar solvents, such as benzene or kerosene. The forces between the solute molecules are replaced by the forces between the solute and solvent molecules. In general, ionic solutes and polar covalent solutes both dissolve in polar solvents. Non-polar solutes dissolve in non-polar solvents. The phrase like dissolves like summarizes these observations. It means that solutes and solvents that have similar properties form solutions. If a compound has both polar and non-polar components, it may dissolve in both polar and non-polar solvents. For example, acetic acid, CH3COOH, is a liquid that forms hydrogen bonds with water. It is fully miscible with water. Acetic acid also dissolves in non-polar solvents, such as benzene and carbon tetrachloride, because the CH3 component is non-polar. 294 MHR Unit 3 Solutions and Solubility Factors That Affect Solubility You have taken a close look at the attractive forces between solute and solvent particles. Now that you understand why solutes dissolve, it is time to examine the three factors that affect solubility: molecule size, temperature, and pressure. Notice that these three factors are similar to the factors that affect the rate of dissolving. Be careful not to confuse them. Molecule Size and Solubility Small molecules are often more soluble than larger molecules. Methanol, CH3OH, and ethanol, CH3CH2OH, are both completely miscible with water. These compounds have OH groups that form hydrogen bonds with water. Larger molecules with the same OH group but more carbon atoms, such as pentanol, CH3CH2CH2CH2CH2OH , are far less soluble. All three compounds form hydrogen bonds with water, but the larger pentanol is less polar overall, making it less soluble. Table 8.2 compares five molecules by size and solubility. Table 8.2 Solubility and Molecule Size Name of compound methanol ethanol propanol butanol pentanol Chemical formula CH3OH CH3CH2OH CH3CH2CH2OH CH3(CH2)3OH CH3(CH2)4OH Solubility infinitely soluble infinitely soluble very soluble 9 g/100 mL (at 25°C) 3 g/100 mL (at 25°C) Temperature and Solubility At the beginning of this section, you learned that temperature affects CHEM the rate of dissolving. Temperature also affects solubility. You may have FA C T noticed that solubility data always include temperature. The solubility The link between cigarette of a solute in water, for example, is usually given as the number of grams smoking and lung cancer is of solute that dissolve in 100 mL of water at a specific temperature. (See well known. Other cancers are Table 8.2 for two examples.) Specifying temperature is essential, since the also related to smoking. It is solubility of a substance is very different at different temperatures. possible that a smoker who When a solid dissolves in a liquid, energy is needed to break the consumes alcohol may be at strong bonds between particles in the solid. At higher temperatures, more greater risk of developing energy is present. Thus, the solubility of most solids increases with tem- stomach cancer. When a person smokes, a thin film of perature. For example, caffeine’s solubility in water is only 2.2 g/100 mL tar forms inside the mouth and at 25˚C. At 100˚C, however, caffeine’s solubility increases to 40 g/100 mL. throat. The tar from cigarette The bonds between particles in a liquid are not as strong as the bonds smoke contains many carcino- between particles in a solid. When a liquid dissolves in a liquid, addition- genic (cancer-causing) com- al energy is not needed. Thus, the solubility of most liquids is not greatly pounds. These compounds are affected by temperature. non-polar and do not dissolve Gas particles move quickly and have a great deal of kinetic energy. in saliva. They are more solu- ble in alcohol, however. As a When a gas dissolves in a liquid, it loses some of this energy. At higher result, if a smoker drinks temperatures, the dissolved gas gains energy again. As a result, the gas alcohol, carcinogenic com- comes out of solution and is less soluble. Thus, the solubility of gases pounds can be washed into decreases with higher temperatures. the stomach. In the next investigation, you will observe and graph the effect of temperature on the solubility of a solid dissolved in a liquid solvent, water. As you have learned, most solid solutes become more soluble at higher temperatures. By determining the solubility of