{"id":4362,"date":"2018-12-22T21:54:58","date_gmt":"2018-12-23T05:54:58","guid":{"rendered":"http:\/\/www.wou.edu\/chemistry\/?page_id=4362"},"modified":"2019-03-05T13:12:10","modified_gmt":"2019-03-05T21:12:10","slug":"ch103-chapter-9-homeostasis-and-cellular-function","status":"publish","type":"page","link":"https:\/\/wou.edu\/chemistry\/courses\/online-chemistry-textbooks\/ch103-allied-health-chemistry\/ch103-chapter-9-homeostasis-and-cellular-function\/","title":{"rendered":"CH103 &#8211; Chapter 8: Homeostasis and Cellular Function"},"content":{"rendered":"<header>\n<h2 id=\"title\"><span><strong>Chapter 8: Homeostasis and Cellular Function<\/strong><\/span><\/h2>\n<p><span style=\"color: #000000\">This text is published under creative commons licensing. For referencing this work, please click<span style=\"color: #ff0000\"> <a href=\"https:\/\/wou.edu\/chemistry\/courses\/online-chemistry-textbooks\/ch103-allied-health-chemistry\/\" style=\"color: #ff0000\"><strong><em>here<\/em><\/strong><\/a>.<\/span><\/span><\/p>\n<h3><a href=\"#stasis\"><span><strong>8.1 The Concept of Homeostasis<\/strong><\/span><\/a><\/h3>\n<h3 class=\"editable\"><a href=\"#disease\"><strong>8.2\u00a0<\/strong><strong><span>Disease as a Homeostatic Imbalance<\/span><\/strong><\/a><a href=\"#typessolutions\"><strong><\/strong><\/a><\/h3>\n<h3 class=\"editable\"><a href=\"#7intro\"><strong>8.3 <\/strong><span><strong>Measuring Homeostasis to Evaluate Health<\/strong><\/span><\/a><\/h3>\n<h3 id=\"solubility\" class=\"editable\"><a href=\"#solubility2\"><strong>8.4 Solubility<\/strong><\/a><\/h3>\n<h3><a href=\"#solutionconcentration\"><strong><span>8.5 Solution Concentration<\/span><\/strong><\/a><\/h3>\n<h4 class=\"title editable block\"><span style=\"color: #000000\"><a href=\"#molarity\" style=\"color: #000000\"><strong>8.5.1 Molarity<\/strong><\/a><\/span><\/h4>\n<h4><span style=\"color: #000000\"><a href=\"#partspermillion\" style=\"color: #000000\"><strong>8.5.2 Parts Per Solutions<\/strong><\/a><\/span><\/h4>\n<h4 class=\"editable\"><a href=\"#equivalents\"><span style=\"color: #000000\"><strong>8.5.3 Equivalents<\/strong><\/span><\/a><\/h4>\n<h3 class=\"para editable block\"><a href=\"#dilutions\"><span><strong>8.6 Dilutions<\/strong><\/span><\/a><\/h3>\n<h3 class=\"para\" id=\"averill_1.0-ch04_s02_s02_p41\"><a href=\"#ionconcentration\"><span><strong>8.7 Ion Concentrations in Solution<\/strong><\/span><\/a><\/h3>\n<h3><a href=\"#movement\"><strong>8.8 Movement of Molecules Across the Membrane<\/strong><\/a><\/h3>\n<h3><a href=\"#7summary\"><strong><span>8.9 Summary<\/span><\/strong><\/a><\/h3>\n<h3><a href=\"#7refs\"><span><strong>8.10 References<\/strong><\/span><\/a><\/h3>\n<p>&nbsp;<\/p>\n<hr \/>\n<h3 id=\"stasis\"><strong>8.1 The Concept of Homeostasis<\/strong><\/h3>\n<p><span style=\"color: #000000\"><strong>Homeostasis <\/strong>refers to the body&#8217;s ability to physiologically regulate its inner environment to ensure its stability in response to fluctuations in external or internal conditions. The liver, the pancreas, the kidneys, and the brain (hypothalamus, the autonomic nervous system and the endocrine system) help maintain homeostasis. The liver is responsible for metabolizing toxic substances and with signaling from the pancreas maintains carbohydrate metabolism. The liver also helps to regulate lipid metabolism and is the primary site of cholesterol production. The kidneys are responsible for regulating blood water levels, re-absorption of substances into the blood, maintenance of salt and ion levels in the blood, regulation of blood pH, and excretion of urea and other waste products. The hypothalamus is involved in the regulation of body temperature, heart rate, blood pressure, and circadian rhythms (which include wake\/sleep cycles).