{"id":3502,"date":"2017-05-10T14:46:54","date_gmt":"2017-05-10T21:46:54","guid":{"rendered":"http:\/\/www.wou.edu\/chemistry\/?page_id=3502"},"modified":"2018-09-05T08:37:09","modified_gmt":"2018-09-05T15:37:09","slug":"chapter-7-solutions","status":"publish","type":"page","link":"https:\/\/wou.edu\/chemistry\/courses\/online-chemistry-textbooks\/ch150-preparatory-chemistry\/chapter-7-solutions\/","title":{"rendered":"CH150: Chapter 7 &#8211; Solutions"},"content":{"rendered":"<header>\n<h2 id=\"title\"><span style=\"color: #000000\"><strong>Chapter 7: Solutions A<\/strong><\/span><span style=\"color: #000000\"><strong>nd Solution Stoichiometry<\/strong><\/span><\/h2>\n<h3><a href=\"#7intro\"><span><strong>7.1 Introduction<\/strong><\/span><\/a><\/h3>\n<h3 class=\"editable\"><a href=\"#typessolutions\"><strong>7.2 Types of Solutions<\/strong><\/a><\/h3>\n<h3 class=\"editable\"><a href=\"#solubility\"><strong>7.3 Solubility<\/strong><\/a><\/h3>\n<h3><a href=\"#tempsol\"><strong>7.4 Temperature and Solubility<\/strong><\/a><\/h3>\n<h3><a href=\"#pressuresol\"><span><strong>7.5 Effects of Pressure on the Solubility of Gases: Henry&#8217;s Law<\/strong><\/span><\/a><\/h3>\n<h3><a href=\"#solidhydrates\"><strong><span>7.6 Solid Hydrates<\/span><\/strong><\/a><\/h3>\n<h3><a href=\"#solutionconcentration\"><strong><span>7.7 Solution Concentration<\/span><\/strong><\/a><\/h3>\n<h4 class=\"title editable block\"><a href=\"#molarity\"><span style=\"color: #000000\"><strong>7.7.1 Molarity<\/strong><\/span><\/a><\/h4>\n<h4><a href=\"#partspermillion\"><span style=\"color: #000000\"><strong>7.7.2 Parts Per Solutions<\/strong><\/span><\/a><\/h4>\n<h3 class=\"para editable block\"><a href=\"#dilutions\"><span><strong>7.8 Dilutions<\/strong><\/span><\/a><\/h3>\n<h3 class=\"para\" id=\"averill_1.0-ch04_s02_s02_p41\"><a href=\"#ionconcentration\"><span><strong>7.9 Ion Concentrations in Solution<\/strong><\/span><\/a><\/h3>\n<h3><a href=\"#7summary\"><strong><span style=\"color: #ff0000\">7.10 Summary<\/span><\/strong><\/a><\/h3>\n<h3><a href=\"#7refs\"><span style=\"color: #ff0000\"><strong>7.11 References<\/strong><\/span><\/a><\/h3>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<hr \/>\n<h3 id=\"7intro\"><span style=\"color: #ff0000\"><strong>7.1 Introduction:<\/strong><\/span><\/h3>\n<p><span style=\"color: #000000\">Recall from Chapter 1 that<em><strong> solutions<\/strong><\/em> are defined as homogeneous mixtures that are mixed so thoroughly that neither component\u00a0can be observed independently of the other. <\/span><span style=\"color: #000000\">Solutions are all around us. Air, for example, is a solution. If you live near a lake, a river, or an ocean, that body of water is not pure H<sub>2<\/sub>O but most probably a solution. Much of what we drink\u2014for example, soda, coffee, tea, and milk are solutions. Solutions are a large part of everyday life. A lot of the chemistry occurring around us happens in solution. In fact, much of the chemistry that occurs in our own bodies takes place in solution, and many solutions\u2014such as the Ringer\u2019s lactate IV solution\u2014are important in healthcare. In our understanding of chemistry, we need to understand a little bit about solutions. In this chapter, you will learn about the special characteristics of solutions, how solutions are characterized, and some of their properties.<\/span><\/p>\n<\/header>\n<section class=\"mt-content-container\">\n<div id=\"skills\">\n<p class=\"boxtitle\"><span style=\"color: #000000\">Skills to Develop<\/span><\/p>\n<ul>\n<li><span style=\"color: #000000\">Define these terms: solution, solute, and solvent.<\/span><\/li>\n<li><span style=\"color: #000000\">Distinguish solutions, mixtures, and colloids.<\/span><\/li>\n<li><span style=\"color: #000000\">Describe various types of solutions.<\/span><\/li>\n<li><span style=\"color: #000000\">Distinguish unsaturated, saturated, and supersaturated solutions.<\/span><\/li>\n<\/ul>\n<\/div>\n<p><span style=\"color: #000000\">The major component of the solution is called <em><strong>solvent<\/strong><\/em>, and the minor component(s) are called<em> <strong>solute<\/strong><\/em>. If both components in a solution are 50%, the term solute can be assigned to either component. When a gaseous or solid material dissolves in a liquid, the gas or solid material is called the solute. When two liquids dissolve in each other, the major component is called the <em><strong>solvent<\/strong> <\/em>and the minor component is called the <em><strong>solute<\/strong>.<\/em><\/span><\/p>\n<p><span style=\"color: #000000\">Many chemical reactions are carried out in solutions, and solutions are also closely related to our everyday lives. The air we breathe, the liquids we drink, and the fluids in our body are all solutions. Furthermore, we are surrounded by solutions such as the air and waters (in rivers, lakes and oceans).<\/span><\/p>\n<p><span style=\"color: #000000\">On the topic of solutions, we include the following sections.<\/span><\/p>\n<ol>\n<li><span style=\"color: #000000\">Types of Solutions: gaseous, liquid and solid solutions are based on the states of the solution.<\/span><\/li>\n<li><span style=\"color: #000000\">Solution Stoichiometry: expressing concentration in various units (mass per unit volume, moles per unit volume, percentage and fractions), reaction stoichiometry calculations involving solutions.<\/span><\/li>\n<li><span style=\"color: #000000\">Solutions of Electrolytes: solutions of acids, bases, and salts in which the solutes dissociate into positive and negative hydrated ions.<\/span><\/li>\n<li><span style=\"color: #000000\">Metathesis or Exchange Reactions<a class=\"external\" href=\"http:\/\/www.science.uwaterloo.ca\/%7Ecchieh\/cact\/c120\/metathes.html\" target=\"_blank\" rel=\"external nofollow noopener noreferrer\"><\/a>: reaction of electrolytes leading to neutral molecules, gases, and solids.<\/span><\/li>\n<\/ol>\n<p><span style=\"color: #000000\">Solving problems of solution stoichiometry requires the concepts introduced in stoichiometry in Chapter 6, which also provides the basis for the discussion on reactions.<\/span><\/p>\n<h4><a href=\"#title\"><span style=\"color: #ff0000\"><em><strong>(Back to the Top)<\/strong><\/em><\/span><\/a><\/h4>\n<div class=\"mt-section\" id=\"section_1\">\n<hr \/>\n<h3 id=\"typessolutions\" class=\"editable\"><strong>7.2 Types of Solutions<\/strong><\/h3>\n<p><span style=\"color: #000000\">In Chapter 1, you were introduced to the concept of a <em><strong>mixture<\/strong><\/em>, which is a substance that is composed of two or more substances. Recall that mixtures can be of two types: Homogeneous and Heterogeneous, where homogeneous mixtures combine so intimately that they are observed as a single substance, even though they are not. Heterogeneous mixtures, on the other hand, are non-uniform and have regions of the mixture that look different from other regions of the mixture. Homogeneous mixtures can be further broken down into two classifications: Colloids and Solutions. A colloid is a mixture that contains particles with diameters ranging from 2 to 500 nm. Colloids appear uniform in nature and have the same composition throughout but are cloudy or opaque. Milk is a good example of a colloid. True solutions have particle sizes of a typical ion or small molecule (~0.1 to 2 nm in diameter) and are transparent, although they may be colored. This chapter will focus on the characteristics of true solutions.<br \/>\n<\/span><\/p>\n<p><span style=\"color: #000000\">Material exists in three states: solid, liquid, and gas. Solutions also exist in all these states:<\/span><\/p>\n<ol>\n<li><span style=\"color: #000000\">Gaseous mixtures are usually homogeneous and are commonly <em><strong>gas-gas solutions<\/strong><\/em>. For quantitative treatment of this type of solutions, we will devote a unit to gases. The atmosphere is a gaseous solution that consists of nitrogen, oxygen, argon, carbon dioxide, water, methane, and some other minor components.\u00a0 Some of these components, such as\u00a0 water, oxygen, and carbon dioxide may vary in concentration in different locations on the Earth depending on factors such as temperature and\u00a0 altitude.<\/span><\/li>\n<li><span style=\"color: #000000\">When molecules of gas, solid or liquid are dispersed and mixed with those of liquid, the homogeneous (uniform) states are called<em> <strong>liquid solutions<\/strong><\/em>. Solids, liquids and gases dissolve in a liquid solvent to form liquid solutions.\u00a0 In this chapter, most of the chemistry that we will discuss occurs in liquid solutions where water is the solvent.<\/span><\/li>\n<li><span style=\"color: #000000\">Many alloys, ceramics, and polymer blends are <em><strong>solid solutions<\/strong>.<\/em> Within a certain range, copper and zinc dissolve in each other and harden to give solid solutions called brass. Silver, gold, and copper form many different alloys with unique colors and appearances. Alloys and other solid solutions are important in the world of materials chemistry.<\/span><\/li>\n<\/ol>\n<h4><a href=\"#title\"><span><em><strong>(Back to the Top)<\/strong><\/em><\/span><\/a><\/h4>\n<\/div>\n<div class=\"mt-section\" id=\"section_2\">\n<hr \/>\n<h3 id=\"solubility\" class=\"editable\"><strong>7.3 Solubility<\/strong><\/h3>\n<p><span style=\"color: #000000\">The maximum amount of a substance that can be dissolved in a given volume of solvent is called<em> <strong>solubility<\/strong><\/em>. Often, the solubility in water is expressed in gram\/100 mL. A solution that has not reached its maximum solubility is called an <strong><em>unsaturated solution. <\/em><\/strong>This means that more solute could still be added to the solvent and dissolving would still occur.<br \/>\n<\/span><\/p>\n<p><span style=\"color: #000000\">A solution that has reached the maximum solubility is called a <em><strong>saturated solution<\/strong><\/em>. If more solute is added at this point, it will not dissolve into the solution. Instead it will remain precipitated as a solid at the bottom of the solution.\u00a0 Thus, one can often tell that a solution is saturated if extra solute is present (this can exist as another phase, such as gas, liquid, or solid). In a saturated solution there is no net change in the amount of solute dissolved, but the system is by no means static. In fact, the solute is constantly being dissolved and deposited at an equal rate. Such a phenomenon is called<em> <strong>equilibrium<\/strong>.<\/em> For example:<br \/>\n<\/span><\/p>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/solution-process.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-3730\" alt=\"\" src=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/solution-process-1024x140.png\" width=\"695\" height=\"95\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/solution-process-1024x140.png 1024w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/solution-process-300x41.png 300w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/solution-process-768x105.png 768w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/solution-process.png 1052w\" sizes=\"(max-width: 695px) 100vw, 695px\" \/><\/a><\/p>\n<p><span style=\"color: #000000\">In special circumstances, a solution may be <em><strong>supersaturated<\/strong><\/em>. Supersaturated solutions are solutions that have dissolved solute beyond the normal saturation point. Usually a condition such as increased temperature or pressure is required to create a supersaturated solution. For example, sodium acetate has a very high solubility at 270 K.\u00a0 When cooled, such a solution stays dissolved in what is called a <strong>meta-stable state<\/strong>. However, when a <em>seeding<\/em> crystal is added to the solution, the extra solute will rapidly solidify. During the crystallization process, heat is evolved, and the solution becomes warm. Common hand warmers use this chemical process to generate heat.<\/span><\/p>\n<div style=\"width: 480px;\" class=\"wp-video\"><video class=\"wp-video-shortcode\" id=\"video-3502-1\" width=\"480\" height=\"360\" preload=\"metadata\" controls=\"controls\"><source type=\"video\/mp4\" src=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Sodium-Acetate-Crystals-Supersaturated-Solution.mp4?_=1\" \/><a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Sodium-Acetate-Crystals-Supersaturated-Solution.mp4\">https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Sodium-Acetate-Crystals-Supersaturated-Solution.mp4<\/a><\/video><\/div>\n<p><span style=\"color: #000000\"><em>Video showing the crystallization of a supersaturated solution of sodium acetate. Video by : <a href=\"https:\/\/www.youtube.com\/watch?v=9XMfR0taF4M\">North Carolina School of Science and Mathematics<\/a><\/em><\/span><\/p>\n<hr \/>\n<h4><strong><span style=\"color: #000000\">So how can we predict the solubility of a substance?<\/span><\/strong><\/h4>\n<p><span style=\"color: #000000\">One useful classification of materials is polarity. As you read about covalent and ionic compounds in Chapters 3 and 4, you learned that ionic compounds have the highest polarity forming full cations and anions within each molecule as electrons are donated from one atom to another. You also learned that covalent bonds could be polar or nonpolar in nature depending on whether or not the atoms involved in the bond share the electrons unequally or equally, respectively. Recall that the electronegativity difference can be used to determine the polarity of a substance.\u00a0 Typically an ionic bond has an electronegativity difference of 1.8 or above, whereas a polar covalent bond is between 0.4 to 1.8, and a nonpolar covalent bond is 0.4 or below.<\/span><\/p>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/01\/Picture1.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1895\" alt=\"\" src=\"https:\/\/wou.edu\/chemistry\/files\/2017\/01\/Picture1-1024x392.png\" width=\"700\" height=\"268\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2017\/01\/Picture1.png 1024w, https:\/\/wou.edu\/chemistry\/files\/2017\/01\/Picture1-300x115.png 300w, https:\/\/wou.edu\/chemistry\/files\/2017\/01\/Picture1-768x294.png 768w\" sizes=\"(max-width: 700px) 100vw, 700px\" \/><\/a><\/p>\n<p><span><span style=\"color: #000000\"><strong>Figure 7.1 Electronegativity Difference Diagram.<\/strong> The diagram above is a guide for discerning what type of bond forms between two different atoms. By taking the difference between the electronegativity values for each of the atoms involved in the bond, the bond type and\u00a0polarity can be predicted. Note that full ionic character is rarely reached, however when metals and nonmetals form bonds, they are named using the rules for ionic bonding.<\/span><\/span><\/p>\n<hr \/>\n<p><span style=\"color: #000000\">Substances with zero or low electronegativity difference such as <span class=\"MathJax\" id=\"MathJax-Element-1-Frame\" role=\"presentation\"><span class=\"math\" id=\"MathJax-Span-1\"><span class=\"mrow\" id=\"MathJax-Span-2\"><span class=\"msubsup\" id=\"MathJax-Span-3\"><span class=\"mtext\" id=\"MathJax-Span-4\">H<\/span><span class=\"texatom\" id=\"MathJax-Span-5\"><span class=\"mrow\" id=\"MathJax-Span-6\"><span class=\"mspace\" id=\"MathJax-Span-7\"><\/span><\/span><\/span><sub><span class=\"texatom\" id=\"MathJax-Span-8\"><span class=\"mrow\" id=\"MathJax-Span-9\"><span class=\"mn\" id=\"MathJax-Span-10\">2<\/span><\/span><\/span><\/sub><\/span><\/span><\/span><\/span>, <span class=\"MathJax\" id=\"MathJax-Element-2-Frame\" role=\"presentation\"><span class=\"math\" id=\"MathJax-Span-11\"><span class=\"mrow\" id=\"MathJax-Span-12\"><span class=\"msubsup\" id=\"MathJax-Span-13\"><span class=\"mtext\" id=\"MathJax-Span-14\">O<\/span><span class=\"texatom\" id=\"MathJax-Span-15\"><span class=\"mrow\" id=\"MathJax-Span-16\"><span class=\"mspace\" id=\"MathJax-Span-17\"><\/span><\/span><\/span><sub><span class=\"texatom\" id=\"MathJax-Span-18\"><span class=\"mrow\" id=\"MathJax-Span-19\"><span class=\"mn\" id=\"MathJax-Span-20\">2<\/span><\/span><\/span><\/sub><\/span><\/span><\/span><\/span>, <span class=\"MathJax\" id=\"MathJax-Element-3-Frame\" role=\"presentation\"><span class=\"math\" id=\"MathJax-Span-21\"><span class=\"mrow\" id=\"MathJax-Span-22\"><span class=\"msubsup\" id=\"MathJax-Span-23\"><span class=\"mtext\" id=\"MathJax-Span-24\">N<\/span><span class=\"texatom\" id=\"MathJax-Span-25\"><span class=\"mrow\" id=\"MathJax-Span-26\"><span class=\"mspace\" id=\"MathJax-Span-27\"><\/span><\/span><\/span><sub><span class=\"texatom\" id=\"MathJax-Span-28\"><span class=\"mrow\" id=\"MathJax-Span-29\"><span class=\"mn\" id=\"MathJax-Span-30\">2<\/span><\/span><\/span><\/sub><\/span><\/span><\/span><\/span>, <span class=\"MathJax\" id=\"MathJax-Element-4-Frame\" role=\"presentation\"><span class=\"math\" id=\"MathJax-Span-31\"><span class=\"mrow\" id=\"MathJax-Span-32\"><span class=\"msubsup\" id=\"MathJax-Span-33\"><span class=\"mtext\" id=\"MathJax-Span-34\">CH<\/span><span class=\"texatom\" id=\"MathJax-Span-35\"><span class=\"mrow\" id=\"MathJax-Span-36\"><span class=\"mspace\" id=\"MathJax-Span-37\"><\/span><\/span><\/span><sub><span class=\"texatom\" id=\"MathJax-Span-38\"><span class=\"mrow\" id=\"MathJax-Span-39\"><span class=\"mn\" id=\"MathJax-Span-40\">4<\/span><\/span><\/span><\/sub><\/span><\/span><\/span><\/span>, <span class=\"MathJax\" id=\"MathJax-Element-5-Frame\" role=\"presentation\"><span class=\"math\" id=\"MathJax-Span-41\"><span class=\"mrow\" id=\"MathJax-Span-42\"><span class=\"msubsup\" id=\"MathJax-Span-43\"><span class=\"mtext\" id=\"MathJax-Span-44\">CCl<\/span><span