a solute at various temperatures, you can make a graph of solubility against temperature. The curve of best fit, drawn through the points, is called the solubility curve. You can use a solubility curve to determine the solubility of a solute at any temperature in the range shown on the graph. Chapter 8 Solutions and Their Concentrations MHR 295  S K I L L F O C U S Predicting Performing and recording Analyzing and interpreting Plotting Solubility Curves In this investigation, you will determine the tem- Procedure perature at which a certain amount of potassium 1. Read through the steps in this Procedure. nitrate is soluble in water. You will then dilute Prepare a data table to record the mass of the solution and determine the solubility again. the solute, the initial volume of water, the By combining your data with other students’ total volume of water after step 9, and the data, you will be able to plot a solubility curve. temperatures at which the solutions begin to crystallize. Question 2. Put the test tube inside a beaker for support. What is the solubility curve of KNO3 ? Place the beaker on a balance pan. Set the reading on the balance to zero. Then measure Prediction 14.0 g of potassium nitrate into the test tube. Draw a sketch to show the shape of the curve 3. Add one of the following volumes of distilled you expect for the solubility of a typical solid water to the test tube, as assigned by your dissolving in water at different temperatures. teacher: 10.0 mL, 15.0 mL, 20.0 mL, 25.0 mL, Plot solubility on the y-axis and temperature 30.0 mL. (If you use a graduated cylinder, on the x-axis. remember to read the volume from the bottom of the water meniscus. You can make a more Safety Precautions accurate volume measurement using either a burette or a pipette.) Before lighting the Bunsen burner, check that 4. Pour about 300 mL of tap water into the there are no flammable solvents nearby. If you beaker. Set up a hot-water bath using a hot are using a Bunsen burner, tie back long hair plate, retort stand, and thermometer clamp. and loose clothing. Be careful of the open flame. Alternatively, use a Bunsen burner, retort After turning it on, be careful not to touch the stand, ring clamp, thermometer clamp, and hot plate. wire gauze. 5. Put the stirring wire through the second Materials hole of the stopper. Insert the stopper, large test tube thermometer, and wire into the test tube. balance Make sure that the thermometer bulb is stirring wire below the surface of the solution. (Check two-hole stopper to fit the test tube, with a the diagram on the next page to make sure thermometer inserted in one hole that you have set up the apparatus properly 400 mL beaker thus far.) graduated cylinder or pipette or burette 6. Place the test tube in the beaker. Secure the hot plate or Bunsen burner with ring clamps test tube and thermometer to the retort stand, and wire gauze using clamps. Begin heating the water bath retort stand and thermometer clamp gently. potassium nitrate, KNO3 7. Using the stirring wire, stir the mixture until distilled water the solute completely dissolves. Turn the heat source off, and allow the solution to cool. 296 MHR Unit 3 Solutions and Solubility  This equation represents the solubility of thermometer stirring wire KNO3 at the temperature at which you recorded the first appearance of crystals. stopper Repeat your calculation to determine the solu- large test tube bility after the solution was diluted. Your teacher will collect and display all the class data for this investigation. 2. Some of your classmates were assigned the water same volume of water that you were assigned. undissolved hot water bath Compare the temperatures they recorded for solid their solutions with the temperatures you recorded. Comment on the precision of the data. Should any data be removed before averaging? 3. Average the temperatures at which crystal hot plate formation occurs for solutions that contain the same volume of water. Plot these data on graph paper. Set up your graph sideways on the graph paper (landscape orientation). Plot 8. Continue stirring. Record the temperature at solubility on the vertical axis. (The units are which crystals begin to appear in the solution. grams of solute per 100 mL of water.) Plot 9. Remove the stopper from the test tube. temperature on the horizontal axis. Carefully add 5.0 mL of distilled water. The 4. Draw the best smooth curve through the solution is now more dilute and therefore points. (Do not simply join the points.) Label more soluble. Crystals will appear at a lower each axis. Give the graph a suitable title. temperature. 