<br \/>\n<\/span><\/p>\n<p><span style=\"color: #000000\">Homeostasis can be influenced by either internal or existing conditions <strong><em>(instrinsic factors<\/em><\/strong>) or external or environmental conditions (<strong><em>extrinsic factors<\/em><\/strong>) and is maintained by many different mechanisms. All homeostatic control mechanisms have at least three interdependent components for the variable being regulated:<\/span><\/p>\n<ul>\n<li><span style=\"color: #000000\">A <strong><em>sensor<\/em><\/strong> or receptor that detects changes in the internal or external environment. An example is peripheral chemoreceptors, which detect changes in blood pH.<\/span><\/li>\n<li><span style=\"color: #000000\">The<strong><em> integrating center<\/em><\/strong> or control center receives information from the sensors and initiates the response to maintain homeostasis. The most important example is the hypothalamus, a region of the brain that controls everything from body temperature to heart rate, blood pressure, satiety (fullness), and circadian rhythms (including, sleep and wake cycles).<\/span><\/li>\n<li><span style=\"color: #000000\">An <em><strong>effector <\/strong><\/em>is any organ or tissue that receives information from the integrating center and acts to bring about the changes needed to maintain homeostasis. One example is the kidney, which retains water if blood pressure is too low.<\/span><\/li>\n<\/ul>\n<p><span style=\"color: #000000\">The sensors, integrating center, and effectors are the basic components of every homeostatic response. Positive and negative feedback are more complicated mechanisms that enable these three basic components to maintain homeostasis for more complex physiological processes.<\/span><\/p>\n<h4><span><strong><em>Negative Feedback<\/em><\/strong><\/span><\/h4>\n<p><span style=\"color: #000000\"><strong><em>Negative feedback <\/em><\/strong>mechanisms use one of the products of the reaction to reduce the output or activity of the process for the purpose of returning an organ or system to its normal range of functioning. Most homeostatic processes use negative feedback regulation to maintain a specific parameter around a setpoint range that supports life Figure 8.1.\u00a0 However, it should be noted that negative feedback processes are also used for other processes that are not homeostatic. <\/span><\/p>\n<p><span style=\"color: #000000\">Within the realm of homeostasis, temperature control is a good example that uses negative feedback. Nerve cells (the sensors) relay information about body temperature to the hypothalamus (the integrating center). The hypothalamus then signals several effectors to return the body temperature to 37<sup>o<\/sup>C (the set point). Two effectors activated in the process when core temperature is too high are the sweat glands which serve to cool the skin and the blood vessels which undergo <strong><em>vasodilation<\/em> <\/strong>(or enlarging) so the body can give off more heat. Once the core temperature is brought back into normal range, the sensor will send negative feedback messages to the integrating center to turn off the process (ie turn off the sweat glands and inhibit further vasodilation). <\/span><span style=\"color: #000000\">Both internal and external events can induce negative feedback mechanisms. The two examples above represent internal mechanisms utilized to return the body within the normal temperature range.\u00a0 However, we can also mediate the cooling of the body through external factors, such as removing a warm hat and gloves or pouring a cool glass of water over our head. Both external and internal mechanisms for cooling can return the temperature of the body to within the normal range and elicit the negative feedback response. <\/span><span style=\"color: #000000\">Similarly, if body temperature is below the set point, muscles shiver to generate heat and the constriction of the blood vessels helps the body retain heat. <\/span><\/p>\n<p><span style=\"color: #000000\">Homeostatic processes are very complex because the setpoint or normal range might change depending on the circumstance. For example, the hypothalamus can change the body\u2019s temperature set point, such as raising it during a fever to help fight an infection.