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11Properties of Solutions 11 Properties of Solutions Contents 11.1 Solution Composition 11.2 The Energies of Solution Formation 11.3 Factors Affecting Solubility • Structure Effects • Pressure Effects • Temperature Effects (for Aqueous Solutions) 11.4 The Vapor Pressures of Solutions • Nonideal Solutions 11.5 Boiling-Point Elevation and Freezing-Point Depression • Boiling-Point Elevation • Freezing-Point Depression 11.6 Osmotic Pressure • Reverse Osmosis 11.7 Colligative Properties of Electrolyte Solutions 11.8 Colloids Opals are formed from colloidal suspensions of silica when the liquid evaporates. 484 Most of the substances we encounter in daily life are mixtures: Wood, milk, gasoline, champagne, seawater, shampoo, steel, and air are common examples. When the components of a mixture are uniformly intermingled—that is, when a mixture is homogeneous—it is called a solution. Solutions can be gases, liquids, or solids, as shown in Table 11.1. However, we will be concerned in this chapter with the properties of liquid solutions, particularly those containing water. As we saw in Chapter 4, many essential chemical reactions occur in aqueous solutions because water is capable of dissolving so many substances. 11.1 Solution Composition Because a mixture, unlike a chemical compound, has a variable composition, the relative amounts of substances in a solution must be specified. The qualitative terms dilute (rela- A solute is the substance being dissolved. tively little solute present) and concentrated (relatively large amount of solute) are often The solvent is the dissolving medium. used to describe solution content, but we need to define solution composition more pre- cisely to perform calculations. For example, in dealing with the stoichiometry of solution reactions in Chapter 4, we found it useful to describe solution composition in terms of moles of solute Molarity ϭ molarity, or the number of moles of solute per liter of solution (symbolized by M). liters of solution Other ways of describing solution composition are also useful. Mass percent (some- times called weight percent) is the percent by mass of the solute in the solution: mass of solute Mass percent ϭ a b ϫ 100% mass of solution Another way of describing solution composition is the mole fraction (symbolized by x When liquids are mixed, the liquid the Greek lowercase letter chi, ), the ratio of the number of moles of a given component present in the largest amount is called to the total number of moles of solution. For a two-component solution, where nA and nB the solvent. represent the number of moles of the two components, n ϭ x ϭ A Mole fraction of component A A ϩ nA nB TABLE 11.1 Various Types of Solutions State of State of State of Example Solution Solute Solvent Air, natural gas Gas Gas Gas Vodka in water, Liquid Liquid Liquid antifreeze Brass Solid Solid Solid Carbonated water Liquid Gas Liquid (soda) Seawater, sugar Liquid Solid Liquid solution Hydrogen in Solid Gas Solid platinum 485 486 Chapter Eleven Properties of Solutions In very dilute aqueous solutions, the Still another way of describing solution composition is molality (symbolized by m), magnitude of the molality and the molar- the number of moles of solute per kilogram of solvent: ity are almost the same. moles of solute Molality ϭ kilogram of solvent Sample Exercise 11.1 Various Methods for Describing Solution Composition A solution is prepared by mixing 1.00 g ethanol (C2H5OH) with 100.0 g water to give a final volume of 101 mL. Calculate the molarity, mass percent, mole fraction, and molal- ity of ethanol in this solution. Solution Since molarity depends on the volume Molarity: The moles of ethanol can be obtained from its molar mass (46.07 g/mol): of the solution, it changes slightly with temperature. Molality is independent of 1 mol C H OH 1.00 g C H OH ϫ 2 5 ϭ 2.17 ϫ 10Ϫ2 mol C H OH temperature because it depends only 2 5 46.07 g C H OH 2 5 on mass. 2 5 1 L Volume ϭ 101 mL ϫ ϭ 0.101 L 1000 mL moles of C H OH 2.17 ϫ 10Ϫ2 mol Molarity of C H OH ϭ 2 5 ϭ 2 5 liters of solution 0.101 L ϭ 0.215 M Mass percent: mass of C H OH Mass percent C H OH ϭ a 2 5 b ϫ 100% 2 5 mass of solution 1.00 g C H OH ϭ a 2 5 b ϫ ϩ 100% 100.0 g H2O 1.00 g C2H5OH ϭ 0.990% C2H5OH Mole fraction: nC H OH Mole fraction of C H OH ϭ 2 5 2 5 n ϩ n C2H5OH H2O 1 mol H O ϭ ϫ 2 ϭ nH2O 100.0 g H2O 5.56 mol 18.0 g H2O ϫ Ϫ2 x ϭ 2.17 10 mol C2H5OH 2.17 ϫ 10Ϫ2 mol ϩ 5.56 mol 2.17 ϫ 10Ϫ2 ϭ ϭ 0.00389 5.58 Molality: moles of C H OH 2.17 ϫ 10Ϫ2 mol Molality of C H OH ϭ 2 5 ϭ 2 5 kilogram of H O 1 kg 2 100.0 g ϫ 1000 g ϫ Ϫ2 ϭ 2.17 10 mol 0.1000 kg ϭ 0.217 m See Exercises 11.25 through 11.27. 