class=\"texatom\" id=\"MathJax-Span-45\"><span class=\"mrow\" id=\"MathJax-Span-46\"><span class=\"mspace\" id=\"MathJax-Span-47\"><\/span><\/span><\/span><sub><span class=\"texatom\" id=\"MathJax-Span-48\"><span class=\"mrow\" id=\"MathJax-Span-49\"><span class=\"mn\" id=\"MathJax-Span-50\">4<\/span><\/span><\/span><\/sub><\/span><\/span><\/span><\/span> are <strong>nonpolar compounds<\/strong>, whereas <span class=\"MathJax\" id=\"MathJax-Element-6-Frame\" role=\"presentation\"><span class=\"math\" id=\"MathJax-Span-51\"><span class=\"mrow\" id=\"MathJax-Span-52\"><span class=\"msubsup\" id=\"MathJax-Span-53\"><span class=\"mtext\" id=\"MathJax-Span-54\">H<\/span><span class=\"texatom\" id=\"MathJax-Span-55\"><span class=\"mrow\" id=\"MathJax-Span-56\"><span class=\"mspace\" id=\"MathJax-Span-57\"><\/span><\/span><\/span><sub><span class=\"texatom\" id=\"MathJax-Span-58\"><span class=\"mrow\" id=\"MathJax-Span-59\"><span class=\"mn\" id=\"MathJax-Span-60\">2<\/span><\/span><\/span><\/sub><\/span><span class=\"mtext\" id=\"MathJax-Span-61\">O<\/span><\/span><\/span><\/span>, <span class=\"MathJax\" id=\"MathJax-Element-7-Frame\" role=\"presentation\"><span class=\"math\" id=\"MathJax-Span-62\"><span class=\"mrow\" id=\"MathJax-Span-63\"><span class=\"msubsup\" id=\"MathJax-Span-64\"><span class=\"mtext\" id=\"MathJax-Span-65\">NH<\/span><span class=\"texatom\" id=\"MathJax-Span-66\"><span class=\"mrow\" id=\"MathJax-Span-67\"><span class=\"mspace\" id=\"MathJax-Span-68\"><\/span><\/span><\/span><sub><span class=\"texatom\" id=\"MathJax-Span-69\"><span class=\"mrow\" id=\"MathJax-Span-70\"><span class=\"mn\" id=\"MathJax-Span-71\">3<\/span><\/span><\/span><\/sub><\/span><\/span><\/span><\/span>, <span class=\"MathJax\" id=\"MathJax-Element-8-Frame\" role=\"presentation\"><span class=\"math\" id=\"MathJax-Span-72\"><span class=\"mrow\" id=\"MathJax-Span-73\"><span class=\"msubsup\" id=\"MathJax-Span-74\"><span class=\"mtext\" id=\"MathJax-Span-75\">CH<\/span><span class=\"texatom\" id=\"MathJax-Span-76\"><span class=\"mrow\" id=\"MathJax-Span-77\"><span class=\"mspace\" id=\"MathJax-Span-78\"><\/span><\/span><\/span><sub><span class=\"texatom\" id=\"MathJax-Span-79\"><span class=\"mrow\" id=\"MathJax-Span-80\"><span class=\"mn\" id=\"MathJax-Span-81\">3<\/span><\/span><\/span><\/sub><\/span><span class=\"mtext\" id=\"MathJax-Span-82\">OH<\/span><\/span><\/span><\/span>, <span class=\"MathJax\" id=\"MathJax-Element-9-Frame\" role=\"presentation\"><span class=\"math\" id=\"MathJax-Span-83\"><span class=\"mrow\" id=\"MathJax-Span-84\"><span class=\"mtext\" id=\"MathJax-Span-85\">NO<\/span><\/span><\/span><\/span>, <span class=\"MathJax\" id=\"MathJax-Element-10-Frame\" role=\"presentation\"><span class=\"math\" id=\"MathJax-Span-86\"><span class=\"mrow\" id=\"MathJax-Span-87\"><span class=\"mtext\" id=\"MathJax-Span-88\">CO<\/span><\/span><\/span><\/span>, <span class=\"MathJax\" id=\"MathJax-Element-11-Frame\" role=\"presentation\"><span class=\"math\" id=\"MathJax-Span-89\"><span class=\"mrow\" id=\"MathJax-Span-90\"><span class=\"mtext\" id=\"MathJax-Span-91\">HCl<\/span><\/span><\/span><\/span>, <span class=\"MathJax\" id=\"MathJax-Element-12-Frame\" role=\"presentation\"><span class=\"math\" id=\"MathJax-Span-92\"><span class=\"mrow\" id=\"MathJax-Span-93\"><span class=\"msubsup\" id=\"MathJax-Span-94\"><span class=\"mtext\" id=\"MathJax-Span-95\">H<\/span><span class=\"texatom\" id=\"MathJax-Span-96\"><span class=\"mrow\" id=\"MathJax-Span-97\"><span class=\"mspace\" id=\"MathJax-Span-98\"><\/span><\/span><\/span><sub><span class=\"texatom\" id=\"MathJax-Span-99\"><span class=\"mrow\" id=\"MathJax-Span-100\"><span class=\"mn\" id=\"MathJax-Span-101\">2<\/span><\/span><\/span><\/sub><\/span><span class=\"mtext\" id=\"MathJax-Span-102\">S<\/span><\/span><\/span><\/span>, <span class=\"MathJax\" id=\"MathJax-Element-13-Frame\" role=\"presentation\"><span class=\"math\" id=\"MathJax-Span-103\"><span class=\"mrow\" id=\"MathJax-Span-104\"><span class=\"msubsup\" id=\"MathJax-Span-105\"><span class=\"mtext\" id=\"MathJax-Span-106\">PH<\/span><span class=\"texatom\" id=\"MathJax-Span-107\"><span class=\"mrow\" id=\"MathJax-Span-108\"><span class=\"mspace\" id=\"MathJax-Span-109\"><\/span><\/span><\/span><span class=\"texatom\" id=\"MathJax-Span-110\"><span class=\"mrow\" id=\"MathJax-Span-111\"><span class=\"mn\" id=\"MathJax-Span-112\"><sub>3<\/sub> higher electronegativity difference<\/span><\/span><\/span><\/span><\/span><\/span><\/span> are <strong>polar compounds<\/strong>. Typically compounds that have similar polarity are soluble in one another. This can be described by the rule: <\/span><\/p>\n<h4><span style=\"color: #000000\"><strong><em>Like Dissolves Like.<\/em><\/strong><br \/>\n<\/span><\/h4>\n<section class=\"mt-content-container\">\n<div class=\"mt-section\" id=\"section_1\">\n<p class=\"Love\" id=\"gob-ch09_s03_p02\"><span style=\"color: #000000\">This means that substances must have similar intermolecular forces to form solutions. When a soluble solute is introduced into a solvent, the particles of solute can interact with the particles of solvent. In the case of a solid or liquid solute, the interactions between the solute particles and the solvent particles are so strong that the individual solute particles separate from each other and, surrounded by solvent molecules, enter the solution. (Gaseous solutes already have their constituent particles separated, but the concept of being surrounded by solvent particles still applies.) This process is called <strong><em>solvatio<\/em><em>n<\/em><\/strong> and is illustrated in Figure <span class=\"MathJax\" id=\"MathJax-Element-1-Frame\" role=\"presentation\"><span class=\"math\" id=\"MathJax-Span-1\"><span class=\"mrow\" id=\"MathJax-Span-2\"><span class=\"texatom\" id=\"MathJax-Span-3\"><span class=\"mrow\" id=\"MathJax-Span-4\"><span class=\"mn\" id=\"MathJax-Span-5\">7.2. <\/span><span class=\"mn\" id=\"MathJax-Span-6\"><\/span><\/span><\/span><\/span><\/span><\/span>When the solvent is water, the word <strong><em>hydration<\/em><\/strong>, rather than solvation, is used.<\/span><\/p>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<div class=\"mt-section\" id=\"section_2\">\n<p><span style=\"color: #000000\">In general polar solvents dissolve polar solutes whereas nonpolar solvents will dissolve nonpolar solutes. Overall, the solution process depends on the strength of the attraction between the solute particles and the solvent particles.\u00a0 For example, water is a highly polar solvent that is capable of dissolving many ionic salts. Figure 7.2 shows the solution process, where water act as the solvent to dissolve the crystalline salt, sodium chloride (NaCl). Note that when ionic compounds dissolve in a solvent they break apart into free floating ions in solution. This enables the compound to interact with the solvent. In the case of water dissolving sodium chloride, the sodium ion is attracted to the partial negative charge of the oxygen atom in the water molecule, whereas the chloride ion is attracted to the partial positive hydrogen atoms.<br \/>\n<\/span><\/p>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/dissolving-salt.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-3729\" alt=\"\" src=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/dissolving-salt-1024x509.png\" width=\"702\" height=\"349\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/dissolving-salt-1024x509.png 1024w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/dissolving-salt-300x149.png 300w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/dissolving-salt-768x382.png 768w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/dissolving-salt.png 1767w\" sizes=\"(max-width: 702px) 100vw, 702px\" \/><\/a><\/p>\n<p><span style=\"color: #000000\"><strong>Figure 7.2: The Process of Dissolving.<\/strong> When an ionic salt, such as sodium chloride, shown in (A), comes into contact with water, the water molecules dissociate the ion molecules of the sodium chloride into their ionic state, shown as a molecular model in (B) the solid crystalline lattice of sodium chloride, and (C) the sodium chloride dissolved in the water solvent. (Photo of sodium chloride provided by <a href=\"https:\/\/commons.wikimedia.org\/w\/index.php?curid=11084\">Chris 73<\/a> ).<br \/>\n<\/span><\/p>\n<hr \/>\n<p><span style=\"color: #000000\">Many ionic compounds are soluble in water, however, not all ionic compounds are soluble. Ionic compounds that are soluble in water exist in their ionic state within the solution. You will notice in Figure 7.2 that the sodium chloride breaks apart into the sodium ion and the chloride ion as it dissolves and interacts with the water molecules. For ionic compounds that are not soluble in water, the ions are so strongly attracted to one another that they cannot be broken apart by the partial charges of the water molecules. The following table can be used to help you predict which ionic compounds will be soluble in water.<br \/>\n<\/span><\/p>\n<h4><span style=\"color: #000000\"><strong>Table 7.1 Solubility Rules<\/strong><\/span><\/h4>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/solubility_rules_2.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-3743\" alt=\"\" src=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/solubility_rules_2-1024x745.png\" width=\"700\" height=\"509\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/solubility_rules_2-1024x745.png 1024w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/solubility_rules_2-300x218.png 300w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/solubility_rules_2-768x559.png 768w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/solubility_rules_2.png 1187w\" sizes=\"(max-width: 700px) 100vw, 700px\" \/><\/a><\/p>\n<p><span style=\"color: #000000\">The dissociation of soluble ionic compounds gives solutions of these compounds an interesting property: they conduct electricity. Because of this property, soluble ionic compounds are referred to as <em><strong>electrolytes<\/strong><\/em>. Many ionic compounds dissociate completely and are therefore called <strong><em>strong electrolytes<\/em><\/strong>. Sodium chloride is an example of a strong electrolyte. Some compounds dissolve but dissociate only partially, and solutions of such solutes may conduct electricity only weakly. These solutes are called <strong><em>weak electrolytes<\/em><\/strong>. Acetic acid (CH<sub class=\"subscript\">3<\/sub>COOH), the compound in vinegar, is a weak electrolyte. Solutes that dissolve into individual neutral molecules without dissociation do not impart additional electrical conductivity to their solutions and are called <em><strong>nonelectrolytes<\/strong><\/em>. Polar covalent compounds, such as table sugar (C<sub class=\"subscript\">12<\/sub>H<sub class=\"subscript\">22<\/sub>O<sub class=\"subscript\">11<\/sub>), are good examples of <strong><em>nonelectrolytes<\/em><\/strong>. <\/span><\/p>\n<p><span style=\"color: #000000\">The term <strong><em class=\"emphasis\">electrolyte<\/em><\/strong> is used in medicine to mean any of the important ions that are dissolved in aqueous solution in the body. Important physiological electrolytes include Na<sup class=\"superscript\">+<\/sup>, K<sup class=\"superscript\">+<\/sup>, Ca<sup class=\"superscript\">2<\/sup><sup class=\"superscript\">+<\/sup>, Mg<sup class=\"superscript\">2<\/sup><sup class=\"superscript\">+<\/sup>, and Cl<sup class=\"superscript\">\u2212<\/sup>. Sports drinks such as Gatoraid have combinations of these key electrolytes, to help replenish electrolyte loss following a hard workout.<\/span><\/p>\n<p><span style=\"color: #000000\">Similarly, solutions can also be made by mixing two compatible liquids together. The liquid in the lower concentration is termed the <em><strong>solute,<\/strong><\/em> and the one in higher concentration the <strong><em>solvent<\/em><\/strong>. For example, grain alcohol (CH<sub>3<\/sub>CH<sub>2<\/sub>OH) is a polar covalent molecule that can mix with water. When two similar solutions are placed together and are able to mix into a solution, they are said to be <strong><em>miscible<\/em><\/strong>. Liquids that do not share similar characteristics and cannot mix together, on the other hand, are termed <strong><em>immiscible<\/em><\/strong>. For example, the oils found in olive oil, such as oleic acid (<span class=\"_Xbe kno-fv\">C<sub>18<\/sub>H<sub>34<\/sub>O<sub>2<\/sub><\/span>) have mainly nonpolar covalent bonds which do not have intermolecular forces that are strong enough to break the hydrogen bonding between the water molecules. Thus, water and oil do not mix and are said to be <strong><em>immiscible<\/em><\/strong>.<br \/>\n<\/span><\/p>\n<p><span style=\"color: #000000\">Other factor such as temperature and pressure also affects the solubility of a solvent. Thus, in specifying solubility, one should also be aware of these other factors.<\/span><\/p>\n<h4><a href=\"#title\"><span><em><strong>(Back to the Top)<\/strong><\/em><\/span><\/a><\/h4>\n<hr \/>\n<\/div>\n<div class=\"mt-section\" id=\"section_3\">\n<h3 id=\"tempsol\"><strong>7.4 Temperature and Solubility<\/strong><\/h3>\n<p><span style=\"color: #000000\">When considering the solubility solids, the relationship of temperature and solubility is not simple or predictable. Figure 7.3 shows plots of the solubilities of several organic and inorganic compounds in water as a function of temperature. Although the solubility of a solid generally increases with increasing temperature, there is no simple relationship between the structure of a substance and the temperature dependence of its solubility. Many compounds (such as glucose and CH<sub class=\"subscript\">3<\/sub>CO<sub class=\"subscript\">2<\/sub>Na) exhibit a dramatic increase in solubility with increasing temperature. Others (such as NaCl and K<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>) exhibit little variation, and still others (such as Li<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>) become less soluble with increasing temperature.<\/span><\/p>\n<div class=\"figure large medium-height editable block\" id=\"averill_1.0-ch13_s04_s01_f01\">\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/2012books.lardbucket.org\/books\/principles-of-general-chemistry-v1.0\/section_17\/1688d1d5941ac728ff289a871c57b432.jpg\" width=\"699\" height=\"443\" \/><\/p>\n<p class=\"para\"><span style=\"color: #000000\"><strong><span class=\"title-prefix\">Figure 7.3<\/span> Solubilities of Several Inorganic and Organic Solids in Water as a Function of Temperature.<\/strong> Solubility may increase or decrease with temperature; the magnitude of this temperature dependence varies widely among compounds.<\/span><\/p>\n<hr \/>\n<\/div>\n<p class=\"para editable block\" id=\"averill_1.0-ch13_s04_s01_p03\"><span style=\"color: #000000\">The variation of solubility with temperature has been measured for a wide range of compounds, and the results are published in many standard reference books. Chemists are often able to use this information to separate the components of a mixture by<strong> <em><span class=\"margin_term\"><a class=\"glossterm\">fractional crystallization<\/a><\/span><\/em><\/strong>, the separation of compounds on the basis of their solubilities in a given solvent. For example, if we have a mixture of 150 g of sodium acetate (CH<sub class=\"subscript\">3<\/sub>CO<sub class=\"subscript\">2<\/sub>Na) and 50 g of KBr, we can separate the two compounds by dissolving the mixture in 100 g of water at 80\u00b0C and then cooling the solution slowly to 0\u00b0C. According to the temperature curves in Figure 7.3, both compounds dissolve in water at 80\u00b0C, and all 50 g of KBr remains in solution at 0\u00b0C. Only about 36 g of CH<sub class=\"subscript\">3<\/sub>CO<sub class=\"subscript\">2<\/sub>Na are soluble in 100 g of water at 0\u00b0C, however, so approximately 114 g (150 g\u00a0\u2212\u00a036 g) of CH<sub class=\"subscript\">3<\/sub>CO<sub class=\"subscript\">2<\/sub>Na crystallizes out on cooling. The crystals can then be separated by filtration. Thus fractional crystallization allows us to recover about 75% of the original CH<sub class=\"subscript\">3<\/sub>CO<sub class=\"subscript\">2<\/sub>Na in essentially pure form in only one step.<\/span><\/p>\n<p class=\"para editable block\" id=\"averill_1.0-ch13_s04_s01_p04\"><span style=\"color: #000000\">Fractional crystallization is a common technique for purifying compounds as diverse as those shown in Figure 7.3 and from antibiotics to enzymes. For the technique to work properly, the compound of interest must be more soluble at high temperature than at low temperature, so that lowering the temperature causes it to crystallize out of solution. In addition, the impurities must be <em class=\"emphasis\">more<\/em> soluble than the compound of interest (as was KBr in this example) and preferably present in relatively small amounts.<\/span><\/p>\n<p class=\"para editable block\" id=\"averill_1.0-ch13_s04_s02_p01\"><span style=\"color: #000000\">The solubility of gases in liquids is much more predictable.\u00a0 The solubility of gases in liquids decreases with increasing temperature, as shown in Figure 7.4. Attractive intermolecular interactions in the gas phase are essentially zero for most substances, because the molecules are so far apart when in the gaseous form. When a gas dissolves, it does so because its molecules interact with solvent molecules. Heat is released when these new attractive forces form.