<br \/>\n<\/span><\/p>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2019\/01\/Temperature-regulation-in-homeostasis-with-feedback.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/wou.edu\/chemistry\/files\/2019\/01\/Temperature-regulation-in-homeostasis-with-feedback-1024x640.png\" alt=\"\" class=\"alignnone wp-image-4545\" width=\"709\" height=\"443\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2019\/01\/Temperature-regulation-in-homeostasis-with-feedback-1024x640.png 1024w, https:\/\/wou.edu\/chemistry\/files\/2019\/01\/Temperature-regulation-in-homeostasis-with-feedback-300x187.png 300w, https:\/\/wou.edu\/chemistry\/files\/2019\/01\/Temperature-regulation-in-homeostasis-with-feedback-768x480.png 768w, https:\/\/wou.edu\/chemistry\/files\/2019\/01\/Temperature-regulation-in-homeostasis-with-feedback.png 1532w\" sizes=\"(max-width: 709px) 100vw, 709px\" \/><\/a><\/p>\n<p><span style=\"color: #000000\"><strong>Figure 8.1 Homeostatic Regulation of Temperature in Humans.<\/strong> Core body temperature is maintained at a normal setpoint of 37<sup>o<\/sup>C.\u00a0 If the core temperature rises above (right hand side) or drops below (left hand side) the setpoint, internal biological responses are initiated to return the core temperature back to the setpoint range. Once this is achieved, negative feedback loops are initiated to down regulate the internal biological responses so that the core temperature doesn&#8217;t overshoot the required change.<\/span><\/p>\n<p><span style=\"color: #000000\">This Figure is adapted from:<\/span> <a href=\"https:\/\/www.khanacademy.org\/science\/high-school-biology\/hs-human-body-systems\/hs-body-structure-and-homeostasis\/a\/homeostasis\">The Kahn Academy<\/a><\/p>\n<hr \/>\n<h4><strong><em><span style=\"color: #ff0000\">Positive Feedback<\/span><\/em><\/strong><\/h4>\n<p><span style=\"color: #000000\"><strong><em>Positive feedback<\/em><\/strong> is a mechanism in which an activated component enhances or further upregulates the process that gave rise to itself in order to create an even stronger response. Positive feedback mechanisms are designed to accelerate or enhance the output created by a stimulus that has already been activated. Positive feedback mechanisms are designed to push levels out of normal ranges and are not used as often in homeostatic responses. To achieve positive feedback, a series of events initiates a cascading process that builds to increase the effect of the stimulus.<\/span><\/p>\n<p><span style=\"color: #000000\">An example of a positive feedback loop is the blood clotting cascade which is originally initiated by external damage to the vasculature (Figure 8.2). During a damage event, extrinsic factors begin the initiation of the blood clotting cascade.\u00a0 The proteins involved in this process are usually held inactive by being produced in a much larger form than is required.\u00a0 To activate the protein, the protein needs to be cleaved into a smaller, active complex.\u00a0 When a protein is held in a large inactive state and cleaved to yield the active component, it is called a <strong><em>zymogen<\/em><\/strong>.\u00a0 The blood clotting cascade contains many zymogens. The first zymogen to be activated is Factor X.\u00a0 When Factor X is cleaved, it becomes active and proceeds to cleave the next downstream target, Prothrombin II. This produces the active component, Thrombin IIa which has multiple effects.\u00a0 First, it cleaves the protein Fibrinogen to produce Fibrin. Fibrin then begins to form a clotting complex with itself.\u00a0 This is referred to as the loose mesh network. Activated Thrombin IIa also cleaves the inactive form of Factor XIII.