10. Put the stopper, with the thermometer and Conclusions stirring wire, back in the test tube. If crystals 5. Go back to the sketch you drew to predict the have already started to appear in the solution, solubility of a typical solid dissolving in begin warming the water bath again. Repeat water at different temperatures. Compare the steps 7 and 8. shape of your sketch with the shape of your 11. If no crystals are present, stir the solution graph. while the water bath cools. Record the 6. Use your graph to interpolate the solubility of temperature at which crystals first begin potassium nitrate at to appear. (a) 60˚C (b) 40˚C 12. Dispose of the aqueous solutions of potassium 7. Use your graph to extrapolate the solubility of nitrate into the labelled waste container. potassium nitrate at (a) 80˚C (b) 20˚C Analysis 1. Use the volume of water assigned by your Application teacher to calculate how much solute 8. At what temperature can 40 mL of water dissolved in 100 mL of water. Use the dissolve the following quantities of following equation to help you: potassium nitrate? xg 14.0 g (a) 35.0 g (b) 20.0 g = 100 mL your volume Chapter 8 Solutions and Their Concentrations MHR 297  Heat Pollution: A Solubility Problem For most solids, and almost all ionic substances, solubility increases as the temperature of the solution increases. Gases, on the other hand, always become less soluble as the temperature increases. This is why a refrigerated soft drink tastes fizzier than the same drink at room tempera- ture. The warmer drink contains less dissolved carbon dioxide than the cooler drink. This property of gases makes heat pollution a serious problem. Many industries and power plants use water to cool down overheated machin- ery. The resulting hot water is then returned to local rivers or lakes. Figure 8.12 shows steam rising from a “heat-polluted” river. Adding warm water into a river or lake does not seem like actual pollution. The heat from the water, however, increases the temperature of the body of water. As the temperature increases, the dissolved oxygen in the water decreases. Fish and other aquatic wildlife and plants may not have enough oxygen to breathe. The natural heating of water in rivers and lakes can pose problems, too. Fish in warmer lakes and rivers are particularly vulnerable in the Figure 8.12 This image shows summer. When the water warms up even further, the amount of dissolved the result of heat pollution. oxygen decreases. Warmer water contains less dissolved oxygen. ExpressLab The Effect of Temperature on Soda Water In this Express Lab, you will have a chance to 3. Measure and record the mass of each beaker. see how a change in temperature affects the Measure and record the temperature of the dissolved gas in a solution. You will be looking soda water. at the pH of soda water. A low pH (1–6) indicates 4. Place one beaker on a heat source. Heat it to that the solution is acidic. You will learn more about 50˚C. Compare the rate of formation of about pH in Chapter 10. the bubbles with the rate of formation in the beaker of cool soda water. Record any change Safety Precautions in colour in the heated solution. Estimate its pH. If using a hot plate, avoid touching it when it 5. Allow the heated solution to cool. Again is hot. record any change in colour in the solution. If using a Bunsen burner, check that there are Estimate its pH. no flammable solvents nearby. 6. Measure and record the masses of both beakers. Determine any change in mass by Procedure comparing the final and initial masses. 1. Open a can of cool soda water. (Listen for the sound of excess carbon dioxide escaping.) Analysis Pour about 50 mL into each of two 100 mL 1. Which sample of soda water lost the most beakers. Note the rate at which bubbles form. mass? Explain your observation. Record your observations. 