\u00a0 Thus, if external heat is added to the system, it overcomes the attractive forces between the gas and the solvent molecules and decreases the solubility of the gas. <\/span><\/p>\n<div class=\"figure medium editable block\" id=\"averill_1.0-ch13_s04_s02_f01\">\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/2012books.lardbucket.org\/books\/principles-of-general-chemistry-v1.0\/section_17\/67558bdc4beb64e06b29db7b4c8d74bb.jpg\" width=\"697\" height=\"794\" \/><\/p>\n<p class=\"para\"><span style=\"color: #000000\"><strong><span class=\"title-prefix\">Figure 7.4<\/span> Solubilities of Several Common Gases in Water as a Function of Temperature at Partial Pressure of 1 atm.<\/strong> The solubilities of gases decrease with increasing temperature.<\/span><\/p>\n<hr \/>\n<\/div>\n<p class=\"para editable block\" id=\"averill_1.0-ch13_s04_s02_p02\"><span style=\"color: #000000\">The decrease in the solubilities of gases at higher temperatures has both practical and environmental implications. Anyone who routinely boils water in a teapot or electric kettle knows that a white or gray deposit builds up on the inside and must eventually be removed. The same phenomenon occurs on a much larger scale in the giant boilers used to supply hot water or steam for industrial applications, where it is called <strong><em>\u201cboiler scale,\u201d<\/em><\/strong> a deposit that can seriously decrease the capacity of hot water pipes (Figure 7.5). The problem is not a uniquely modern one: aqueducts that were built by the Romans 2000 years ago to carry cold water from alpine regions to warmer, drier regions in southern France were clogged by similar deposits. The chemistry behind the formation of these deposits is moderately complex, but the driving force is the loss of dissolved carbon dioxide (CO<sub class=\"subscript\">2<\/sub>) from solution. Hard water contains dissolved Ca<sup class=\"superscript\">2+<\/sup> and HCO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup> (bicarbonate) ions. Calcium bicarbonate [Ca(HCO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>] is rather soluble in water, but calcium carbonate (CaCO<sub class=\"subscript\">3<\/sub>) is quite insoluble. A solution of bicarbonate ions can react to form carbon dioxide, carbonate ion, and water:<\/span><\/p>\n<div class=\"equation block\" id=\"averill_1.0-ch13_s04_s02_eq01\"><span class=\"mathphrase\" style=\"color: #000000\">2HCO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq)\u00a0\u2192\u00a0CO<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">2\u2212<\/sup>(aq)\u00a0+\u00a0H<sub class=\"subscript\">2<\/sub>O(l)\u00a0+\u00a0CO<sub class=\"subscript\">2<\/sub>(aq)<\/span><\/div>\n<p class=\"para editable block\" id=\"averill_1.0-ch13_s04_s02_p03\"><span style=\"color: #000000\">Heating the solution decreases the solubility of CO<sub class=\"subscript\">2<\/sub>, which escapes into the gas phase above the solution. In the presence of calcium ions, the carbonate ions precipitate as insoluble calcium carbonate, the major component of boiler scale.<\/span><\/p>\n<div class=\"figure small editable block\" id=\"averill_1.0-ch13_s04_s02_f02\">\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/2012books.lardbucket.org\/books\/principles-of-general-chemistry-v1.0\/section_17\/6c9bc40f1fc303541d0062d946b951f8.jpg\" width=\"557\" height=\"539\" \/><\/p>\n<p class=\"para\"><span style=\"color: #000000\"><strong><span class=\"title-prefix\">Figure 7.5<\/span> Boiler Scale in a Water Pipe.<\/strong> Calcium carbonate (CaCO<sub class=\"subscript\">3<\/sub>) deposits in hot water pipes can significantly reduce pipe capacity. These deposits, called boiler scale, form when dissolved CO<sub class=\"subscript\">2<\/sub> is driven into the gas phase at high temperatures.<\/span><\/p>\n<\/div>\n<p class=\"para editable block\" id=\"averill_1.0-ch13_s04_s02_p04\"><span style=\"color: #000000\">In <em class=\"emphasis\">thermal pollution<\/em>, lake or river water that is used to cool an industrial reactor or a power plant is returned to the environment at a higher temperature than normal. Because of the reduced solubility of O<sub class=\"subscript\">2<\/sub> at higher temperatures (Figure 7.4), the warmer water contains less dissolved oxygen than the water did when it entered the plant. Fish and other aquatic organisms that need dissolved oxygen to live can literally suffocate if the oxygen concentration of their habitat is too low. Because the warm, oxygen-depleted water is less dense, it tends to float on top of the cooler, denser, more oxygen-rich water in the lake or river, forming a barrier that prevents atmospheric oxygen from dissolving. Eventually even deep lakes can be suffocated if the problem is not corrected. Additionally, most fish and other nonmammalian aquatic organisms are cold-blooded, which means that their body temperature is the same as the temperature of their environment. Temperatures substantially greater than the normal range can lead to severe stress or even death. Cooling systems for power plants and other facilities must be designed to minimize any adverse effects on the temperatures of surrounding bodies of water. In the Pacific Northwest, salmonid populations are extremely susceptible to changes in water temperature. Within these population, optimal water temperatures are between 12.8 and 17.8 <sup>o<\/sup>C (55-65 <sup>o<\/sup>F). In addition to reduced oxygen levels, salmon populations are much more susceptible to disease, predation, and parasite infections at higher water temperatures. Thus, thermal pollution and global climate change are creating real challenges to the survival and maintenance of these species. For more information on the effects of rising temperatures on salmonid populations visit the State of Washington&#8217;s <a href=\"https:\/\/fortress.wa.gov\/ecy\/publications\/publications\/0010046.pdf\">Focus Publication.<\/a><br \/>\n<\/span><\/p>\n<p class=\"para editable block\" id=\"averill_1.0-ch13_s04_s02_p05\"><span style=\"color: #000000\">A similar effect is seen in the rising temperatures of bodies of water such as the Chesapeake Bay, the largest estuary in North America, where global warming has been implicated as the cause. For each 1.5\u00b0C that the bay\u2019s water warms, the capacity of water to dissolve oxygen decreases by about 1.1%. Many marine species that are at the southern limit of their distributions have shifted their populations farther north. In 2005, the eelgrass, which forms an important nursery habitat for fish and shellfish, disappeared from much of the bay following record high water temperatures. Presumably, decreased oxygen levels decreased populations of clams and other filter feeders, which then decreased light transmission to allow the eelsgrass to grow. The complex relationships in ecosystems such as the Chesapeake Bay are especially sensitive to temperature fluctuations that cause a deterioration of habitat quality.<\/span><\/p>\n<h4><a href=\"#title\"><span><em><strong>(Back to the Top)<\/strong><\/em><\/span><\/a><\/h4>\n<hr \/>\n<\/div>\n<h3 id=\"pressuresol\"><span style=\"color: #ff0000\"><strong>7.5 Effects of Pressure on the Solubility of Gases: Henry&#8217;s Law<\/strong><\/span><\/h3>\n<p class=\"para editable block\" id=\"averill_1.0-ch13_s04_s03_p01\"><span style=\"color: #000000\">External pressure has very little effect on the solubility of liquids and solids. In contrast, the solubility of gases increases as the partial pressure of the gas above a solution increases. This point is illustrated in Figure 7.6<a class=\"xref\" href=\"https:\/\/2012books.lardbucket.org\/books\/principles-of-general-chemistry-v1.0\/s17-solutions.html#averill_1.0-ch13_s04_s03_f01\"><\/a>, which shows the effect of increased pressure on the dynamic equilibrium that is established between the dissolved gas molecules in solution and the molecules in the gas phase above the solution. Because the concentration of molecules in the gas phase increases with increasing pressure, the concentration of dissolved gas molecules in the solution at equilibrium is also higher at higher pressures.<\/span><\/p>\n<div class=\"figure large medium-height editable block\" id=\"averill_1.0-ch13_s04_s03_f01\">\n<p class=\"title\"><span class=\"title-prefix\"><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/2012books.lardbucket.org\/books\/principles-of-general-chemistry-v1.0\/section_17\/e21684c864731d8f2e572dfcbf18b54f.jpg\" width=\"696\" height=\"242\" \/><\/p>\n<p class=\"para\"><span style=\"color: #000000\"><strong><span class=\"title-prefix\">Figure 7.6<\/span> A Model Depicting Why the Solubility of a Gas Increases as the Partial Pressure Increases at Constant Temperature.<\/strong> (a) When a gas comes in contact with a pure liquid, some of the gas molecules (purple spheres) collide with the surface of the liquid and dissolve. When the concentration of dissolved gas molecules has increased so that the rate at which gas molecules escape into the gas phase is the same as the rate at which they dissolve, a dynamic equilibrium has been established, as depicted here.\u00a0 (b) Increasing the pressure of the gas increases the number of molecules of gas per unit volume, which increases the rate at which gas molecules collide with the surface of the liquid and dissolve. (c) As additional gas molecules dissolve at the higher pressure, the concentration of dissolved gas increases until a new dynamic equilibrium is established.<\/span><\/p>\n<hr \/>\n<\/div>\n<p class=\"para block\" id=\"averill_1.0-ch13_s04_s03_p02\"><span style=\"color: #000000\">The relationship between pressure and the solubility of a gas is described quantitatively by <span class=\"margin_term\"><a class=\"glossterm\">Henry\u2019s law<\/a><\/span>, which is named for its discoverer, the English physician and chemist, William Henry (1775\u20131836):<\/span><\/p>\n<div class=\"equation block\" id=\"averill_1.0-ch13_s04_s03_eq01\"><strong><span class=\"mathphrase\" style=\"color: #000000\"><em class=\"emphasis\">C<\/em>\u00a0=\u00a0<em class=\"emphasis\">kP<\/em><\/span><\/strong><\/div>\n<p class=\"para editable block\" id=\"averill_1.0-ch13_s04_s03_p03\"><span style=\"color: #000000\">where <em class=\"emphasis\">C<\/em> is the concentration of dissolved gas at equilibrium, <em class=\"emphasis\">P<\/em> is the partial pressure of the gas, and <em class=\"emphasis\">k<\/em> is the <em class=\"emphasis\">Henry\u2019s law constant<\/em>, which must be determined experimentally for each combination of gas, solvent, and temperature. Although the gas concentration may be expressed in any convenient units, we will use molarity exclusively. The units of the Henry\u2019s law constant are therefore mol\/(L\u00b7atm)\u00a0=\u00a0M\/atm. Values of the Henry\u2019s law constants for solutions of several gases in water at 20\u00b0C are listed in Table 7.2<\/span><\/p>\n<hr \/>\n<h4><strong><a href=\"https:\/\/www.khanacademy.org\/science\/health-and-medicine\/respiratory-system\/gas-exchange-jv\/v\/henry-s-law\">Video Tutorial on Henry&#8217;s Law from Kahn Academy<\/a><\/strong><\/h4>\n<p><em><span style=\"color: #000000\"><b>All Khan Academy content is available for free at www.khanacademy.org<\/b><\/span><\/em><\/p>\n<hr \/>\n<p class=\"para editable block\" id=\"averill_1.0-ch13_s04_s03_p04\"><span style=\"color: #000000\">As the data in Table 7.2 demonstrate, the concentration of a dissolved gas in water at a given pressure depends strongly on its physical properties. For a series of related substances, London dispersion forces increase as molecular mass increases. Thus among the elements of group 18, the Henry\u2019s law constants increase smoothly from He to Ne to Ar. The table also shows that O<sub class=\"subscript\">2<\/sub> is almost twice as soluble as N<sub class=\"subscript\">2<\/sub>. Although London dispersion forces are too weak to explain such a large difference, O<sub class=\"subscript\">2<\/sub> is <strong><em>paramagnetic<\/em> <\/strong>and hence more polarizable than N<sub class=\"subscript\">2<\/sub>, which explains its high solubility. (Note: When a substance is <strong><em>paramagnetic<\/em> <\/strong>it is very weakly attracted by the poles of a magnet, but does not retain any permanent magnetism).<\/span><\/p>\n<div class=\"table block\" id=\"averill_1.0-ch13_s04_s03_t01\">\n<h4 class=\"title\"><span style=\"color: #000000\"><strong><span class=\"title-prefix\">Table 7.2<\/span> Henry\u2019s Law Constants for Selected Gases in Water at 20\u00b0C<\/strong><\/span><\/h4>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/pressure_and_solubility.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-3751\" alt=\"\" src=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/pressure_and_solubility-1024x579.png\" width=\"704\" height=\"398\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/pressure_and_solubility-1024x579.png 1024w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/pressure_and_solubility-300x170.png 300w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/pressure_and_solubility-768x434.png 768w\" sizes=\"(max-width: 704px) 100vw, 704px\" \/><\/a><\/p>\n<p><span style=\"color: #000000\">The partial pressure of a gas can be expressed as concentration by writing Henry&#8217;s Law as <em>P<\/em><sub>gas<\/sub> = C<em>\/k.\u00a0 <\/em>This is important in many aspects of life including medicine where blood gases, like oxygen and carbon dioxide are commonly measured. Since partial pressure and concentration are directly proportional, if the partial pressure of a gas changes while the temperature remains constant, the new concentration of the gas within the liquid can be easily calculated using the following equation:<\/span><\/p>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Henrys-Law-2.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-3767\" alt=\"\" src=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Henrys-Law-2.png\" width=\"246\" height=\"157\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Henrys-Law-2.png 326w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Henrys-Law-2-300x191.png 300w\" sizes=\"(max-width: 246px) 100vw, 246px\" \/><\/a><\/p>\n<p><span style=\"color: #000000\">Where<em> C<sub>1<\/sub><\/em> and <em>P<sub>1<\/sub><\/em> are the concentration and partial pressure, respectively, of the gas at the initial condition, and<em> C<sub>2<\/sub> <\/em>and <em>P<sub>2<\/sub><\/em> are the concentration and partial pressure, respectively, of the gas at the final condition. For example:<\/span><\/p>\n<h4><span style=\"color: #000000\"><span style=\"color: #008000\"><em><strong>Practice Problem:\u00a0<\/strong><\/em><\/span> <strong>The concentration of CO<sub>2<\/sub> in a solution is 0.032 M at 3.0 atm.\u00a0 What is the concentration of CO<sub>2<\/sub> at 5.0 atm of pressure?<\/strong><\/span><\/h4>\n<h4><span style=\"color: #000000\"><em><strong><span style=\"color: #008000\">Solution:<\/span> <\/strong><\/em><strong>To address this problem, first we must identify what we want to find.\u00a0 This is the concentration of CO<sub>2<\/sub> at 5.0 atm of pressure.\u00a0 These two values represent <em>C<sub>2<\/sub> = ?? <\/em>and <em>P<sub>2<\/sub><\/em> = 5.0 atm. At this point it will be easiest to rearrange our equation above to solve for <em>C<sub>2<\/sub><\/em>.<\/strong><\/span><span style=\"color: #000000\"><strong> Next we need to identify the starting conditions, <em>C<sub>1<\/sub> <\/em>= 0.032 M and <em>P<sub>1<\/sub><\/em> = 3.0 atm.\u00a0 Then we can plug that values into the equation and solve for<em> C<sub>2<\/sub> <\/em>:<\/strong><\/span><\/h4>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/partial-pressure-example.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-3772\" alt=\"\" src=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/partial-pressure-example-606x1024.png\" width=\"455\" height=\"769\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/partial-pressure-example-606x1024.png 606w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/partial-pressure-example-178x300.png 178w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/partial-pressure-example.png 745w\" sizes=\"(max-width: 455px) 100vw, 455px\" \/><\/a><\/p>\n<\/div>\n<p class=\"para editable block\" id=\"averill_1.0-ch13_s04_s03_p05\"><span style=\"color: #000000\">Gases that react chemically with water, such as HCl and the other hydrogen halides, H<sub class=\"subscript\">2<\/sub>S, and NH<sub class=\"subscript\">3<\/sub>, do not obey Henry\u2019s law; all of these gases are much more soluble than predicted by Henry\u2019s law. For example, HCl reacts with water to give H<sup class=\"superscript\">+<\/sup>(aq) and Cl<sup class=\"superscript\">\u2212<\/sup>(aq), <em class=\"emphasis\">not<\/em> dissolved HCl molecules, and its dissociation into ions results in a much higher solubility than expected for a neutral molecule. Overall, gases that react with water do not obey Henry&#8217;s Law.<br \/>\n<\/span><\/p>\n<div class=\"callout editable block\" id=\"averill_1.0-ch13_s04_s03_n01\">\n<h4 class=\"title\"><em><span style=\"color: #000000\"><strong>Note the Pattern<\/strong><\/span><\/em><\/h4>\n<\/div>\n<p class=\"para editable block\" id=\"averill_1.0-ch13_s04_s03_p07\"><span style=\"color: #000000\">Henry\u2019s law has important applications. For example, bubbles of CO<sub class=\"subscript\">2<\/sub> form as soon as a carbonated beverage is opened because the drink was bottled under CO<sub class=\"subscript\">2<\/sub> at a pressure greater than 1 atm. When the bottle is opened, the pressure of CO<sub class=\"subscript\">2<\/sub> above the solution drops rapidly, and some of the dissolved gas escapes from the solution as bubbles. Henry\u2019s law also explains why scuba divers have to be careful to ascend to the surface slowly after a dive if they are breathing compressed air. At the higher pressures under water, more N<sub class=\"subscript\">2<\/sub> from the air dissolves in the diver\u2019s internal fluids. If the diver ascends too quickly, the rapid pressure change causes small bubbles of N<sub class=\"subscript\">2<\/sub> to form throughout the body, a condition known as \u201cthe bends.