\u00a0 Activated Factor XIIIa causes crosslinks to form in the loose mesh network creating the finalized stable mesh that forms the blood clot. To accelerate this process further, Thrombin IIa also has two positive feedback effects.\u00a0 It can also cleave Inactive Factor X creating more activated Factor X and ultimately more activated Throbmin IIa.\u00a0 It also increases the activity of the instrinsic blood clotting cascade, which further upregulates the activation of Factor X.<\/span><\/p>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2019\/01\/Blood-Clotting-Cascade.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/wou.edu\/chemistry\/files\/2019\/01\/Blood-Clotting-Cascade-1024x870.png\" alt=\"\" class=\"alignnone wp-image-4542\" width=\"700\" height=\"595\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2019\/01\/Blood-Clotting-Cascade-1024x870.png 1024w, https:\/\/wou.edu\/chemistry\/files\/2019\/01\/Blood-Clotting-Cascade-300x255.png 300w, https:\/\/wou.edu\/chemistry\/files\/2019\/01\/Blood-Clotting-Cascade-768x653.png 768w, https:\/\/wou.edu\/chemistry\/files\/2019\/01\/Blood-Clotting-Cascade.png 1440w\" sizes=\"(max-width: 700px) 100vw, 700px\" \/><\/a><\/p>\n<p><span style=\"color: #000000\"><strong>Figure 8.2 The Positive Feedback Mechanism of the Blood Clotting Cascade.<\/strong> Extrinsic factors such as damage or injury activated the cleavage of zymogen proteins in the blood clotting cascade. Activation of the zymogen, Thrombin IIa begins the formation of the fibrin clotting network and also elicits positive feedback that further upregulates the entire clotting cascade.<\/span><\/p>\n<p><span style=\"color: #000000\">This Figure is adapted from: <\/span><a href=\"https:\/\/en.wikipedia.org\/wiki\/File:D-dimer_production.pdf\">MPT-Matthew<\/a><\/p>\n<hr \/>\n<p><span style=\"color: #000000\">Many parameters are regulated within the body within a narrow homeostatic window to maintain proper functioning and balance within biological systems. Some examples of homeostatic parameters include:<\/span><\/p>\n<h4><span style=\"color: #ff0000\"><strong>Temperature<\/strong><\/span><\/h4>\n<p><span style=\"color: #000000\">Humans are warm-blooded or <strong><em>endothermic<\/em><\/strong>, maintaining a near-constant body temperature. Thermoregulation is an important aspect of human homeostasis. Heat is mainly produced by the liver and muscle contractions. Humans have been able to adapt to a great diversity of climates, including hot humid and hot arid environments. High temperatures pose serious stresses for the human body, placing it in great danger of injury or even death. In order to deal with these climatic conditions, humans have developed physiologic and cultural modes of adaptation. When internal temperature reaches extremes of 45\u00b0C (113\u00b0F), <strong><em>hyperthermia<\/em><\/strong>,\u00a0a condition where an individual&#8217;s body temperature is elevated beyond normal, occurs and cellular proteins will denature, causing metabolism to stop and ultimately lead to death. <strong><em>Hypothermia<\/em> <\/strong>is the opposite condition, where internal body temperature falls below homeostatic norms. Hypothermia occurs when body core temperatures fall below 35.0\u00a0\u00b0C (95.0\u00a0\u00b0F). Symptoms depend on the temperature. In mild hypothermia there is shivering and mental confusion. In moderate hypothermia shivering stops and confusion increases. In severe hypothermia, there may be paradoxical undressing, in which a person removes their clothing, as well as an increased risk of the heart stopping. Hypothermia has two main types of causes. It classically occurs from exposure to extreme cold. It may also occur from any condition that decreases heat production or increases heat loss. Commonly this includes alcohol intoxication but may also include low blood sugar, anorexia, and advanced age.