2. Did the heated soda water become more or 2. Add a few drops of universal indicator to both less acidic when it was heated? Explain why beakers. Record the colour of the solutions. you think this change happened. Then estimate the pH. 298 MHR Unit 3 Solutions and Solubility Pressure and Solubility The final factor that affects solubility is pressure. Changes in pressure have hardly any effect on solid and liquid solutions. Such changes do affect the solubility of a gas in a liquid solvent, however. The solubility of the gas is directly proportional to the pressure of the gas above the liquid. For example, the solubility of oxygen in lake water depends on the air pressure above the lake. When you open a carbonated drink, you can observe the effect of pres- sure on solubility. Figure 8.13 shows this effect. Inside a soft drink bottle, the pressure of the carbon dioxide gas is very high: about 400 kPa. When you open the bottle, you hear the sound of escaping gas as the pressure is reduced. Carbon dioxide gas escapes quickly from the bottle, since the pressure of the carbon dioxide in the atmosphere is much lower: only about 0.03 kPa. The solubility of the carbon dioxide in the liquid soft drink decreases greatly. Bubbles begin to rise in the liquid as gas comes out of solution and escapes. It takes a while for all the gas to leave the solution, so you have time to enjoy the taste of the soft drink before it goes “flat.” Figure 8.14 illustrates another example of dissolved gases and pressure. As a scuba diver goes deeper underwater, the water pressure Figure 8.13 What happens increases. The solubility of nitrogen gas, which is present in the lungs, when the pressure of the carbon also increases. Nitrogen gas dissolves in the diver’s blood. As the diver dioxide gas in a soft drink bottle returns to the surface, the pressure acting on the diver decreases. The is released? The solubility of the nitrogen gas in the blood comes out of solution. If the diver surfaces too gas in the soft drink solution quickly, the effect is similar to opening a soft drink bottle. Bubbles of decreases. nitrogen gas form in the blood. This leads to a painful and sometimes fatal condition known as “the bends.” You will learn more about gases and deep-sea diving in Chapter 11. CHEM FA C T Do you crack your knuckles? The sound you hear is another example of the effect of pressure on solubility. Joints contain fluid. When a joint is suddenly pulled or stretched, the cavity that holds the fluid gets larger. This causes the pressure to decrease. A bubble of gas forms, making the sound you hear. You cannot repeatedly crack your knuckles because it takes some time for Figure 8.14 Scuba divers must heed the effects of decreasing water pressure on the gas to re-dissolve. dissolved nitrogen gas in their blood. They must surface slowly to avoid “the bends.” Chapter 8 Solutions and Their Concentrations MHR 299  Chemistry Bulletin Solvents and Coffee: What’s the Connection? In the common Swiss Water Process, The story of coffee starts with the coffee berry. coffee beans are soaked in hot water. This First the pulp of the berry is removed. This dissolves the caffeine and the flavouring leaves two beans, each containing 1% to 2% compounds from the beans. The liquid is caffeine. The beans are soaked in water and passed through activated carbon filters. The natural enzymes to remove the outer parch- filters retain the caffeine, but let the flavouring ment husk and to start a slight fermentation compounds pass through. The filtered liquid, process. Once the beans have been fermented, now caffeine-free, is sprayed back onto the they are dried and roasted. Then the coffee is beans. The beans reabsorb the flavouring ready for grinding. Grinding increases the sur- compounds. Now they are ready for roasting. face area of the coffee. Thus, finer grinds make Carbon dioxide gas is a normal component it easier to dissolve the coffee in hot water. of air. In the carbon dioxide decaffeination Decaffeinated coffee satisfies people who process, the gas is raised to a temperature of at like the smell and taste of coffee but cannot least 32˚C. Then it is compressed to a pressure tolerate the caffeine. How is caffeine removed of about 7400 kPa. At this pressure, it resem- from coffee? bles a liquid but can flow like a gas. The All the methods of extracting caffeine take carbon dioxide penetrates the coffee beans place before the beans are roasted. Caffeine and and dissolves the caffeine. When the pressure the other organic compounds that give coffee returns to normal, the carbon dioxide reverts its taste are mainly