\u201d These bubbles can block the flow of blood through the small blood vessels, causing great pain and even proving fatal in some cases.<\/span><\/p>\n<p class=\"para editable block\" id=\"averill_1.0-ch13_s04_s03_p08\"><span style=\"color: #000000\">Due to the low Henry\u2019s law constant for O<sub class=\"subscript\">2<\/sub> in water, the levels of dissolved oxygen in water are too low to support the energy needs of multicellular organisms, including humans. To increase the O<sub class=\"subscript\">2<\/sub> concentration in internal fluids, organisms synthesize highly soluble carrier molecules that bind O<sub class=\"subscript\">2<\/sub> reversibly. For example, human red blood cells contain a protein called hemoglobin that specifically binds O<sub class=\"subscript\">2<\/sub> and facilitates its transport from the lungs to the tissues, where it is used to oxidize food molecules to provide energy. The concentration of hemoglobin in normal blood is about 2.2 mM, and each hemoglobin molecule can bind four O<sub class=\"subscript\">2<\/sub> molecules. Although the concentration of dissolved O<sub class=\"subscript\">2<\/sub> in blood serum at 37\u00b0C (normal body temperature) is only 0.010 mM, the total dissolved O<sub class=\"subscript\">2<\/sub> concentration is 8.8 mM, almost a thousand times greater than would be possible without hemoglobin. Synthetic oxygen carriers based on fluorinated alkanes have been developed for use as an emergency replacement for whole blood. Unlike donated blood, these \u201cblood substitutes\u201d do not require refrigeration and have a long shelf life. Their very high Henry\u2019s law constants for O<sub class=\"subscript\">2<\/sub> result in dissolved oxygen concentrations comparable to those in normal blood.<\/span><\/p>\n<h4><a href=\"#title\"><span><em><strong>(Back to the Top)<\/strong><\/em><\/span><\/a><\/h4>\n<hr \/>\n<h3 id=\"solidhydrates\"><strong><span style=\"color: #ff0000\">7.6 Solid Hydrates:<\/span><\/strong><\/h3>\n<p><span style=\"color: #000000\">Some ionic solids will accept a small number of water molecules into their crystal lattice structure and remain in a solid state.\u00a0 These solids are called<em><strong> solid hydrates<\/strong><\/em>. Solid hydrates contain water molecules combined in a definite ratio as an integral part of the crystal that are either bound to a metal center or that have crystallized with the metal complex. Such hydrates are also said to contain <strong><i>water of crystallization<a title=\"Water of crystallization\" href=\"https:\/\/en.wikipedia.org\/wiki\/Water_of_crystallization\"><\/a><\/i><\/strong> or <strong><i>water of hydration<\/i><\/strong>. <\/span><\/p>\n<p><span style=\"color: #000000\">A colorful example is cobalt(II) chloride, which turns from blue to red upon hydration, and can therefore be used as a water indicator.<\/span><\/p>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/cobalt-chloride.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-3759\" alt=\"\" src=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/cobalt-chloride-1024x741.png\" width=\"699\" height=\"506\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/cobalt-chloride-1024x741.png 1024w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/cobalt-chloride-300x217.png 300w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/cobalt-chloride-768x556.png 768w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/cobalt-chloride.png 1108w\" sizes=\"(max-width: 699px) 100vw, 699px\" \/><\/a><\/p>\n<p><span style=\"color: #000000\"><strong>Figure 7.7: Cobalt chloride as an example of a solid hydrate.<\/strong> Anhydrous cobalt chloride (upper left) and it&#8217;s crystal lattice structure (lower left) compared with cobalt chloride hexahydrate (upper right) and it&#8217;s crystal lattice (lower right). Notice that the water molecules shown in red (oxygen) and white (hydrogen) are integrated into the crystal lattice of the cobalt (II) chloride, shown in blue (cobalt) and green (chloride),\u00a0 based on polarity.\u00a0 The partially negative oxygen atoms are attracted to the positively charged cobalt while the partially positive hydrogen atoms are attracted to the negatively charged chloride ions.\u00a0 Images provided by Wikipedia Commons (<a href=\"https:\/\/commons.wikimedia.org\/w\/index.php?curid=95431\">upper left<\/a> and <a href=\"https:\/\/upload.wikimedia.org\/wikipedia\/commons\/a\/a2\/Cobalt(II)-chloride-layer-3D-balls.png\">lower left<\/a>), Benjah-bmm27 (<a href=\"https:\/\/commons.wikimedia.org\/w\/index.php?curid=2174311\">upper right<\/a>), and Smokefoot <a href=\"https:\/\/commons.wikimedia.org\/w\/index.php?curid=55917556\">(lower right<\/a>)<\/span><\/p>\n<p><span style=\"color: #000000\">The notation used to represent a solid hydrate is: &#8220;<i>hydrated compound<\/i><b>\u22c5<i>n<\/i>H<sub>2<\/sub>O<\/b>&#8220;, where <i>n<\/i> is the number of water molecules per formula unit of the salt. The <i>n<\/i> is usually a low integer, though it is possible for fractional values to occur. For example, in a <b>monohydrate<\/b> n is one, and in a <b>hexahydrate<\/b> n is 6. For the example in Figure 7.7, the hydrated cobalt chloride would be designated: <strong>&#8220;cobalt (II) chloride\u22c5<i>6<\/i>H<sub>2<\/sub>O&#8221;.<\/strong> Numerical prefixes of Greek origin that are used to designate solid hydrates are:<\/span><\/p>\n<div class=\"div-col columns column-count column-count-2\">\n<ul>\n<li><span style=\"color: #000000\">Hemi \u2013 1\/2<\/span><\/li>\n<li><span style=\"color: #000000\">Mono \u2013 1<\/span><\/li>\n<li><span style=\"color: #000000\">Sesqui &#8211; 1\u00bd<\/span><\/li>\n<li><span style=\"color: #000000\">Di \u2013 2<\/span><\/li>\n<li><span style=\"color: #000000\">Tri \u2013 3<\/span><\/li>\n<li><span style=\"color: #000000\">Tetra \u2013 4<\/span><\/li>\n<li><span style=\"color: #000000\">Penta \u2013 5<\/span><\/li>\n<li><span style=\"color: #000000\">Hexa \u2013 6<\/span><\/li>\n<li><span style=\"color: #000000\">Hepta \u2013 7<\/span><\/li>\n<li><span style=\"color: #000000\">Octa \u2013 8<\/span><\/li>\n<li><span style=\"color: #000000\">Nona \u2013 9<\/span><\/li>\n<li><span style=\"color: #000000\">Deca \u2013 10<\/span><\/li>\n<li><span style=\"color: #000000\">Undeca &#8211; 11<\/span><\/li>\n<li><span style=\"color: #000000\">Dodeca &#8211; 12<\/span><\/li>\n<\/ul>\n<\/div>\n<p><span style=\"color: #000000\">A hydrate which has lost water is referred to as an<strong><em> anhydride<\/em><\/strong><\/span><span style=\"color: #000000\">; the remaining water, if any exists, can only be removed with very strong heating. A substance that does not contain any water is referred to as <strong><em>anhydrous<\/em><\/strong><\/span><span style=\"color: #000000\">. Some anhydrous compounds are hydrated so easily that they will pull water out of the atmosphere and become hydrated. These substances are said to be<strong><em> hygroscopic<\/em><\/strong> and can be used as drying agents or <strong><em>desiccants<\/em><\/strong>.<\/span><\/p>\n<h4><a href=\"#title\"><span><em><strong>(Back to the Top)<\/strong><\/em><\/span><\/a><\/h4>\n<hr \/>\n<h3 id=\"solutionconcentration\"><strong><span style=\"color: #ff0000\">7.7 Solution Concentration<\/span><\/strong><\/h3>\n<p class=\"para editable block\" id=\"averill_1.0-ch04_s02_p01\"><span style=\"color: #000000\">In chemistry,<em><strong> concentration<\/strong><\/em> is defined as the abundance of a constituent divided by the total volume of a mixture. All of us have a qualitative idea of what is meant by <strong><em class=\"emphasis\">concentration<\/em><\/strong>. Anyone who has made instant coffee or lemonade knows that too much powder gives a strongly flavored, highly concentrated drink, whereas too little results in a dilute solution that may be hard to distinguish from water. Quantitatively, the <span class=\"margin_term\"><a class=\"glossterm\">concentration<\/a><\/span> of a solution describes the quantity of a solute that is contained in a particular quantity of that solution. Knowing the concentration of solutes is important in controlling the stoichiometry of reactants for reactions that occur in solution, and are critical for many aspects of our lives, from measuring the correct dose of medicine to detecting chemical pollutants like lead and arsenic. Chemists use many different ways to define concentrations. In this section, we will cover the most common ways of presenting solution concentration.\u00a0 These include: Molarity and Parts Per Solutions.<br \/>\n<\/span><\/p>\n<div class=\"section\" id=\"averill_1.0-ch04_s02_s01\">\n<h4 id=\"molarity\" class=\"title editable block\"><span style=\"color: #000000\"><strong>7.7.1 Molarity<\/strong><\/span><\/h4>\n<p class=\"para block\" id=\"averill_1.0-ch04_s02_s01_p01\"><span style=\"color: #000000\">The most common unit of concentration is<strong> <em class=\"emphasis\">molarity<\/em><\/strong>, which is also the most useful for calculations involving the stoichiometry of reactions in solution. The <strong><span class=\"margin_term\"><a class=\"glossterm\"><em>molarity<\/em> (M)<\/a><\/span><\/strong> of a solution is the number of moles of solute present in exactly 1 L of solution.\u00a0<\/span><\/p>\n<div class=\"equation block\" id=\"averill_1.0-ch04_s02_s01_eq01\">\n<div class=\"MathJax_Display\"><span class=\"MathJax\" id=\"MathJax-Element-2-Frame\" role=\"presentation\"><span class=\"math\" id=\"MathJax-Span-11\"><span><span class=\"mrow\" id=\"MathJax-Span-12\"><span class=\"semantics\" id=\"MathJax-Span-13\"><span class=\"mrow\" id=\"MathJax-Span-14\"><span class=\"mfrac\" id=\"MathJax-Span-23\"><span class=\"mrow\" id=\"MathJax-Span-26\"><span class=\"mtext\" id=\"MathJax-Span-27\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span><span><\/span><\/span><\/span><\/div>\n<\/div>\n<\/div>\n<div class=\"equation block\" id=\"averill_1.0-ch04_s02_s01_eq01\"><a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Molarity-equations.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-3794\" alt=\"\" src=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Molarity-equations-1024x537.png\" width=\"385\" height=\"202\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Molarity-equations-1024x537.png 1024w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Molarity-equations-300x157.png 300w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Molarity-equations-768x403.png 768w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Molarity-equations.png 1353w\" sizes=\"(max-width: 385px) 100vw, 385px\" \/><\/a><\/div>\n<p class=\"para editable block\" id=\"averill_1.0-ch04_s02_s01_p02\"><span style=\"color: #000000\">The units of molarity are therefore moles per liter of solution (mol\/L), abbreviated as M. Note that the volume indicated is the total volume of the solution and includes both the solute and the solvent.\u00a0 For example, an aqueous solution that contains 1 mol (342 g) of sucrose in enough water to give a final volume of 1.00 L has a sucrose concentration of 1.00 mol\/L or 1.00 M. In chemical notation, square brackets around the name or formula of the solute represent the concentration of a solute. So<\/span><\/p>\n<p><span class=\"informalequation block\" style=\"color: #000000\"> <span class=\"mathphrase\">[sucrose] = 1.00 M<\/span> <\/span><\/p>\n<p class=\"para editable block\" id=\"averill_1.0-ch04_s02_s01_p03\"><span style=\"color: #000000\">is read as \u201cthe concentration of sucrose is 1.00 molar.\u201d The equation above can be used to calculate how much solute is required to make any amount of a desired solution.\u00a0<\/span><\/p>\n<div class=\"exercises block\" id=\"averill_1.0-ch04_s02_s01_n01\">\n<h4 class=\"title\"><strong><em><span style=\"color: #008000\">Example Problem:<\/span><\/em><\/strong><\/h4>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s01_p06\"><span style=\"color: #000000\">Calculate the number of moles of sodium hydroxide (NaOH) needed to make 2.50 L of 0.100 M NaOH.<\/span><\/p>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s01_p07\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">Given: (1) <\/strong>identity of solute = NaOH,\u00a0<strong> (2)<\/strong> volume = 2.50 L,\u00a0 and <strong>(3)<\/strong> molarity of solution = 0.100 mol\/L (Note: when calculating problems always write out the units of molarity as mol\/L, rather than M. This will allow you to cancel out your units when doing the calculation.)<\/span><\/p>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s01_p08\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">Asked for: <\/strong>amount of solute in moles<\/span><\/p>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s01_p09\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">Strategy:\u00a0 (1) <\/strong>Rearrange the equation above to solve for the desired unit, in this case for moles.<strong> (2)<\/strong> Double check all the units in the equation and make sure they match. Perform any conversions that are needed so that the units match.<strong> (3)<\/strong> Fill in values appropriately and do the math.<\/span><\/p>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s01_p11\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">Solution:<\/strong><\/span><\/p>\n<p><span style=\"color: #000000\"><strong class=\"emphasis bold\">(1) <\/strong>Rearrange the equation above to solve for moles.<\/span><\/p>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/rearrange-molarity.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-3797\" alt=\"\" src=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/rearrange-molarity.png\" width=\"477\" height=\"120\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/rearrange-molarity.png 879w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/rearrange-molarity-300x75.png 300w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/rearrange-molarity-768x193.png 768w\" sizes=\"(max-width: 477px) 100vw, 477px\" \/><\/a><\/p>\n<p><span style=\"color: #000000\"><strong>(2)\u00a0<\/strong>Double check all the units in the equation and make sure they match. <\/span><\/p>\n<p><span style=\"color: #000000\">The given values for this equation are the volume 2.50 L and the molarity 0.100 mol\/L.\u00a0 The volume units for both of these numbers are in Liters (L) and thus, match. Therefore, no conversions need to be made.<\/span><\/p>\n<p><span style=\"color: #000000\"><strong>(3)<\/strong> \u00a0Fill in values appropriately and do the math.<\/span><\/p>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/worked-molarity-problem.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-3798\" alt=\"\" src=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/worked-molarity-problem.png\" width=\"510\" height=\"534\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/worked-molarity-problem.png 974w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/worked-molarity-problem-286x300.png 286w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/worked-molarity-problem-768x804.png 768w\" sizes=\"(max-width: 510px) 100vw, 510px\" \/><\/a><\/p>\n<div><\/div>\n<h4 class=\"MathJax_Display\"><span style=\"color: #000000\"><strong>Preparation of Solutions<\/strong><\/span><\/h4>\n<\/div>\n<div>Note that in the example above, we still don&#8217;t have enough information to actually make the solution in the laboratory. There is no piece of equipment that can measure out the moles of a substance.\u00a0 For this, we need to convert the number of moles of the sample into the number of grams represented by that number. We can then easily use a balance to weigh the amount of substance needed for the solution. For the example above:<\/div>\n<div><\/div>\n<div><a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/mol-to-gram-conversion-1.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-3802\" alt=\"\" src=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/mol-to-gram-conversion-1-1024x769.png\" width=\"558\" height=\"419\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/mol-to-gram-conversion-1-1024x769.png 1024w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/mol-to-gram-conversion-1-300x225.png 300w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/mol-to-gram-conversion-1-768x577.png 768w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/mol-to-gram-conversion-1.png 1146w\" sizes=\"(max-width: 558px) 100vw, 558px\" \/><\/a><\/div>\n<div>To actually make the solution, it is typical to dissolve the solute in a small amount of the solvent and then once the solute is dissolved, the final volume can be brought up to 2.50 L.\u00a0 If you were to add 10 g of NaOH directly to 2.50 L, the final volume would be larger than 2.50 L and the solution concentration would be less than 0.100 M. Remember that the final volume must include both the solute and the solvent.<\/div>\n<div class=\"section\" id=\"averill_1.0-ch04_s02_s02\">\n<p class=\"para editable block\" id=\"averill_1.0-ch04_s02_s02_p01\"><span style=\"color: #000000\">Figure 7.8 illustrates the procedure for making a solution of cobalt(II) chloride dihydrate in ethanol. Note that the volume of the <em class=\"emphasis\">solvent<\/em> is not specified. Since the solute occupies space in the solution, the volume of the solvent needed is <em class=\"emphasis\">less<\/em> than the desired total volume of solution.