<\/span><span style=\"color: #000000\">\u00a0<\/span><\/p>\n<h4><strong><span style=\"color: #ff0000\">Iron<\/span><\/strong><\/h4>\n<p><span style=\"color: #000000\">Iron is an essential element for human beings. The control of this necessary but potentially toxic substance is an important part of many aspects of human health and disease. Hematologists have been especially interested in the system of iron metabolism because iron is essential to red blood cells. In fact, most of the human body&#8217;s iron is contained in red blood cells&#8217; hemoglobin protein where it aids in the binding and transport of oxygen for cellular respiration, and iron deficiency is the most common cause of anemia.<\/span><\/p>\n<p><span style=\"color: #000000\">When body levels of iron are too low, an iron-sensitive hormone called hepcidin is decreased in the duodenal epithelium (lining of the small intestine). This causes an increase in ferroportin activity, an iron-selective protein channel embedded in the membrane of intestinal cells. Activation of this channel stimulates iron uptake in the digestive system. An iron surplus will stimulate the reverse of this process.<\/span><\/p>\n<h4><strong><span style=\"color: #ff0000\">Sugar<\/span><\/strong><\/h4>\n<p><span style=\"color: #000000\">Blood glucose is regluated with two hormones, <strong><em>insulin <\/em><\/strong>and <strong><em>glucagon<\/em><\/strong>, both released from the pancreas.<\/span><\/p>\n<p><span style=\"color: #000000\">When blood sugar levels become too high, insulin is released from the pancreas. Glucose, or sugar, is taken up by cells (especially liver and muscle tissue) where it is stored as glycogen. This results in a lowering of the blood sugar levels. On the other hand, when blood sugar levels become too low, glucagon is released by the pancreas. It promotes the breakdown of glycogen into the glucose monomers within liver cells. The liver cells then release free glucose back into the blood stream and restore blood sugar levels.<\/span><span style=\"color: #000000\"><\/span><\/p>\n<p><span style=\"color: #000000\">Improper glucagon functioning results in <strong><em>hypoglycemia<\/em><\/strong>, a condition where blood sugar is too low.\u00a0 This can be life threatening leading to coma and death if not treated promptly. Improper insulin function results in <strong><em>hyperglycemia<\/em><\/strong> or increased blood sugar levels. If this state is prolonged the disease called diabetes results. Diabetes will be discussed in more detail in section 8.2 below.<\/span><\/p>\n<h4><span style=\"color: #ff0000\"><strong>Osmoregulation<\/strong><\/span><\/h4>\n<p><span style=\"color: #000000\">Osmoregulation is the active regulation of the osmotic pressure of bodily fluids to maintain the homeostasis of the body&#8217;s water content; that is it keeps the body&#8217;s fluids from becoming too dilute or too concentrated. Osmotic pressure is a measure of the tendency of water to move into one solution from another by osmosis. The higher the osmotic pressure of a solution the more water wants to go into the solution.<\/span><\/p>\n<p><span style=\"color: #000000\">The kidneys are used to remove excess ions (such as Na<sup>+<\/sup>, K<sup>+<\/sup> and Ca<sup>2+<\/sup>) from the blood, thus affecting the osmotic pressure. These are then expelled as urine. The kidneys are also important for maintaining acid\/base levels, such that the pH of the blood remains close to the neutral point.<br \/>\n<\/span><\/p>\n<h4><strong><span style=\"color: #ff0000\">Water Volume<\/span><\/strong><\/h4>\n<p><span style=\"color: #000000\">The kidneys also determine the overall water volume maintained within the body. The hormones Anti-Diuretic Hormone (ADH), also known as vasopressin, and Aldosterone play a major role in regulating kidney function.