non-polar. (Caffeine does to a gaseous state. The caffeine is left behind. contain some polar bonds, however, which What happens to the caffeine that is allows it to dissolve in hot water.) Non-polar removed by decaffeination? Caffeine is so solvents, such as benzene and trichloroethene, valuable that it is worth more than the cost of were once used to dissolve and remove taking it out of the beans. It is extensively used caffeine from the beans. These chemicals are in the pharmaceutical industry, and for colas now considered to be too hazardous. Today and other soft drinks. most coffee manufacturers use water or carbon Making Connections dioxide as solvents. 1. As you have read, water is a polar liquid and the soluble fractions of the coffee grounds are non-polar. Explain, in chemical terms, how caffeine and the coffee flavour and aroma are transferred to hot coffee. 2. Why does hot water work better in the brewing process than cold water? 3. In chemical terms, explain why fine grinds of coffee make better coffee. How can caffeine form hydrogen bonds with water? 300 MHR Unit 3 Solutions and Solubility Section Wrap-up In this section, you examined the factors that affect the rate of dissolving: temperature, agitation, and particle size. Next you looked at the forces between solute and solvent particles. Finally, you considered three main factors that affect solubility: molecule size, temperature, and pressure. In section 8.3, you will learn about the effects of differing amounts of solute dissolved in a certain amount of solvent. Section Review 1 K/U Describe the particle attractions that occur as sodium chloride dissolves in water. 2 K/U When water vaporizes, which type of attraction, intramolecular or intermolecular, is broken? Explain. 3 K/U Describe the effect of increasing temperature on the solubility of (a) a typical solid in water (b) a gas in water 4 K/U Sugar is more soluble in water than salt. Why does a salt solution (brine) conduct electricity, while a sugar solution does not? 5 K/U Dissolving a certain solute in water releases heat. Dissolving a different solute in water absorbs heat. Explain why. 6 I The graph below shows the solubility of various substances plotted against the temperature of the solution. (a) Which substance decreases in solubility as the 100 temperature increases? NaNO3 90 Solubility (g/100 g H2O) (b) Which substance is least soluble at room 80 temperature? Which substance is most soluble KNO3 70 at room temperature? 60 (c) The solubility of which substance is least affected KCI 50 by a change in temperature? 40 (d) At what temperature is the solubility of potassium NaCI 30 chlorate equal to 40 g/100 g of water? 20 KCIO3 (e) 20 mL of a saturated solution of potassium nitrate at 50˚C is cooled to 20˚C. Approximately what mass 10 Ce2(SO3)3 of solid will precipitate from the solution? Why is 0 it not possible to use the graph to interpolate an 10 20 30 40 50 60 70 80 90 100 accurate value? Temperature (˚C) 7 I A saturated solution of potassium nitrate was prepared at 70˚C and then cooled to 55˚C. Use your graph from Investigation 8-A to predict the fraction of the dissolved solute that crystallized out of the solution. Unit Issue Prep Think about how the properties 8 MC Would you expect to find more mineral deposits near a thermal of water affect its behaviour in spring or near a cool mountain spring? Explain. the environment. Look ahead to the Unit 3 Issue. How could water’s excellent ability as a solvent become a problem? Chapter 8 Solutions and Their Concentrations MHR 301  8.3 The Concentration of Solutions Section Preview/ Specific Expectations Material Safety Data Sheet In this section, you will Component Name CAS Number solve problems involving the PHENOL, 100% Pure 108952 concentration of solutions SECTION III: Hazards Identification express concentration as Very hazardous in case of ingestion, inhalation, grams per 100 mL, mass and skin contact, or eye contact. volume percents, parts per Product is corrosive to internal membranes when million and billion, and moles ingested. per litre Inhalation of vapours may damage central nervous system. Symptoms: nausea, headache, dizziness. communicate your under- Skin contact may cause itching and blistering. standing of the following Eye-contact may lead to corneal damage or terms: concentration, blindness. mass/volume