<\/span><\/p>\n<div class=\"figure large editable block\" id=\"averill_1.0-ch04_s02_s02_f01\">\n<p class=\"title\"><span class=\"title-prefix\"><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/2012books.lardbucket.org\/books\/principles-of-general-chemistry-v1.0\/section_08\/421a1cdc4a96997bd44d842d079b4e2e.jpg\" class=\"\" width=\"890\" height=\"354\" \/><\/p>\n<\/div>\n<div class=\"figure large editable block\" id=\"averill_1.0-ch04_s02_s02_f02\">\n<p class=\"title\"><span class=\"title-prefix\"><span style=\"color: #000000\"><strong>Figure 7.8: Preparation of a Solution of Known Concentration Using a Solid Solute.<\/strong> To make a solution, start by addition a portion of the solvent to the flask.\u00a0 Next, weigh out the appropriate amount of solute and slowly add it to the solvent.\u00a0 Once it is dissolved in the solvent, the volume of the solution can be brought up to the final solution volume. For the volumetric flask shown, this is indicated by the black line in the neck of the flask. In this case, it indicates 500 mL of solution. Volumetric flasks exist in many different sizes to accommodate different solution volumes. Graduated cylinders can also be used to accurately bring a solution to its final volume. Other glassware, including beakers and Erlenmeyer flasks are not accurate enough to make most solutions.<\/span> \u00a0 <\/span><\/p>\n<hr \/>\n<\/div>\n<div class=\"exercises block\" id=\"averill_1.0-ch04_s02_s02_n01\">\n<h4 class=\"title\"><em><strong><span style=\"color: #008000\">Example Molarity Calculation<br \/>\n<\/span><\/strong><\/em><\/h4>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s02_p02\"><span style=\"color: #000000\">The solution in Figure 7.8 contains 10.0 g of cobalt(II) chloride dihydrate, CoCl<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>\u00b72H<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>O, in enough ethanol to make exactly 500 mL of solution. What is the molar concentration of CoCl<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>\u00b72H<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>O?<\/span><\/p>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s02_p03\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">Given: <\/strong>mass of solute and volume of solution<\/span><\/p>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s02_p04\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">Asked for: <\/strong>concentration (M)<\/span><\/p>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s02_p05\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">Strategy:<\/strong><\/span><\/p>\n<p><span style=\"color: #000000\">1. We know that Molarity equals moles\/Liter<\/span><\/p>\n<p><span class=\"title-prefix\"><a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Molarity-defined.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Molarity-defined.png\" alt=\"\" class=\"alignnone wp-image-3812\" width=\"192\" height=\"109\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Molarity-defined.png 738w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Molarity-defined-300x170.png 300w\" sizes=\"(max-width: 192px) 100vw, 192px\" \/><\/a><\/span><\/p>\n<p><span style=\"color: #000000\">2. To calculate Molarity, we need to express:<\/span><\/p>\n<ul>\n<li><span style=\"color: #000000\">the mass in the form of moles<\/span><\/li>\n<li><span style=\"color: #000000\">the volume in the form of Liters<\/span><\/li>\n<li><span style=\"color: #000000\">Plug both into the equation above and calculate<\/span><\/li>\n<\/ul>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s02_p07\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">Solution:<\/strong><\/span><\/p>\n<ol>\n<li class=\"para\" id=\"averill_1.0-ch04_s02_s02_p08\"><span style=\"color: #000000\">Converting the mass into moles. We can use the molar mass to convert the grams of CoCl<sub class=\"subscript\">2<\/sub>\u00b72H<sub class=\"subscript\">2<\/sub>O to moles.<\/span><\/li>\n<\/ol>\n<ul>\n<li><span style=\"color: #000000\">The molar mass of CoCl<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>\u00b72H<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>O is 165.87 g\/mol (and includes the two water molecules as they are part of the crystal lattice structure of this solid hydrate!)<\/span><\/li>\n<\/ul>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/calculation-of-moles-ii.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/calculation-of-moles-ii-1024x188.png\" alt=\"\" class=\"alignnone wp-image-3816\" width=\"697\" height=\"128\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/calculation-of-moles-ii-1024x188.png 1024w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/calculation-of-moles-ii-300x55.png 300w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/calculation-of-moles-ii-768x141.png 768w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/calculation-of-moles-ii.png 1549w\" sizes=\"(max-width: 697px) 100vw, 697px\" \/><\/a><\/p>\n<p><span style=\"color: #000000\">\u00a0\u00a0 2. Convert the volume into Liters<\/span><\/p>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/conversion-to-liters.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/conversion-to-liters-1024x242.png\" alt=\"\" class=\"alignnone wp-image-3814\" width=\"554\" height=\"131\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/conversion-to-liters-1024x242.png 1024w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/conversion-to-liters-300x71.png 300w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/conversion-to-liters-768x182.png 768w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/conversion-to-liters.png 1214w\" sizes=\"(max-width: 554px) 100vw, 554px\" \/><\/a><\/p>\n<div>\u00a0\u00a0\u00a0 3. Plug values into the Molarity equation:<\/div>\n<div><\/div>\n<div><a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/final-molarity-calculation.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/final-molarity-calculation-1024x230.png\" alt=\"\" class=\"alignnone wp-image-3817\" width=\"704\" height=\"158\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/final-molarity-calculation-1024x230.png 1024w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/final-molarity-calculation-300x67.png 300w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/final-molarity-calculation-768x172.png 768w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/final-molarity-calculation.png 1440w\" sizes=\"(max-width: 704px) 100vw, 704px\" \/><\/a><\/div>\n<\/div>\n<\/div>\n<div class=\"MathJax_Display\"><span class=\"MathJax\" id=\"MathJax-Element-8-Frame\" role=\"presentation\"><span class=\"math\" id=\"MathJax-Span-181\"><span><span class=\"mrow\" id=\"MathJax-Span-182\"><span class=\"semantics\" id=\"MathJax-Span-183\"><span class=\"mrow\" id=\"MathJax-Span-184\"><span class=\"mtext\" id=\"MathJax-Span-206\"><\/span><\/span><\/span><\/span><\/span><span><\/span><\/span><\/span><\/div>\n<div class=\"exercises block\" id=\"averill_1.0-ch04_s02_s02_n01\">\n<p><span class=\"informalequation\"> <\/span><\/p>\n<hr \/>\n<\/div>\n<h4 id=\"partspermillion\"><span style=\"color: #000000\"><strong>7.7.2 Parts Per Solutions<\/strong><\/span><\/h4>\n<p><span style=\"color: #000000\">In the consumer and industrial world, the most common method of expressing the concentration is based on the quantity of solute in a fixed quantity of solution. The \u201cquantities\u201d referred to here can be expressed in mass, in volume, or both (i.e., the <i>mass<\/i> of solute in a given <i>volume<\/i> of solution.) In order to distinguish among these possibilities, the abbreviations (m\/m), (v\/v) and (m\/v) are used.<\/span><\/p>\n<p><span style=\"color: #000000\">In most applied fields of Chemistry, (m\/m) measure is often used, whereas in clinical chemistry, (m\/v) is commonly used, with <strong><em>mass expressed in grams<\/em><\/strong> and <strong><em>volume in mL.<\/em><\/strong><\/span><\/p>\n<p><span style=\"color: #000000\">One of the more common ways to express such concentrations as &#8220;<strong>parts per\u00a0100<\/strong>&#8220;, which we all know as &#8220;<strong>percent<\/strong>&#8220;.\u00a0 &#8220;<i>Cent<\/i>&#8221; is the Latin-derived prefix relating to the number 100<\/span><br \/>\n<span style=\"color: #000000\"> (L. <i>centum<\/i>), as in <i>century<\/i> or <i>centennial<\/i>. It also denotes 1\/100th (from L. <i>centesimus<\/i>) as in <i>centimeter<\/i> and the monetary unit <i>cent<\/i>. <\/span><span style=\"color: #000000\">Percent solutions define the quantity of a solute that is dissolved in a quantity of solution multiplied by 100.\u00a0 Percent solutions can be expressed in terms of mass solute per mass solution (m\/m%), or mass solute per volume of solution (m\/v%), or volume of solute per volume of solution (v\/v%). When making a percent solution, it is important to indicate what units are being used, so that others can also make the solution properly. Also, recall that the solution is the sum of both the solvent and the solute when you are performing percent calculations.<\/span><\/p>\n<p><span style=\"color: #000000\">Solution = Solute + Solvent<\/span><\/p>\n<p><span style=\"color: #000000\">Thus, the following equation can be used when calculating percent solutions:<\/span><\/p>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2018\/02\/percent-solution.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/wou.edu\/chemistry\/files\/2018\/02\/percent-solution.png\" alt=\"\" class=\"alignnone wp-image-3841\" width=\"573\" height=\"229\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2018\/02\/percent-solution.png 971w, https:\/\/wou.edu\/chemistry\/files\/2018\/02\/percent-solution-300x120.png 300w, https:\/\/wou.edu\/chemistry\/files\/2018\/02\/percent-solution-768x307.png 768w\" sizes=\"(max-width: 573px) 100vw, 573px\" \/><\/a><\/p>\n<h4><span style=\"color: #008000\"><strong><em>Example 1:<\/em> <\/strong><\/span><\/h4>\n<p><span style=\"color: #000000\">As an example, a 7.0% v\/v solution of ethanol in water, would contain 7 mL of ethanol in a total of 100 mL of solution. How much water is in the solution?<\/span><\/p>\n<p><span style=\"color: #000000\">In this problem, we know that the:<\/span><\/p>\n<p><strong><span style=\"color: #000000\">Solution = Solute + Solvent<\/span><\/strong><\/p>\n<p><span style=\"color: #000000\">Thus, we can fill in the values and then solve for the unknown.<\/span><\/p>\n<p><strong><span style=\"color: #000000\">100 mL = 7 mL + X mL of Solvent (in this case water)<\/span><\/strong><\/p>\n<p><span style=\"color: #000000\">shifting the 7 over to the other side, we can see that:<\/span><\/p>\n<p><strong><span style=\"color: #000000\">100 mL &#8211; 7 mL\u00a0 = 93 mL H<sub>2<\/sub>O<\/span><\/strong><\/p>\n<h4><span style=\"color: #000000\"><span style=\"color: #008000\"><strong><em>Example 2<\/em> <\/strong><\/span><br \/>\n<\/span><\/h4>\n<p><span style=\"color: #000000\">What is the (m\/v)% of a solution if 24.0 g of sucrose is dissolved in a total solution of 243 mL?<\/span><\/p>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2018\/02\/percent-example-2.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/wou.edu\/chemistry\/files\/2018\/02\/percent-example-2.png\" alt=\"\" class=\"alignnone wp-image-3842\" width=\"430\" height=\"395\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2018\/02\/percent-example-2.png 827w, https:\/\/wou.edu\/chemistry\/files\/2018\/02\/percent-example-2-300x276.png 300w, https:\/\/wou.edu\/chemistry\/files\/2018\/02\/percent-example-2-768x706.png 768w\" sizes=\"(max-width: 430px) 100vw, 430px\" \/><\/a><\/p>\n<h4><span style=\"color: #008000\"><strong><em>Example 3<\/em> <\/strong><\/span><\/h4>\n<p><span style=\"color: #000000\">How many grams of NaCl are required to make 625 mL of a 13.5% solution?<\/span><\/p>\n<p><span style=\"color: #000000\"><a href=\"https:\/\/wou.edu\/chemistry\/files\/2018\/02\/example-3.png\" style=\"color: #000000\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/wou.edu\/chemistry\/files\/2018\/02\/example-3.png\" alt=\"\" class=\"alignnone wp-image-3843\" width=\"430\" height=\"459\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2018\/02\/example-3.png 897w, https:\/\/wou.edu\/chemistry\/files\/2018\/02\/example-3-281x300.png 281w, https:\/\/wou.edu\/chemistry\/files\/2018\/02\/example-3-768x819.png 768w\" sizes=\"(max-width: 430px) 100vw, 430px\" \/><\/a><\/span><\/p>\n<hr \/>\n<p><span style=\"color: #000000\">For more dilute solutions, parts per million (10<sup>6<\/sup> ppm) and parts per billion (10<sup>9<\/sup>; ppb) are used. These terms are widely employed to express the amounts of trace pollutants in the environment.<\/span><\/p>\n<p id=\"fs-idm39037584\"><span style=\"color: #000000\">Like percentage (\u201cpart per hundred\u201d) units, ppm and ppb may be defined in terms of masses, volumes, or mixed mass-volume units. There are also ppm and ppb units defined with respect to numbers of atoms and molecules.<\/span><\/p>\n<p id=\"fs-idm18786352\"><span style=\"color: #000000\">The mass-based definitions of ppm and ppb are given here:<\/span><\/p>\n<div class=\"equation\" id=\"fs-idp17696176\"><a href=\"https:\/\/wou.edu\/chemistry\/files\/2018\/02\/ppm-ppb.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/wou.edu\/chemistry\/files\/2018\/02\/ppm-ppb.png\" alt=\"\" class=\"alignnone wp-image-3845\" width=\"540\" height=\"281\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2018\/02\/ppm-ppb.png 672w, https:\/\/wou.edu\/chemistry\/files\/2018\/02\/ppm-ppb-300x156.png 300w\" sizes=\"(max-width: 540px) 100vw, 540px\" \/><\/a><\/div>\n<p id=\"fs-idm37568416\"><span style=\"color: #000000\">Both ppm and ppb are convenient units for reporting the concentrations of pollutants and other trace contaminants in water. Concentrations of these contaminants are typically very low in treated and natural waters, and their levels cannot exceed relatively low concentration thresholds without causing adverse effects on health and wildlife. For example, the EPA has identified the maximum safe level of fluoride ion in tap water to be 4 ppm. Inline water filters are designed to reduce the concentration of fluoride and several other trace-level contaminants in tap water (Figure 7.9).\u00a0<\/span><\/p>\n<figure id=\"CNX_Chem_03_05_faucet\"><figcaption>\n<div class=\"wp-caption alignleft\" style=\"width: 699px\">\n<p><a href=\"https:\/\/opentextbc.ca\/chemistry\/wp-content\/uploads\/sites\/150\/2016\/05\/CNX_Chem_03_05_faucet.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/opentextbc.ca\/chemistry\/wp-content\/uploads\/sites\/150\/2016\/05\/CNX_Chem_03_05_faucet.jpg\" alt=\"Two pictures are shown labeled a and b. Picture a is a close-up shot of water coming out of a faucet. Picture b shows a machine with the words, \u201cFiltered Water Dispenser.\u201d This machine appears to be inside a refrigerator.\" class=\"\" width=\"699\" height=\"355\" \/><\/a><\/p>\n<p class=\"wp-caption-text\"><span style=\"color: #000000\"><strong>Figure 7.9.<\/strong> (a) In some areas, trace-level concentrations of contaminants can render unfiltered tap water unsafe for drinking and cooking. (b) Inline water filters reduce the concentration of solutes in tap water. (credit a: modification of work by Jenn Durfey; credit b: modification of work by \u201cvastateparkstaff\u201d\/Wikimedia commons<\/span><\/p>\n<\/div>\n<\/figcaption><\/figure>\n<hr \/>\n<hr \/>\n<h4><\/h4>\n<p class=\"para editable block\" id=\"averill_1.0-ch04_s02_s02_p13\"><span style=\"color: #000000\"><span style=\"color: #000000\">When reporting contaminants like lead in drinking water, ppm and ppb concentrations are often reported in mixed unit values of mass\/volume.\u00a0 This can be very useful as it is easier for us to think about water in terms of its volume, rather than by its mass. In addition,\u00a0 the density of water is 1.0 g\/mL or 1.0 mg\/0.001 mL\u00a0 which makes the conversion between the two units easier. For example, if we find that there is lead contamination in water of 4 ppm, this would mean that\u00a0 there are:<\/span><strong><span style=\"color: #000000\"><br \/>\n<\/span><\/strong><\/span><\/p>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2018\/02\/ppm-example.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/wou.edu\/chemistry\/files\/2018\/02\/ppm-example-869x1024.png\" alt=\"\" class=\"alignnone wp-image-3846\" width=\"600\" height=\"707\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2018\/02\/ppm-example-869x1024.png 869w, https:\/\/wou.edu\/chemistry\/files\/2018\/02\/ppm-example-255x300.png 255w, https:\/\/wou.edu\/chemistry\/files\/2018\/02\/ppm-example-768x905.png 768w, https:\/\/wou.edu\/chemistry\/files\/2018\/02\/ppm-example.png 916w\" sizes=\"(max-width: 600px) 100vw, 600px\" \/><\/a><\/p>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2018\/02\/ppm-and-ppb-m-to-vol.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/wou.edu\/chemistry\/files\/2018\/02\/ppm-and-ppb-m-to-vol.png\" alt=\"\" class=\"alignnone wp-image-3849\" width=\"701\" height=\"305\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2018\/02\/ppm-and-ppb-m-to-vol.png 885w, https:\/\/wou.edu\/chemistry\/files\/2018\/02\/ppm-and-ppb-m-to-vol-300x131.png 300w, https:\/\/wou.edu\/chemistry\/files\/2018\/02\/ppm-and-ppb-m-to-vol-768x334.png 768w\" sizes=\"(max-width: 701px) 100vw, 701px\" \/><\/a><\/p>\n<h4 class=\"editable\"><strong><span style=\"color: #000000\">7.74 Equivalents<\/span><\/strong><\/h4>\n<p class=\"Love\" id=\"gob-ch09_s02_s05_p01\"><span style=\"color: #000000\">Concentrations of ionic solutes are occasionally expressed in units called equivalents (Eq). One equivalent equals 1 mol of positive or negative charge. Thus, 1 mol\/L of Na<sup class=\"superscript\">+<\/sup>(aq) is also 1 Eq\/L because sodium has a 1+ charge. A 1 mol\/L solution of Ca<sup class=\"superscript\">2<\/sup><sup class=\"superscript\">+<\/sup>(aq) ions has a concentration of 2 Eq\/L because calcium has a 2+ charge. Dilute solutions may be expressed in milliequivalents (mEq)\u2014for example, human blood plasma has a total concentration of about 150 mEq\/L.