11.1 Solution Composition 487 TABLE 11.2 The Molar Mass, Equivalent Mass, and Relationship of Molarity and Normality for Several Acids and Bases Molar Equivalent Relationship of Molarity Acid or Base Mass Mass and Normality HCl 36.5 36.5 1 M ϭ 1 N 98 H SO 98 ϭ 49 1 M ϭ 2 N 2 4 2 NaOH 40 40 1 M ϭ 1 N 74 Ca(OH) 74 ϭ 37 1 M ϭ 2 N 2 2 The definition of an equivalent depends Another concentration measure sometimes encountered is normality (symbolized on the reaction taking place in the by N). Normality is defined as the number of equivalents per liter of solution, where the solution. definition of an equivalent depends on the reaction taking place in the solution. For an acid–base reaction, the equivalent is the mass of acid or base that can furnish or accept exactly 1 mole of protons (Hϩ ions). In Table 11.2 note, for example, that the equivalent The quantity we call equivalent mass mass of sulfuric acid is the molar mass divided by 2, since each mole of H2SO4 can fur- here traditionally has been called equiva- nish 2 moles of protons. The equivalent mass of calcium hydroxide is also half the mo- lent weight. Ϫ lar mass, since each mole of Ca(OH)2 contains 2 moles of OH ions that can react with 2 moles of protons. The equivalent is defined so that 1 equivalent of acid will react with exactly 1 equivalent of base. Oxidation–reduction half-reactions were For oxidation–reduction reactions, the equivalent is defined as the quantity of oxidizing discussed in Section 4.10. or reducing agent that can accept or furnish 1 mole of electrons. Thus 1 equivalent of reducing agent will react with exactly 1 equivalent of oxidizing agent. The equivalent mass of an ox- idizing or reducing agent can be calculated from the number of electrons in its half-reaction. Ϫ 2ϩ For example, MnO4 reacting in acidic solution absorbs five electrons to produce Mn : Ϫ ϩ Ϫ ϩ ϩ 2ϩ ϩ MnO4 5e 8H ¡ Mn 4H2O Ϫ Since the MnO4 ion present in 1 mole of KMnO4 consumes 5 moles of electrons, the equivalent mass is the molar mass divided by 5: molar mass 158 g Equivalent mass of KMnO ϭ ϭ ϭ 31.6 g 4 5 5 Calculating Various Methods of Solution Composition Sample Exercise 11.2 from the Molarity The electrolyte in automobile lead storage batteries is a 3.75 M sulfuric acid solution that has a density of 1.230 g/mL. Calculate the mass percent, molality, and normality of the sulfuric acid. Solution The density of the solution in grams per liter is g 1000 mL 1.230 ϫ ϭ 1.230 ϫ 103 g/L mL 1 L Thus 1 liter of this solution contains 1230. g of the mixture of sulfuric acid and water. Since the solution is 3.75 M, we know that 3.75 mol H2SO4 is present per liter of solu- tion. The number of grams of H2SO4 present is 98.1 g H SO A modern 12-volt lead storage battery of 3.75 mol ϫ 2 4 ϭ 368 g H SO the type used in automobiles. 1 mol 2 4 488 Chapter Eleven Properties of Solutions CHEMICAL IMPACT Electronic Ink he printed page has been a primary means of commu- page. The key to this self-printing “paper” is microencap- T nication for over 3000 years, and researchers at the sulation technology—the same technology that is used in Massachusetts Institute of Technology (MIT) believe they “carbonless” carbon paper and “scratch-and-sniff” cologne have discovered why. It seems that the brain responds pos- and perfume advertisements in magazines. Jacobson’s sys- itively to fixed images on a sheet of paper, particularly tem involves the use of millions of transparent fluid-filled those areas of the brain that store and process “spatial capsules containing microscopic particles. These particles maps.” In comparison, information displayed on computer are colored and positively charged on one side and white screens or TV screens seems to lack some of the visual and negatively charged on the other. When an electric field signals that stimulate the learning centers of the brain to is selectively applied to the capsules, the white side of the retain knowledge.
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