percent, Severe over-exposure may lead to lung-damage, mass/mass percent, Figure 8.15 Should phenol be choking, or coma. volume/volume percent, banned from drugstores? parts per million, parts per billion, molar concentration Phenol is a hazardous liquid, especially when it is at room temperature. It is a volatile chemical. Inhaling phenol adversely affects the central nervous system, and can lead to a coma. Inhalation is not the only danger. Coma and death have been known to occur within 10 min after phenol has contacted the skin. Also, as little as 1 g of phenol can be fatal if swallowed. Would you expect to find such a hazardous chemical in over-the- counter medications? Check your medicine cabinet at home. You may find phenol listed as an ingredient in throat sprays and in lotions to relieve itching. You may also find it used as an antiseptic or disinfectant. Is phenol a hazard or a beneficial ingredient in many medicines? This depends entirely on concentration: the amount of solute per quantity of solvent. At high concentrations, phenol can kill. At low concentrations, it is a safe component of certain medicines. Modern analytical tests allow chemists to detect and measure almost any chemical at extremely low concentrations. In this section, you will learn about various ways that chemists use to express the concentration of a solution. As well, you will find the concentration of a solution by experiment. Concentration as a Mass/Volume Percent Recall that the solubility of a compound at a certain temperature is often expressed as the mass of the solute per 100 mL of solvent. For example, you know that the solubility of sodium chloride is 36 g/100 mL of water at room temperature. The final volume of the sodium chloride solution may or may not be 100 mL. It is the volume of the solvent that is important. Chemists often express the concentration of an unsaturated solution as the mass of solute dissolved per volume of the solution. This is differ- ent from solubility. It is usually expressed as a percent relationship. A mass/volume percent gives the mass of solute dissolved in a volume of solution, expressed as a percent. The mass/volume percent is also referred to as the percent (m/v). 302 MHR Unit 3 Solutions and Solubility  PROBEWARE Mass of solute (in g) Mass/volume percent = × 100% Volume of solution (in mL) If you have access to probe- ware, do the Chemistry 11 lab, Suppose that a hospital patient requires an intravenous drip to replace “Concentration of Solutions” lost body fluids. The intravenous fluid may be a saline solution that now. contains 0.9 g of sodium chloride dissolved in 100 mL of solution, or 0.9% (m/v). Notice that the number of grams of solute per 100 mL of solution is numerically equal to the mass/volume percent. Explore this idea further in the following problems. Sample Problem Solving for a Mass/Volume Percent Problem A pharmacist adds 2.00 mL of distilled water to 4.00 g of a powdered drug. The final volume of the solution is 3.00 mL. What is the concentration of the drug in g/100 mL of solution? What is the percent (m/v) of the solution? What Is Required? You need to calculate the concentration of the solution, in grams of solute dissolved in 100 mL of solution. Then you need to express this concentration as a mass/volume percent. What Is Given? The mass of the dissolved solute is 4.00 g. The volume of the solution is 3.00 mL. Plan Your Strategy There are two possible methods for solving this problem. Method 1 Use the formula Mass of solute (in g) Mass/volume percent = × 100% Volume of solution (in mL) Method 2 Let x represent the mass of solute dissolved in 100 mL of solution. The ratio of the dissolved solute, x, in 100 mL of solution must be the same as the ratio of 4.00 g of solute dissolved in 3.00 mL of solution. The concentration, expressed in g/100 mL, is numerically equal to the percent (m/v) of the solution. Act on Your Strategy Method 1 4.00 g Percent (m/v) = × 100% 3.00 mL = 133% Continued... Chapter 8 Solutions and Their Concentrations MHR 303  Continued... FROM PAGE 303 Method 2 x 4.00 g = 100 mL 3.00 mL x = 1.33 g/mL 100 mL x = 100 mL × 1.33 g/mL = 133 g The concentration of the drug is 133 g/100 mL of solution, or 133% (m/v). Check Your Solution The units are correct. The numerical answer is large, but this is reasonable for an extremely soluble