<\/span><\/p>\n<p><span style=\"color: #000000\">In a more formal definition, the <strong><i>equivalent<\/i><\/strong> is the amount of a substance needed to do one of the following:<\/span><\/p>\n<ul>\n<li><span style=\"color: #000000\">react with or supply one mole of hydrogen ions (H<sup>+<\/sup>) in an acid\u2013base reaction<\/span><\/li>\n<li><span style=\"color: #000000\">react with or supply one mole of electrons in a redox reaction.<\/span><\/li>\n<\/ul>\n<p><span style=\"color: #000000\">By this definition, an <i>equivalent<\/i> is the number of moles of an ion in a solution, multiplied by the valence of that ion. If 1\u00a0mol of NaCl and 1\u00a0mol of CaCl<sub>2<\/sub> dissolve in a solution, there is 1\u00a0equiv Na, 2 \u00a0equiv Ca, and 3\u00a0equiv Cl in that solution. (The valence of calcium is 2, so for that ion you have 1 mole and 2 equivalents.)<\/span><\/p>\n<h4><a href=\"#title\"><span><em><strong>(Back to the Top)<\/strong><\/em><\/span><\/a><\/h4>\n<hr \/>\n<h3 id=\"dilutions\" class=\"para editable block\"><span style=\"color: #000000\"><strong><span style=\"color: #000000\"><span style=\"color: #ff0000\">7.8 Dilutions<\/span><\/span><\/strong><\/span><\/h3>\n<div class=\"exercises block\" id=\"averill_1.0-ch04_s02_s02_n02\">\n<p class=\"para editable block\" id=\"averill_1.0-ch04_s02_s02_p25\"><span style=\"color: #000000\">A solution of a desired concentration can also be prepared by diluting a small volume of a more concentrated solution with additional solvent. A <span class=\"margin_term\"><a class=\"glossterm\" style=\"color: #000000\">stock solution<\/a><\/span>, which is a prepared solution of known concentration, is often used for this purpose. Diluting a stock solution is preferred when making solutions of very weak concentrations,\u00a0 because the alternative method, weighing out tiny amounts of solute, can be difficult to carry out with a high degree of accuracy. Dilution is also used to prepare solutions from substances that are sold as concentrated aqueous solutions, such as strong acids.<\/span><\/p>\n<\/div>\n<p class=\"para editable block\" id=\"averill_1.0-ch04_s02_s02_p26\"><span style=\"color: #000000\">The procedure for preparing a solution of known concentration from a stock solution is shown in Figure 7.10. It requires calculating the amount of solute desired in the final volume of the more dilute solution and then calculating the volume of the stock solution that contains this amount of solute. Remember that diluting a given quantity of stock solution with solvent does <em class=\"emphasis\">not<\/em> change the amount of solute present, only the volume of the solution is changing. The relationship between the volume and concentration of the stock solution and the volume and concentration of the desired diluted solution can therefore be expressed mathematically as:<br \/>\n<\/span><\/p>\n<div class=\"equation block\" id=\"averill_1.0-ch04_s02_s02_eq01\">\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Dilution-equation-match.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Dilution-equation-match.png\" alt=\"\" class=\"alignnone wp-image-3821\" width=\"439\" height=\"87\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Dilution-equation-match.png 620w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Dilution-equation-match-300x60.png 300w\" sizes=\"(max-width: 439px) 100vw, 439px\" \/><\/a><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #000000\">Where M<sub>s<\/sub> is the concentration of the stock solution, V<sub>s<\/sub> is the volume of the stock solution, M<sub>d<\/sub> is the concentration of the diluted solution, and V<sub>d<\/sub> is the volume of the diluted solution.<\/span><\/p>\n<\/div>\n<div class=\"figure large editable block\" id=\"averill_1.0-ch04_s02_s02_f03\">\n<p class=\"title\"><span class=\"title-prefix\"><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/2012books.lardbucket.org\/books\/principles-of-general-chemistry-v1.0\/section_08\/d30bd057f60c6bf73ed4f18305fa177c.jpg\" class=\"\" width=\"701\" height=\"345\" \/><\/p>\n<p class=\"para\"><span style=\"color: #000000\"><strong><span class=\"title-prefix\">Figure 7.10<\/span> Preparation of a Solution of Known Concentration by Diluting a Stock Solution.<\/strong> (a) A volume (<em class=\"emphasis\">V<\/em><sub class=\"subscript\"><span style=\"font-size: small\">s<\/span><\/sub>) containing the desired amount of solute (M<sub class=\"subscript\"><span style=\"font-size: small\">s<\/span><\/sub>) is measured from a stock solution of known concentration. (b) The measured volume of stock solution is transferred to a second volumetric flask. (c) The measured volume in the second flask is then diluted with solvent up to the volumetric mark [(<em class=\"emphasis\">V<\/em><sub class=\"subscript\"><span style=\"font-size: small\">s<\/span><\/sub>)(M<sub class=\"subscript\"><span style=\"font-size: small\">s<\/span><\/sub>) = (<em class=\"emphasis\">V<\/em><sub class=\"subscript\"><span style=\"font-size: small\">d<\/span><\/sub>)(M<sub class=\"subscript\"><span style=\"font-size: small\">d<\/span><\/sub>)].<\/span><\/p>\n<hr \/>\n<\/div>\n<div class=\"exercises block\" id=\"averill_1.0-ch04_s02_s02_n03\">\n<h4 class=\"title\"><span style=\"color: #008000\"><em><strong>Example of Dilution Calculations<\/strong><\/em><\/span><\/h4>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s02_p28\"><span style=\"color: #000000\">What volume of a 3.00 M glucose stock solution is necessary to prepare 2500 mL of 0.400 M solution?<\/span><\/p>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s02_p29\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">Given: <\/strong>volume and molarity of dilute solution, and molarity of stock solution<\/span><\/p>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s02_p30\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">Asked for: <\/strong>volume of stock solution<\/span><\/p>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s02_p31\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">Strategy and Solution:<\/strong><\/span><\/p>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s02_p32\"><span style=\"color: #000000\">For Dilution problems, as long as you know 3 of the variables, you can solve for the 4th variable.<\/span><\/p>\n<ol>\n<li><span style=\"color: #000000\">Start by rearranging the equation to solve for the variable that you want to find. In this case, you want to find the volume of the stock solution, V<sub>s<\/sub><\/span><\/li>\n<\/ol>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/dilution-rearranged.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/dilution-rearranged.png\" alt=\"\" class=\"alignnone wp-image-3824\" width=\"521\" height=\"133\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/dilution-rearranged.png 819w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/dilution-rearranged-300x77.png 300w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/dilution-rearranged-768x196.png 768w\" sizes=\"(max-width: 521px) 100vw, 521px\" \/><\/a><\/p>\n<p><span style=\"color: #000000\">2. Next, check to make sure that like terms have the same units.\u00a0 For example, Md and Ms are both concentrations, thus, to be able to perform the calculations, they should be in the same unit (in this case they are both listed in Molarity). If the concentrations were different, say one was given in Molarity and the other in percent\u00a0 or one was in Molarity and the other was in Millimolarity, one of the terms would need to be converted so that they match.\u00a0 That way, the units will cancel out and leave you with units of volume, in this case.<\/span><\/p>\n<\/div>\n<p><span style=\"color: #000000\">\u00a0\u00a0 3. Finally, fill in the equation with known values and calculate the final answer.<\/span><\/p>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Dilution-answer.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Dilution-answer-1024x215.png\" alt=\"\" class=\"alignnone wp-image-3825\" width=\"695\" height=\"146\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Dilution-answer-1024x215.png 1024w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Dilution-answer-300x63.png 300w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Dilution-answer-768x161.png 768w, https:\/\/wou.edu\/chemistry\/files\/2017\/05\/Dilution-answer.png 1465w\" sizes=\"(max-width: 695px) 100vw, 695px\" \/><\/a><\/p>\n<div class=\"exercises block\" id=\"averill_1.0-ch04_s02_s02_n03\">\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s02_p33\"><strong class=\"emphasis bold\"><span style=\"color: #000000\">Note that if 333mL of stock solution is needed, that you can also calculate the amount of solvent needed to make the final dilution. (Total volume &#8211; volume of stock solution = volume of solvent needed for the final dilution. In this case 2,500 mL &#8211; 333 mL = 2,167 mL of water needed to make the final dilution (this should be done in a graduated cylinder or volumetric flask).<\/span><br \/>\n<\/strong><\/p>\n<h4><span><em><strong>(Back to the Top)<\/strong><\/em><\/span><\/h4>\n<hr \/>\n<\/div>\n<div class=\"section\" id=\"averill_1.0-ch04_s02_s02\">\n<div class=\"exercises block\" id=\"averill_1.0-ch04_s02_s02_n03\">\n<h3 id=\"ionconcentration\" class=\"para\"><span style=\"color: #ff0000\"><strong>7.9 Ion Concentrations in Solution<\/strong><\/span><\/h3>\n<\/div>\n<\/div>\n<div class=\"section\" id=\"averill_1.0-ch04_s02_s03\">\n<p class=\"para editable block\" id=\"averill_1.0-ch04_s02_s03_p01\"><span style=\"color: #000000\">Thus far, we have been discussing the concentration of the overall solution in terms of total solute divided by the volume of the solution. Let\u2019s consider in more detail exactly what that means when considering ionic and covalent compounds.\u00a0 When ionic compounds dissolve in a solution, they break apart into their ionic state. Cations and anions associate with the polar water molecules. Recall that solutions that contain ions are called<em><strong> electrolyotes<\/strong><\/em>, due to their ability to conduct electricity.\u00a0 For example, ammonium dichromate (NH<sub>4<\/sub>)<sub>2<\/sub>Cr<sub>2<\/sub>O<sub>7<\/sub> is an ionic compound that contains two NH<span style=\"font-size: small\"><sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup><\/span> ions and one Cr<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>O<span style=\"font-size: small\"><sub class=\"subscript\">7<\/sub><sup class=\"superscript\">2\u2212<\/sup><\/span> ion per formula unit. Like other ionic compounds, it is a strong electrolyte that dissociates in aqueous solution to give hydrated NH<span style=\"font-size: small\"><sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup><\/span> and Cr<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>O<span style=\"font-size: small\"><sub class=\"subscript\">7<\/sub><sup class=\"superscript\">2\u2212<\/sup><\/span> ions.\u00a0 If we consider this this solution mathematically, we can see that for every ammonium dichromate molecule that dissolves, there will be three resulting ions that form (the two <span>NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup><\/span> ions and the one <span>Cr<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">7<\/sub><sup class=\"superscript\">2\u2212<\/sup>\u00a0 ion).\u00a0\u00a0 This can also be thought of on a larger molar scale.\u00a0 When 1 mole of (NH<sub>4<\/sub>)<sub>2<\/sub>Cr<sub>2<\/sub>O<sub>7<\/sub> is dissolved, it\u00a0 results in 3 moles of ions (1 mol of Cr<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">7<\/sub><sup class=\"superscript\">2\u2212<\/sup> anions and 2 mol of NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup> cations) within the solution (Figure 7.11).\u00a0 To discuss the relationship between the concentration of a solution and the resulting number of ions, the term <em><strong>equivalents<\/strong><\/em> is used.<\/span><br \/>\n<\/span><\/p>\n<div class=\"equation block\" id=\"averill_1.0-ch04_s02_s03_eq01\">\n<p class=\"MathJax_Display\"><span class=\"MathJax\" id=\"MathJax-Element-14-Frame\" role=\"presentation\"><span class=\"math\" id=\"MathJax-Span-408\"><span><span class=\"mrow\" id=\"MathJax-Span-409\"><span class=\"semantics\" id=\"MathJax-Span-410\"><span class=\"mrow\" id=\"MathJax-Span-411\"><span class=\"mtext\" id=\"MathJax-Span-457\"><span style=\"color: #000000\">One equivalent is defined as the amount of an ionic compound that provides 1 mole of electrical charge (+ or -).\u00a0 It is calculated by dividing the molarity of the solution by the total charge created in the solution.<\/span> <\/span><\/span><\/span><\/span><\/span><span><\/span><\/span><\/span><\/p>\n<\/div>\n<\/div>\n<div class=\"equation block\" id=\"averill_1.0-ch04_s02_s03_eq01\"><\/div>\n<div class=\"figure large editable block\" id=\"averill_1.0-ch04_s02_s03_f01\">\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/2012books.lardbucket.org\/books\/principles-of-general-chemistry-v1.0\/section_08\/11367b0b19bd6682f637472684716c36.jpg\" class=\"\" width=\"789\" height=\"308\" \/><\/p>\n<p><span style=\"color: #000000\"><strong>Figure 7.11 Dissolution of 1 mol of an Ioncic Compound.<\/strong> Dissoliving 1 mol of ammonium dichromate formula units in water produces 1 mol of Cr<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">7<\/sub><sup class=\"superscript\">2\u2212<\/sup> anions and 2 mol of NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup> cations. (Water molecules are omitted from a molecular view of the solution for clarity.)<\/span><\/p>\n<hr \/>\n<\/div>\n<p class=\"para editable block\" id=\"averill_1.0-ch04_s02_s03_p03\"><span style=\"color: #000000\">When we carry out a chemical reaction using a solution of a salt such as ammonium dichromate, we need to know the concentration of each ion present in the solution. If a solution contains 1.43 M (NH<sub class=\"subscript\"><span style=\"font-size: small\">4<\/span><\/sub>)<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>Cr<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>O<sub class=\"subscript\"><span style=\"font-size: small\">7<\/span><\/sub>, then the concentration of Cr<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>O<span style=\"font-size: small\"><sub class=\"subscript\">7<\/sub><sup class=\"superscript\">2\u2212<\/sup><\/span> must also be 1.43 M because there is one Cr<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>O<span style=\"font-size: small\"><sub class=\"subscript\">7<\/sub><sup class=\"superscript\">2\u2212<\/sup><\/span> ion per formula unit. However, there are two NH<span style=\"font-size: small\"><sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup><\/span> ions per formula unit, so the concentration of NH<span style=\"font-size: small\"><sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup><\/span> ions is 2\u00a0\u00d7\u00a01.43 M = 2.86 M. Because each formula unit of (NH<sub class=\"subscript\"><span style=\"font-size: small\">4<\/span><\/sub>)<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>Cr<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>O<sub class=\"subscript\"><span style=\"font-size: small\">7<\/span><\/sub> produces <em class=\"emphasis\">three<\/em> ions when dissolved in water (2NH<span style=\"font-size: small\"><sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup><\/span>\u00a0+\u00a01Cr<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>O<span style=\"font-size: small\"><sub class=\"subscript\">7<\/sub><sup class=\"superscript\">2\u2212<\/sup><\/span>), the <em class=\"emphasis\">total<\/em> concentration of ions in the solution is 3\u00a0\u00d7\u00a01.43 M = 4.29 M. The equivalent value of <span>(NH<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub>Cr<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">7<\/sub><\/span> can then be calculated by dividing 1.43 M by 4.29 M, yielding 0.333 equivalents. Thus, for <span>(NH<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub>Cr<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">7<\/sub><\/span>, dissolving 0.333 moles of the compound will yield 1 mole of ions in the solution.<br \/>\n<\/span><\/p>\n<div class=\"exercises block\" id=\"averill_1.0-ch04_s02_s03_n01\">\n<h4 class=\"title\"><strong><em><span style=\"color: #008000\">Example 1<\/span><\/em><\/strong><\/h4>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s03_p04\"><span style=\"color: #000000\">What are the concentrations of all ionic species derived from the solutes in these aqueous solutions?