solute. Sample Problem Finding Mass for an (m/v) Concentration Problem Many people use a solution of trisodium phosphate, Na3PO4 (commonly called TSP), to clean walls before putting up wallpaper. The recommended concentration is 1.7% (m/v). What mass of TSP is needed to make 2.0 L of solution? What Is Required? You need to find the mass of TSP needed to make 2.0 L of solution. What Is Given? The concentration of the solution should be 1.7% (m/v). The volume of solution that is needed is 2.0 L. Plan Your Strategy There are two different methods you can use. Method 1 Use the formula for (m/v) percent. Rearrange the formula to solve for mass. Then substitute in the known values. Method 2 The percent (m/v) of the solution is numerically equal to the concentration in g/100 mL. Let x represent the mass of TSP dis- solved in 2.0 L of solution. The ratio of dissolved solute in 100 mL of solution must be the same as the ratio of the mass of solute, x, dissolved in 2.0 L (2000 mL) of solution. Continued... 304 MHR Unit 3 Solutions and Solubility Continued... FROM PAGE 304 Act on Your Strategy Method 1 Mass of solute (in g) (m/v) percent = × 100% Volume of solution (in mL) (m/v) percent × Volume of solution ∴ Mass of solute = 100% 1.7% × 2000 mL = 100% = 34 g Method 2 A TSP solution that is 1.7% (m/v) contains 1.7 g of solute dissolved in 100 mL of solution. 1.7 g x = 100 mL 2000 mL x 0.017 g/mL = 2000 mL x = 0.017 g/mL × 2000 mL = 34 g Therefore, 34 g of TSP are needed to make 2.0 L of cleaning solution. Check Your Solution The units are appropriate for the problem. The answer appears to be reasonable. Practice Problems 1. What is the concentration in percent (m/v) of each solution? (a) 14.2 g of potassium chloride, KCl (used as a salt substitute), dissolved in 450 mL of solution (b) 31.5 g of calcium nitrate, Ca(NO3)2 (used to make explosives), dissolved in 1.80 L of solution (c) 1.72 g of potassium permanganate, KMnO4 (used to bleach stone-washed blue jeans), dissolved in 60 mL of solution 2. A solution of hydrochloric acid was formed by dissolving 1.52 g of hydrogen chloride gas in enough water to make 24.1 mL of solution. What is the concentration in percent (m/v) of the solution? 3. At 25˚C, a saturated solution of carbon dioxide gas in water has a concentration of 0.145% (m/v). What mass of carbon dioxide is present in 250 mL of the solution? 4. Ringer’s solution contains three dissolved salts in the same proportions as they are found in blood. The salts and their concentrations (m/v) are as follows: 0.86% NaCl, 0.03% KCl, and 0.033% CaCl2. Suppose that a patient needs to receive 350 mL of Ringer’s solution by an intravenous drip. What mass of each salt does the pharmacist need to make the solution? Chapter 8 Solutions and Their Concentrations MHR 305  Concentration as a Mass/Mass Percent The concentration of a solution that contains a solid solute dissolved in a liquid solvent can also be expressed as a mass of solute dissolved in a mass of solution. This is usually expressed as a percent relationship. A mass/mass percent gives the mass of a solute divided by the mass of solution, expressed as a percent. The mass/mass percent is also referred to as the percent (m/m), or the mass percent. It is often inaccurately referred to as a weight (w/w) percent, as well. Look at your tube of toothpaste, at home. The percent of sodium fluoride in the toothpaste is usually given as a w/w percent. This can be confusing, since weight (w) is not the same as mass (m). In fact, this concentration should be expressed as a mass/mass percent. Mass of solute (in g) Mass/mass percent = × 100% Mass of solution (in g) For example, 100 g of seawater contains 0.129 g of magnesium ion (along with many other substances). The concentration of Mg2+ in seawater is 0.129 (m/m). Notice that the number of grams of solute per 100 g of solution is numerically equal to the mass/mass percent. The concentration of a solid solution, such as an alloy, is usually expressed as a mass/mass percent. Often the concentration of a particular alloy may vary. Table 8.3 gives typical compositions of some common alloys. Table 8.3 The Composition of Some Common Alloys Alloy Uses Typical percent (m/m) composition brass ornaments, musical instruments Cu (85%) Zn (15%) bronze statues, castings Cu (80%) Zn (10%) Sn (10%) cupronickel “silver” coins Cu (75%) Ni (25%) dental amalgam dental fillings Hg (50%) Ag (35%) Sn (15%) duralumin aircraft parts Al (93%) Cu (5%) other (2%) pewter ornaments Sn (85%) Cu (7%) Bi (6%) Sb ((2%) stainless steel cutlery, knives Fe (78%) Cr (15%) Ni (7%) sterling silver jewellery Ag (92.5%) Cu (7.5%) Figure 8.16, on the following page, shows two objects made from brass that have distinctly different colours. The difference in colours reflects the varying concentrations of the copper and zinc that make up the objects. 306 MHR Unit 3 Solutions and Solubility  Figure 8.16 Brass can be made using any percent from 50% to 85% copper, and from 15% to 50% zinc. As a result, two objects made of brass can look very different. Sample Problem Solving for a Mass/Mass Percent Problem Calcium chloride, CaCl2 , can be used instead of road salt to melt the ice on roads during the winter. To determine how much calcium chloride had been used on a nearby road, a student took a sample of slush to analyze. The sample had a mass of 23.47 g. When the solu- tion was evaporated, the residue had a mass of 4.58 g. (Assume that no other solutes were present.) What was the mass/mass percent of calcium chloride in the slush? How many grams of calcium chloride were present in 100 g of solution? What Is Required? You need to calculate the mass/mass percent of calcium chloride in the solution (slush). Then you need to use your answer to find the mass of calcium chloride in 100 g of solution. What Is Given? The mass of the solution is 23.47 g. The mass of calcium chloride that was dissolved in the solution is 4.58 g. Plan Your Strategy There are two methods that you can use to solve this problem. Method 1 Use the formula for mass/mass percent. Mass of solute (in g) Mass/mass percent = × 100% Mass of solution (in g) Continued... Chapter 8 Solutions and Their Concentrations MHR 307  Continued... FROM PAGE 307 The mass of calcium chloride in 100 g of solution will be numerically equal to the mass/mass percent. Method 2 Use ratios, as in the previous Sample Problems. Act on Your Strategy Method 1 4.58 g Mass/mass percent = × 100% 23.47 g = 19.5% Method 2 xg 4.58 g = 100 g 23.47 g xg = 0.195 100 g x = 19.5% The mass/mass percent was 19.5% (m/m). 19.5 g of calcium chloride was dissolved in 100 g of solution. Check Your Solution The mass units divide out properly. The final answer has the correct number of significant digits. It appears to be reasonable. Practice Problems 5. Calculate the mass/mass percent of solute for each solution. (a) 17 g of sulfuric acid in 65 g of solution (b) 18.37 g of sodium chloride dissolved in 92.2 g of water Hint: Remember that a solution consists of both solute and solvent. (c) 12.9 g of carbon tetrachloride dissolved in 72.5 g of benzene 6. If 55 g of potassium hydroxide is dissolved in 100 g of water, what is the concentration of the solution expressed as mass/mass percent? 7. Steel is an alloy of iron and about 1.7% carbon. It also contains small amounts of other materials, such as manganese and phosphorus. What mass of carbon is needed to make a 5.0 kg sample of steel? 8. Stainless steel is a variety of steel that resists corrosion. Your cutlery at home may be made of this material. Stainless steel must contain at least 10.5% chromium. What mass of chromium is needed to make a stainless steel fork with a mass of 60.5 g? 9. 18-carat white gold is an alloy. It contains 75% gold, 12.5% silver, and 12.5% copper. A piece of jewellery, made of 18-carat white gold, has a mass of 20 g. How much pure gold does it contain? 308 MHR Unit 3 Solutions and Solubility Concentration as a Volume/Volume Percent When mixing two liquids to form a solution, it is easier to measure their volumes than their masses. A volume/volume percent gives the volume of solute divided by the volume of solution, expressed as a percent. The volume/volume percent is also referred to as the volume percent concen- tration, volume percent, percent (v/v), or the percent by volume. You can see this type of concentration on a bottle of rubbing alcohol from a drug- store. (See Figure 8.17.) Volume of solute (in mL) Volume/volume percent = × 100% Volume of solution (in mL) Read through the Sample Problem below, and complete the Practice Problems that follow. You will then have a better understanding of how to calculate the volume/volume percent of a solution. Figure 8.17 The concentration