<\/span><\/p>\n<ol class=\"orderedlist\" id=\"averill_1.0-ch04_s02_s03_l01\">\n<li><span style=\"color: #000000\">0.21 M NaOH<\/span><\/li>\n<li><span style=\"color: #000000\">3.7 M (CH<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub>)CHOH<\/span><\/li>\n<li><span style=\"color: #000000\">0.032 M In(NO<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub>)<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub><\/span><\/li>\n<\/ol>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s03_p05\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">Given: <\/strong>molarity<\/span><\/p>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s03_p06\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">Asked for: <\/strong>concentrations<\/span><\/p>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s03_p07\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">Strategy:<\/strong><\/span><span><\/span><\/p>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s03_p08\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">A<\/strong> Classify each compound as either a strong electrolyte or a nonelectrolyte.<\/span><\/p>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s03_p09\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">B<\/strong> If the compound is a nonelectrolyte, its concentration is the same as the molarity of the solution. If the compound is a strong electrolyte, determine the number of each ion contained in one formula unit. Find the concentration of each species by multiplying the number of each ion by the molarity of the solution.<\/span><\/p>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s03_p10\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">Solution:<\/strong><\/span><\/p>\n<p class=\"para\"><strong><span style=\"color: #000000\">1.\u00a0\u00a0 0.21 M NaOH<\/span><\/strong><\/p>\n<p class=\"para\"><span style=\"color: #000000\"><strong>A<\/strong> Sodium hydroxide is an ionic compound that is a strong electrolyte (and a strong base) in aqueous solution:<\/span><\/p>\n<p class=\"para\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">B<\/strong> Because each formula unit of NaOH produces one Na<sup class=\"superscript\"><span style=\"font-size: small\">+<\/span><\/sup> ion and one OH<sup class=\"superscript\"><span style=\"font-size: small\">\u2212<\/span><\/sup> ion, the concentration of each ion is the same as the concentration of NaOH: [Na<sup class=\"superscript\"><span style=\"font-size: small\">+<\/span><\/sup>] = 0.21 M and [OH<sup class=\"superscript\"><span style=\"font-size: small\">\u2212<\/span><\/sup>] = 0.21 <\/span><\/p>\n<p class=\"para\"><span style=\"color: #000000\"><strong>2.\u00a0\u00a0\u00a0 <\/strong><strong>3.7 M (CH<sub class=\"subscript\">3<\/sub>)CHOH<\/strong><\/span><\/p>\n<p class=\"para\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">A<\/strong> The formula (CH<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub>)<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>CHOH represents 2-propanol (isopropyl alcohol) and contains the \u2013OH group, so it is an alcohol. Recall from <a class=\"xref\" href=\"https:\/\/2012books.lardbucket.org\/books\/principles-of-general-chemistry-v1.0\/s08-reactions-in-aqueous-solution.html#averill_1.0-ch04_s01\" style=\"color: #000000\">Section 4.1 &#8220;Aqueous Solutions&#8221;<\/a> that alcohols are covalent compounthat dissolve in water to give solutions of neutral molecules. Thus alcohols are nonelectrolytes<\/span><\/p>\n<p class=\"para\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">B<\/strong> The only solute species in solution is therefore (CH<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub>)<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>CHOH molecules, so [(CH<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub>)<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>CHOH] = 3.7 M<\/span><\/p>\n<p class=\"para\"><span style=\"color: #000000\"><strong>3.\u00a0\u00a0 0.032 M In(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">3<\/sub><\/strong><\/span><\/p>\n<p class=\"para\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">A<\/strong> Indium nitrate is an ionic compound that contains In<sup class=\"superscript\"><span style=\"font-size: small\">3+<\/span><\/sup> ions and NO<span style=\"font-size: small\"><sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup><\/span> ions, so we expect it to behave like a strong electrolyte in aqueous solution<\/span><\/p>\n<p class=\"para\"><span style=\"color: #000000\"><strong class=\"emphasis bold\">B<\/strong> One formula unit of In(NO<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub>)<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub> produces one In<sup class=\"superscript\"><span style=\"font-size: small\">3+<\/span><\/sup> ion and three NO<span style=\"font-size: small\"><sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup><\/span> ions, so a 0.032 M In(NO<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub>)<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub> solution contains 0.032 M In<sup class=\"superscript\"><span style=\"font-size: small\">3+<\/span><\/sup> and 3\u00a0\u00d7\u00a00.032 M = 0.096 M NO<span style=\"font-size: small\"><sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2013<\/sup><\/span>\u2014that is, [In<sup class=\"superscript\"><span style=\"font-size: small\">3+<\/span><\/sup>] = 0.032 M and [NO<span style=\"font-size: small\"><sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup><\/span>] = 0.096 M<\/span><\/p>\n<h4><a href=\"#title\"><span><em><strong>(Back to the Top)<\/strong><\/em><\/span><\/a><\/h4>\n<hr \/>\n<\/div>\n<div class=\"key_takeaways block\" id=\"averill_1.0-ch04_s02_s03_n02\">\n<h3 id=\"7summary\" class=\"para\"><span style=\"color: #ff0000\"><strong>7.10 Summary<\/strong><\/span><\/h3>\n<\/div>\n<div class=\"callout editable block\" id=\"averill_1.0-ch04_s02_s03_n03\">\n<p class=\"para\" id=\"gob-ch09_s05_p01\"><span style=\"color: #000000\"><em class=\"emphasis\">To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter.<\/em><\/span><\/p>\n<p class=\"para\" id=\"gob-ch09_s05_p02\"><span style=\"color: #000000\">A <strong class=\"emphasis bold\">solution<\/strong> is a homogeneous mixture. The major component is the <strong class=\"emphasis bold\">solvent<\/strong>, while the minor component is the <strong class=\"emphasis bold\">solute<\/strong>. Solutions can have any phase; for example, an <strong class=\"emphasis bold\">alloy<\/strong> is a solid solution. Solutes are <strong class=\"emphasis bold\">soluble<\/strong> or <strong class=\"emphasis bold\">insoluble<\/strong>, meaning they dissolve or do not dissolve in a particular solvent. The terms <strong class=\"emphasis bold\">miscible<\/strong> and <strong class=\"emphasis bold\">immiscible<\/strong>, instead of soluble and insoluble, are used for liquid solutes and solvents. The statement <em class=\"emphasis\">like dissolves like<\/em> is a useful guide to predicting whether a solute will dissolve in a given solvent.<\/span><\/p>\n<p><span style=\"color: #000000\">Dissolving occurs by <strong class=\"emphasis bold\">solvation<\/strong>, the process in which particles of a solvent surround the individual particles of a solute, separating them to make a solution. For water solutions, the word <strong class=\"emphasis bold\">hydration<\/strong> is used. If the solute is molecular, it dissolves into individual molecules. If the solute is ionic, the individual ions separate from each other, forming a solution that conducts electricity. Such solutions are called <strong class=\"emphasis bold\">electrolytes<\/strong>. If the dissociation of ions is complete, the solution is a <strong class=\"emphasis bold\">strong electrolyte<\/strong>. If the dissociation is only partial, the solution is a <strong class=\"emphasis bold\">weak electrolyte<\/strong>. Solutions of molecules do not conduct electricity and are called <strong class=\"emphasis bold\">nonelectrolytes<\/strong>.<\/span><\/p>\n<p class=\"para\" id=\"gob-ch09_s05_p03\"><span style=\"color: #000000\">The amount of solute in a solution is represented by the <strong class=\"emphasis bold\">concentration<\/strong> of the solution. The maximum amount of solute that will dissolve in a given amount of solvent is called the <strong class=\"emphasis bold\">solubility<\/strong> of the solute. Such solutions are <strong class=\"emphasis bold\">saturated<\/strong>. Solutions that have less than the maximum amount are <strong class=\"emphasis bold\">unsaturated<\/strong>. Most solutions are unsaturated, and there are various ways of stating their concentrations. <strong class=\"emphasis bold\">Mass\/mass percent<\/strong>, <strong class=\"emphasis bold\">volume\/volume percent<\/strong>, and <strong class=\"emphasis bold\">mass\/volume percent<\/strong> indicate the percentage of the overall solution that is solute. <strong class=\"emphasis bold\">Parts per million (ppm)<\/strong> and <strong class=\"emphasis bold\">parts per billion (ppb)<\/strong> are used to describe very small concentrations of a solute. <strong class=\"emphasis bold\">Molarity<\/strong>, defined as the number of moles of solute per liter of solution, is a common concentration unit in the chemistry laboratory. <strong class=\"emphasis bold\">Equivalents<\/strong> express concentrations in terms of moles of charge on ions. When a solution is diluted, we use the fact that the amount of solute remains constant to be able to determine the volume or concentration of the final diluted solution. Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a <strong class=\"emphasis bold\">stock solution<\/strong>) to the desired final volume.<\/span><\/p>\n<p class=\"para\" id=\"averill_1.0-ch04_s02_s03_p16\"><span style=\"color: #000000\">\u00a0<\/span><\/p>\n<\/div>\n<div class=\"key_takeaways editable block\" id=\"averill_1.0-ch04_s02_s03_n04\">\n<h3 class=\"title\">Key Takeaway<\/h3>\n<ul class=\"itemizedlist\" id=\"averill_1.0-ch04_s02_s03_l05\">\n<li><span style=\"color: #000000\">Solution concentrations are typically expressed as molarity and can be prepared by dissolving a known mass of solute in a solvent or diluting a stock solution.<\/span><\/li>\n<\/ul>\n<\/div>\n<div class=\"qandaset block\" id=\"averill_1.0-ch04_s02_s03_qs01\">\n<h3 class=\"title\">Conceptual Problems<\/h3>\n<ol class=\"qandadiv\" id=\"averill_1.0-ch04_s02_s03_qs01_qd01\">\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs01_qd01_qa01\">\n<div class=\"question\">\n<p class=\"para\"><span style=\"color: #000000\">Which of the representations best corresponds to a 1 M aqueous solution of each compound? Justify your answers.<\/span><\/p>\n<ol class=\"orderedlist\" style=\"list-style-type: lower-alpha\">\n<li><span style=\"color: #000000\">NH<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub><\/span><\/li>\n<li><span style=\"color: #000000\">HF<\/span><\/li>\n<li><span style=\"color: #000000\">CH<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub>CH<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>CH<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>OH<\/span><\/li>\n<li>\n<p class=\"para\"><span style=\"color: #000000\">Na<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>SO<sub class=\"subscript\"><span style=\"font-size: small\">4<\/span><\/sub><\/span><\/p>\n<div class=\"informalfigure medium\"><img decoding=\"async\" src=\"https:\/\/2012books.lardbucket.org\/books\/principles-of-general-chemistry-v1.0\/section_08\/4b8d2b1f623560048a737b8eb7a3ade4.jpg\" \/><\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs01_qd01_qa02\">\n<div class=\"question\">\n<p class=\"para\"><span style=\"color: #000000\">Which of the representations shown in Problem 1 best corresponds to a 1 M aqueous solution of each compound? Justify your answers.<\/span><\/p>\n<ol class=\"orderedlist\" style=\"list-style-type: lower-alpha\">\n<li><span style=\"color: #000000\">CH<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub>CO<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>H<\/span><\/li>\n<li><span style=\"color: #000000\">NaCl<\/span><\/li>\n<li><span style=\"color: #000000\">Na<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>S<\/span><\/li>\n<li><span style=\"color: #000000\">Na<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub>PO<sub class=\"subscript\"><span style=\"font-size: small\">4<\/span><\/sub><\/span><\/li>\n<li><span style=\"color: #000000\">acetaldehyde<\/span><\/li>\n<\/ol>\n<\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs01_qd01_qa03\">\n<div class=\"question\">\n<p class=\"para\"><span style=\"color: #000000\">Would you expect a 1.0 M solution of CaCl<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub> to be a better conductor of electricity than a 1.0 M solution of NaCl? Why or why not?<\/span><\/p>\n<\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs01_qd01_qa04\">\n<div class=\"question\">\n<p class=\"para\"><span style=\"color: #000000\">An alternative way to define the concentration of a solution is <em class=\"emphasis\">molality<\/em>, abbreviated <em class=\"emphasis\">m<\/em>. Molality is defined as the number of moles of solute in 1 kg of <em class=\"emphasis\">solvent<\/em>. How is this different from molarity? Would you expect a 1 M solution of sucrose to be more or less concentrated than a 1 <em class=\"emphasis\">m<\/em> solution of sucrose? Explain your answer.<\/span><\/p>\n<\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs01_qd01_qa05\">\n<div class=\"question\">\n<p class=\"para\"><span style=\"color: #000000\">What are the advantages of using solutions for quantitative calculations?<\/span><\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<div class=\"qandaset block\" id=\"averill_1.0-ch04_s02_s03_qs01_ans\">\n<h3 class=\"title\">Answer<\/h3>\n<ol class=\"qandadiv\">\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs01_qd01_qa01_ans\">\n<div class=\"answer\"><span style=\"color: #000000\">a) NH3 is a weak base, which means that some of the molecules will accept a proton from water molecules causing them to dissociate into H+ and -OH ions.\u00a0 The H+ ion will associate with the NH3 to form NH4+.\u00a0 Thus this would look the most like beaker #2. b) HF is a weak acid even though F is strongly electronegative.\u00a0 This is because the H-F molecule can form strong hydrogen bonds with the water molecules and remain in a covalent bond that is harder to dissociate. Thus, beaker #2 is also a good choice for this molecule, as only some of the H-F will dissociate to H3O+ and F- ions. c) <\/span><span><span style=\"color: #000000\">CH<sub class=\"subscript\">3<\/sub>CH<sub class=\"subscript\">2<\/sub>CH<sub class=\"subscript\">2<\/sub>OHis a covalent compound and will not dissociate to any appreciable extent, thus, beaker #3 is the correct choice. d) Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> is a soluble ionic compound and will fully dissociate into ions looking most like beaker #1.<\/span><br \/>\n<\/span><\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs01_qd01_qa02_ans\">\n<div class=\"answer\"><\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs01_qd01_qa03_ans\">\n<div class=\"answer\"><span style=\"color: #000000\">\u00a0Yes, because when CaCl<sub>2<\/sub> dissociates it will form 3 ions (1 Ca<sup>2+<\/sup> and 2 Cl<sup>&#8211;<\/sup> ions) whereas NaCl will only dissociate into 2 ions (Na<sup>+<\/sup> and a Cl<sup>&#8211;<\/sup>) for each molecule. Thus, CaCl<sub>2<\/sub> will generate more ions per mole than 1 mole of NaCl and be a better conductor of electricity.<\/span><\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs01_qd01_qa04_ans\">\n<div class=\"answer\"><\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs01_qd01_qa05_ans\">\n<div class=\"answer\">\n<p class=\"para\"><span style=\"color: #000000\">If the amount of a substance required for a reaction is too small to be weighed accurately, the use of a solution of the substance, in which the solute is dispersed in a much larger mass of solvent, allows chemists to measure the quantity of the substance more accurately.<\/span><\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<div class=\"qandaset block\" id=\"averill_1.0-ch04_s02_s03_qs02\">\n<h3 class=\"title\">Numerical Problems<\/h3>\n<ol class=\"qandadiv\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01\">\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa02\">\n<div class=\"question\">\n<p class=\"para\"><span style=\"color: #000000\">Calculate the number of grams of solute in 1.000 L of each solution.<\/span><\/p>\n<ol class=\"orderedlist\" style=\"list-style-type: lower-alpha\">\n<li><span style=\"color: #000000\">0.2593 M NaBrO<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub><\/span><\/li>\n<li><span style=\"color: #000000\">1.592 M KNO<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub><\/span><\/li>\n<li><span style=\"color: #000000\">1.559 M acetic acid<\/span><\/li>\n<li><span style=\"color: #000000\">0.943 M potassium iodate<\/span><\/li>\n<\/ol>\n<\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa01\">\n<div class=\"question\">\n<p class=\"para\"><span style=\"color: #000000\">Calculate the number of grams of solute in 1.000 L of each solution.<\/span><\/p>\n<ol class=\"orderedlist\" style=\"list-style-type: lower-alpha\">\n<li><span style=\"color: #000000\">0.1065 M BaI<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub><\/span><\/li>\n<li><span style=\"color: #000000\">1.135 M Na<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>SO<sub class=\"subscript\"><span style=\"font-size: small\">4<\/span><\/sub><\/span><\/li>\n<li><span style=\"color: #000000\">1.428 M NH<sub class=\"subscript\"><span style=\"font-size: small\">4<\/span><\/sub>Br<\/span><\/li>\n<li><span style=\"color: #000000\">0.889 M sodium acetate<\/span><\/li>\n<\/ol>\n<\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa03\">\n<div class=\"question\">\n<p class=\"para\"><span style=\"color: #000000\">If all solutions contain the same solute, which solution contains the greater mass of solute?<\/span><\/p>\n<ol class=\"orderedlist\" style=\"list-style-type: lower-alpha\">\n<li><span style=\"color: #000000\">1.40 L of a 0.334 M solution or 1.10 L of a 0.420 M solution<\/span><\/li>\n<li><span style=\"color: #000000\">25.0 mL of a 0.134 M solution or 10.0 mL of a 0.295 M solution<\/span><\/li>\n<li><span style=\"color: #000000\">250 mL of a 0.489 M solution or 150 mL of a 0.769 M solution<\/span><\/li>\n<\/ol>\n<\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa04\">\n<div class=\"question\">\n<p class=\"para\"><span style=\"color: #000000\">Complete the following table for 500 mL of solution.<\/span><\/p>\n<div class=\"informaltable\">\n<table cellspacing=\"0\" cellpadding=\"0\">\n<thead>\n<tr>\n<th align=\"center\"><span style=\"color: #000000\">Compound<\/span><\/th>\n<th align=\"center\"><span style=\"color: #000000\">Mass (g)<\/span><\/th>\n<th align=\"center\"><span style=\"color: #000000\">Moles<\/span><\/th>\n<th align=\"center\"><span style=\"color: #000000\">Concentration (M)<\/span><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td><span style=\"color: #000000\">calcium sulfate<\/span><\/td>\n<td align=\"right\"><span style=\"color: #000000\">4.86<\/span><\/td>\n<td align=\"right\"><\/td>\n<td align=\"right\"><\/td>\n<\/tr>\n<tr>\n<td><span style=\"color: #000000\">acetic acid<\/span><\/td>\n<td align=\"right\"><\/td>\n<td align=\"right\"><span style=\"color: #000000\">3.62<\/span><\/td>\n<td align=\"right\"><\/td>\n<\/tr>\n<tr>\n<td><span style=\"color: #000000\">hydrogen iodide dihydrate<\/span><\/td>\n<td align=\"right\"><\/td>\n<td align=\"right\"><\/td>\n<td align=\"right\"><span style=\"color: #000000\">1.273<\/span><\/td>\n<\/tr>\n<tr>\n<td><span style=\"color: #000000\">barium bromide<\/span><\/td>\n<td align=\"right\"><span style=\"color: #000000\">3.92<\/span><\/td>\n<td align=\"right\"><\/td>\n<td align=\"right\"><\/td>\n<\/tr>\n<tr>\n<td><span style=\"color: #000000\">glucose<\/span><\/td>\n<td align=\"right\"><\/td>\n<td align=\"right\"><\/td>\n<td align=\"right\"><span style=\"color: #000000\">0.983<\/span><\/td>\n<\/tr>\n<tr>\n<td><span style=\"color: #000000\">sodium acetate<\/span><\/td>\n<td align=\"right\"><\/td>\n<td align=\"right\"><span style=\"color: #000000\">2.42<\/span><\/td>\n<td align=\"right\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa06\">\n<div class=\"question\">\n<p class=\"para\"><span style=\"color: #000000\">What is the concentration of each species present in the following aqueous solutions?<\/span><\/p>\n<ol class=\"orderedlist\" style=\"list-style-type: lower-alpha\">\n<li><span style=\"color: #000000\">0.489 mol of NiSO<sub class=\"subscript\"><span style=\"font-size: small\">4<\/span><\/sub> in 600 mL of solution<\/span><\/li>\n<li><span style=\"color: #000000\">1.045 mol of magnesium bromide in 500 mL of solution<\/span><\/li>\n<li><span style=\"color: #000000\">0.146 mol of glucose in 800 mL of solution<\/span><\/li>\n<li><span style=\"color: #000000\">0.479 mol of CeCl<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub> in 700 mL of solution<\/span><\/li>\n<\/ol>\n<\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa05\">\n<div class=\"question\">\n<p class=\"para\"><span style=\"color: #000000\">What is the concentration of each species present in the following aqueous solutions?<\/span><\/p>\n<ol class=\"orderedlist\" style=\"list-style-type: lower-alpha\">\n<li><span style=\"color: #000000\">0.324 mol of K<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>MoO<sub class=\"subscript\"><span style=\"font-size: small\">4<\/span><\/sub> in 250 mL of solution<\/span><\/li>\n<li><span style=\"color: #000000\">0.528 mol of potassium formate in 300 mL of solution<\/span><\/li>\n<li><span style=\"color: #000000\">0.477 mol of KClO<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub> in 900 mL of solution<\/span><\/li>\n<li><span style=\"color: #000000\">0.378 mol of potassium iodide in 750 mL of solution<\/span><\/li>\n<\/ol>\n<\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa08\">\n<div class=\"question\">\n<p class=\"para\"><span style=\"color: #000000\">What is the molar concentration of each solution?<\/span><\/p>\n<ol class=\"orderedlist\" style=\"list-style-type: lower-alpha\">\n<li><span style=\"color: #000000\">8.7 g of calcium bromide in 250 mL of solution<\/span><\/li>\n<li><span style=\"color: #000000\">9.8 g of lithium sulfate in 300 mL of solution<\/span><\/li>\n<li><span style=\"color: #000000\">12.4 g of sucrose (C<sub class=\"subscript\"><span style=\"font-size: small\">12<\/span><\/sub>H<sub class=\"subscript\"><span style=\"font-size: small\">22<\/span><\/sub>O<sub class=\"subscript\"><span style=\"font-size: small\">11<\/span><\/sub>) in 750 mL of solution<\/span><\/li>\n<li><span style=\"color: #000000\">14.2 g of iron(III) nitrate hexahydrate in 300 mL of solution<\/span><\/li>\n<\/ol>\n<\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa07\">\n<div class=\"question\">\n<p class=\"para\"><span style=\"color: #000000\">What is the molar concentration of each solution?<\/span><\/p>\n<ol class=\"orderedlist\" style=\"list-style-type: lower-alpha\">\n<li><span style=\"color: #000000\">12.8 g of sodium hydrogen sulfate in 400 mL of solution<\/span><\/li>\n<li><span style=\"color: #000000\">7.5 g of potassium hydrogen phosphate in 250 mL of solution<\/span><\/li>\n<li><span style=\"color: #000000\">11.4 g of barium chloride in 350 mL of solution<\/span><\/li>\n<li><span style=\"color: #000000\">4.3 g of tartaric acid (C<sub class=\"subscript\"><span style=\"font-size: small\">4<\/span><\/sub>H<sub class=\"subscript\"><span style=\"font-size: small\">6<\/span><\/sub>O<sub class=\"subscript\"><span style=\"font-size: small\">6<\/span><\/sub>) in 250 mL of solution<\/span><\/li>\n<\/ol>\n<\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa09\">\n<div class=\"question\">\n<p class=\"para\"><span style=\"color: #000000\">Give the concentration of each reactant in the following equations, assuming 20.0 g of each and a solution volume of 250 mL for each reactant.<\/span><\/p>\n<ol class=\"orderedlist\" style=\"list-style-type: lower-alpha\">\n<li><span style=\"color: #000000\">BaCl<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>(aq)\u00a0+\u00a0Na<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>SO<sub class=\"subscript\"><span style=\"font-size: small\">4<\/span><\/sub>(aq) \u2192<\/span><\/li>\n<li><span style=\"color: #000000\">Ca(OH)<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>(aq)\u00a0+\u00a0H<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub>PO<sub class=\"subscript\"><span style=\"font-size: small\">4<\/span><\/sub>(aq) \u2192<\/span><\/li>\n<li><span style=\"color: #000000\">Al(NO<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub>)<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub>(aq)\u00a0+\u00a0H<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>SO<sub class=\"subscript\"><span style=\"font-size: small\">4<\/span><\/sub>(aq) \u2192<\/span><\/li>\n<li><span style=\"color: #000000\">Pb(NO<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub>)<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>(aq)\u00a0+\u00a0CuSO<sub class=\"subscript\"><span style=\"font-size: small\">4<\/span><\/sub>(aq) \u2192<\/span><\/li>\n<li><span style=\"color: #000000\">Al(CH<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub>CO<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>)<sub class=\"subscript\"><span style=\"font-size: small\">3<\/span><\/sub>(aq)\u00a0+\u00a0NaOH(aq) \u2192<\/span><\/li>\n<\/ol>\n<\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa10\">\n<div class=\"question\">\n<p class=\"para\"><span style=\"color: #000000\">An experiment required 200.0 mL of a 0.330 M solution of Na<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>CrO<sub class=\"subscript\"><span style=\"font-size: small\">4<\/span><\/sub>. A stock solution of Na<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>CrO<sub class=\"subscript\"><span style=\"font-size: small\">4<\/span><\/sub> containing 20.0% solute by mass with a density of 1.19 g\/cm<sup class=\"superscript\"><span style=\"font-size: small\">3<\/span><\/sup> was used to prepare this solution. Describe how to prepare 200.0 mL of a 0.330 M solution of Na<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>CrO<sub class=\"subscript\"><span style=\"font-size: small\">4<\/span><\/sub> using the stock solution.<\/span><\/p>\n<\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa11\">\n<div class=\"question\">\n<p class=\"para\"><span style=\"color: #000000\">Calcium hypochlorite [Ca(OCl)<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>] is an effective disinfectant for clothing and bedding. If a solution has a Ca(OCl)<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub> concentration of 3.4 g per 100 mL of solution, what is the molarity of hypochlorite?<\/span><\/p>\n<\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa12\">\n<div class=\"question\">\n<p class=\"para\"><span style=\"color: #000000\">Phenol (C<sub class=\"subscript\"><span style=\"font-size: small\">6<\/span><\/sub>H<sub class=\"subscript\"><span style=\"font-size: small\">5<\/span><\/sub>OH) is often used as an antiseptic in mouthwashes and throat lozenges. If a mouthwash has a phenol concentration of 1.5 g per 100 mL of solution, what is the molarity of phenol?<\/span><\/p>\n<\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa13\">\n<div class=\"question\">\n<p class=\"para\"><span style=\"color: #000000\">If a tablet containing 100 mg of caffeine (C<sub class=\"subscript\"><span style=\"font-size: small\">8<\/span><\/sub>H<sub class=\"subscript\"><span style=\"font-size: small\">10<\/span><\/sub>N<sub class=\"subscript\"><span style=\"font-size: small\">4<\/span><\/sub>O<sub class=\"subscript\"><span style=\"font-size: small\">2<\/span><\/sub>) is dissolved in water to give 10.0 oz of solution, what is the molar concentration of caffeine in the solution?<\/span><\/p>\n<\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa14\">\n<div class=\"question\">\n<p class=\"para\"><span style=\"color: #000000\">A certain drug label carries instructions to add 10.0 mL of sterile water, stating that each milliliter of the resulting solution will contain 0.500 g of medication. If a patient has a prescribed dose of 900.0 mg, how many milliliters of the solution should be administered?<\/span><\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<div class=\"qandaset block\" id=\"averill_1.0-ch04_s02_s03_qs02_ans\">\n<h3 class=\"title\">Answers<\/h3>\n<ol class=\"qandadiv\">\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa02_ans\">\n<div class=\"answer\"><span style=\"color: #000000\">a. 39.13 g\u00a0\u00a0 b. 161.0 g\u00a0\u00a0 c. 93.57 g\u00a0\u00a0 d. 201.8 g<\/span><\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa01_ans\">\n<div class=\"answer\"><\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa03_ans\">\n<div class=\"answer\"><span style=\"color: #000000\">a. 1.40 L of a 0.334 M solution, b. 25.0 mL of a 0.134 M solution, c. 150 mL of a 0.769 M solution<\/span><\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa04_ans\">\n<div class=\"answer\"><\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa06_ans\">\n<div class=\"answer\"><span style=\"color: #000000\">a. 0.815 M, b. 2.09 M, c. 0.182 M, d. 0.684 M<\/span><\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa05_ans\">\n<div class=\"answer\"><\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa08_ans\">\n<div class=\"answer\"><span style=\"color: #000000\">a. 0.174 M, b. 0.297 M, c. 0.048 M, d. 0.135 M<\/span><\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa07_ans\">\n<div class=\"answer\"><\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa09_ans\">\n<div class=\"answer\"><span style=\"color: #000000\">a. BaCl<sub>2<\/sub> = 0.384 M, Na<sub>2<\/sub>SO<sub>4<\/sub> = 0.563 M, b. Ca(OH)<sub>2<\/sub> = 1.08 M, H3PO4 = 0.816 M, c. Al(NO<sub>3<\/sub>)<sub>3<\/sub> = 0.376 M, H<sub>2<\/sub>SO<sub>4<\/sub> = 0.816 M, d. Pb(NO<sub>3<\/sub>)<sub>2<\/sub> = 0.242 M, CuSO<sub>4<\/sub> = 0.501 M, e. Al(CH<sub>3<\/sub>CO<sub>2<\/sub>) = 0.392 M, NaOH = 2.00 M<\/span><\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa10_ans\">\n<div class=\"answer\"><\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa11_ans\">\n<div class=\"answer\">\n<p class=\"para\"><span style=\"color: #000000\">0.48 M ClO<sup class=\"superscript\"><span style=\"font-size: small\">\u2212<\/span><\/sup><\/span><\/p>\n<\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa12_ans\">\n<div class=\"answer\"><\/div>\n<\/li>\n<li class=\"qandaentry\" id=\"averill_1.0-ch04_s02_s03_qs02_qd01_qa13_ans\">\n<div class=\"answer\">\n<p class=\"para\"><span style=\"color: #000000\">1.74\u00a0\u00d7\u00a010<sup class=\"superscript\"><span style=\"font-size: small\">\u22123<\/span><\/sup> M caffeine<\/span><\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<section class=\"mt-content-container\">\n<div class=\"mt-section\" id=\"section_3\">\n<div class=\"mt-section\" id=\"section_5\">\n<ul><\/ul>\n<form name=\"1\">\n<div class=\"mt-section\" id=\"section_6\">\n<h4><a href=\"#title\"><em><strong><span id=\"Contributors\" style=\"color: #ff0000\">(Back to the Top)<\/span><\/strong><\/em><\/a><\/h4>\n<hr \/>\n<h3 id=\"7refs\" class=\"editable\">References<\/h3>\n<ul>\n<li><a title=\"http:\/\/www.science.uwaterloo.ca\/~cchieh\/cact\/\" class=\"external\" href=\"http:\/\/www.science.uwaterloo.ca\/%7Ecchieh\/cact\/\" target=\"_blank\" rel=\"external nofollow noopener noreferrer\">Chung (Peter) Chieh<\/a> <span style=\"color: #000000\">(2016) Inorganic Chemistry. <em>Libretexts<\/em>. Available at:<\/span> <a href=\"https:\/\/chem.libretexts.org\/Core\/Inorganic_Chemistry\/Chemical_Reactions\/Chemical_Reactions_1\/Solutions\">https:\/\/chem.libretexts.org\/Core\/Inorganic_Chemistry\/Chemical_Reactions\/Chemical_Reactions_1\/Solutions<\/a><\/li>\n<li><span style=\"color: #000000\">Ball, D.W., Hill, J.W., and Scott, R.J.\u00a0(2016) <em>MAP: The Basics of General,\u00a0Organic and Biological\u00a0Chemistry<\/em>.\u00a0 Libre Texts. Available at:<\/span> <a href=\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Introductory_Chemistry_Textbook_Maps\/Map%3A_The_Basics_of_GOB_Chemistry_%28Ball_et_al.%29\">https:\/\/chem.libretexts.org\/Textbook_Maps\/Introductory_Chemistry_Textbook_Maps\/Map%3A_The_Basics_of_GOB_Chemistry_(Ball_et_al.)<\/a><\/li>\n<li><span style=\"color: #000000\">Averill, B.A., Eldredge, P. (2012) <em>The Principles of Chemistry<\/em>. Libre Texts. Available at:<\/span>\u00a0<a href=\"https:\/\/2012books.lardbucket.org\/books\/principles-of-general-chemistry-v1.0\/index.html\">https:\/\/2012books.lardbucket.org\/books\/principles-of-general-chemistry-v1.0\/index.html <\/a><\/li>\n<li><span style=\"color: #000000\">Hydrate. (2017, August 30). In <i>Wikipedia, The Free Encyclopedia<\/i>. Retrieved 16:20, September 26, 2017, from<\/span> <a class=\"external free\" href=\"https:\/\/en.wikipedia.org\/w\/index.php?title=Hydrate&amp;oldid=798015169\">https:\/\/en.wikipedia.org\/w\/index.php?title=Hydrate&amp;oldid=798015169<\/a><\/li>\n<li><span style=\"color: #000000\">Lower, S. (2010). Solutions 1: Solutions and their concentrations. <em>In the Online textbook, &#8220;Chem1 Virtual Textbook&#8221;.<\/em> Available at:<\/span>\u00a0<a href=\"http:\/\/www.chem1.com\/acad\/webtext\/solut\/solut-1.html\"> http:\/\/www.chem1.com\/acad\/webtext\/solut\/solut-1.html<\/a><\/li>\n<li>\n<div class=\"citation\"><span style=\"color: #000000\"><span class=\"name\">OpenStax<\/span>, Chemistry. OpenStax CNX. <\/span><span><span style=\"color: #000000\">Jun 20, 2016<\/span> <a href=\"http:\/\/cnx.org\/contents\/85abf193-2bd2-4908-8563-90b8a7ac8df6@9.311\">http:\/\/cnx.org\/contents\/85abf193-2bd2-4908-8563-90b8a7ac8df6@9.311<\/a>.<\/span><\/div>\n<\/li>\n<\/ul>\n<\/div>\n<\/form>\n<\/div>\n<\/div>\n<footer class=\"mt-content-footer\"><\/footer>\n<\/section>\n","protected":false},"excerpt":{"rendered":"<p>Chapter 7: Solutions And Solution Stoichiometry 7.1 Introduction 7.2 Types of Solutions 7.3 Solubility 7.4 Temperature and Solubility 7.5 Effects of Pressure on the Solubility of Gases: Henry&#8217;s Law 7.6 Solid Hydrates 7.7 Solution Concentration 7.7.1 Molarity 7.7.2 Parts Per Solutions 7.8 Dilutions 7.9 Ion Concentrations in Solution 7.10 Summary 7.11 References &nbsp; &nbsp; &nbsp; [&hellip;]<\/p>\n","protected":false},"author":280,"featured_media":0,"parent":2825,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_seopress_robots_primary_cat":"","_seopress_titles_title":"","_seopress_titles_desc":"","_seopress_robots_index":"","_lmt_disableupdate":"","_lmt_disable":"","_et_pb_use_builder":"","_et_pb_old_content":"","_et_gb_content_width":"","footnotes":"","_links_to":"","_links_to_target":""},"class_list":["post-3502","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/wou.edu\/chemistry\/wp-json\/wp\/v2\/pages\/3502","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/wou.edu\/chemistry\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/wou.edu\/chemistry\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/wou.edu\/chemistry\/wp-json\/wp\/v2\/users\/280"}],"replies":[{"embeddable":true,"href":"https:\/\/wou.edu\/chemistry\/wp-json\/wp\/v2\/comments?post=3502"}],"version-history":[{"count":0,"href":"https:\/\/wou.edu\/chemistry\/wp-json\/wp\/v2\/pages\/3502\/revisions"}],"up":[{"embeddable":true,"href":"https:\/\/wou.edu\/chemistry\/wp-json\/wp\/v2\/pages\/2825"}],"wp:attachment":[{"href":"https:\/\/wou.edu\/chemistry\/wp-json\/wp\/v2\/media?parent=3502"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}