<\/span><\/p>\n<ul>\n<li><span style=\"color: #000000\">If the body is becoming fluid-<em>deficient<\/em>, there will be an increase in the secretion of ADH from the pituitary gland. This hormone then travels to the distal tubules or collecting ducts of the kidneys, causing fluid to be retained and urine output to be reduced. Similarly, the hormone aldosterone, a mineralcorticoid hormone with a steroid backbone, is secreted from the adrenal cortex. Aldosterone causes the kidneys to reabsorb Na<sup>+<\/sup>. As Na<sup>+<\/sup> is reabsored, water is reabsorbed as well. Thus, Na<sup>+<\/sup> retention also leads to fluid retention (Figure 8.3).<\/span><\/li>\n<li><span style=\"color: #000000\">Conversely, if fluid levels are <em>excessive<\/em>, secretion of the hormone (aldosterone) is suppressed, resulting in less retention of fluid by the kidneys and a subsequent increase in the volume of urine produced.<\/span><span style=\"color: #000000\"><\/span><\/li>\n<\/ul>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2019\/02\/adh-and-aldo-III.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/wou.edu\/chemistry\/files\/2019\/02\/adh-and-aldo-III-1024x859.jpg\" alt=\"\" class=\"alignnone wp-image-4607 \" width=\"700\" height=\"587\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2019\/02\/adh-and-aldo-III-1024x859.jpg 1024w, https:\/\/wou.edu\/chemistry\/files\/2019\/02\/adh-and-aldo-III-300x252.jpg 300w, https:\/\/wou.edu\/chemistry\/files\/2019\/02\/adh-and-aldo-III-768x644.jpg 768w, https:\/\/wou.edu\/chemistry\/files\/2019\/02\/adh-and-aldo-III.jpg 1239w\" sizes=\"(max-width: 700px) 100vw, 700px\" \/><\/a><\/p>\n<p><span style=\"color: #000000\"><strong>Figure 8.3: Effects of Aldosterone and ADH on Kidney Function.<\/strong> When fluid levels in the body are low, ADH (Vasopressin) is secreted by the pituitary gland and Aldosterone is secreted by the adrenal glands. ADH decreases the loss of water whereas Aldosterone increases the reabsorbtion of Na+ within the collecting duct of the kidneys. Water is reabsorbed with the Na+ causing an increase in fluid retention and decreased urine output.<\/span><\/p>\n<p><span style=\"color: #000000\">This figure has been modified from:<\/span><span style=\"color: #ff0000\"> <a href=\"https:\/\/commons.wikimedia.org\/w\/index.php?curid=1394171\" style=\"color: #ff0000\">EEOC<\/a> <\/span><span style=\"color: #000000\">and<\/span> <a href=\"https:\/\/commons.wikimedia.org\/w\/index.php?title=File:Pituitary_gland_image.png&amp;oldid=63617876\"><span style=\"color: #ff0000\">Wikimedia Commons<\/span><\/a>.<\/p>\n<hr \/>\n<h4><strong><span style=\"color: #ff0000\">Hemostasis<\/span><\/strong><\/h4>\n<p><span style=\"color: #000000\"><em><strong>Hemostasis<\/strong><\/em> is the process whereby bleeding is halted. A major part of this is the coagulation cascade highlighted in Figure 8.2.<\/span><\/p>\n<p><span style=\"color: #000000\">Platelet accumulation causes blood clotting in response to a break or tear in the lining of blood vessels. Unlike the majority of control mechanisms in human body, the hemostasis utilizes positive feedback, for the more the clot grows, the more clotting occurs, until the blood stops.<\/span><\/p>\n<h4 class=\"paragraph\"><strong><span style=\"color: #ff0000\">Sleep<\/span><\/strong><\/h4>\n<p class=\"paragraph\"><span style=\"color: #000000\">Sleep timing depends upon a balance between<em><strong> homeostatic sleep propensity<\/strong><\/em>, the need for sleep as a function of the amount of time elapsed since the last adequate sleep episode, and<strong><em> circadian rhythms<\/em> <\/strong>which determine the ideal timing of a correctly structured and restorative sleep episode. A sleep deficit will elicit a compensatory increase in the intensity and duration of sleep, while excessive sleep reduces sleep propensity.<\/span><\/p>\n<p><a href=\"#title\"><span style=\"color: #ff0000\"><em><strong><span style=\"color: #ff0000\">(Back to the Top)<\/span><\/strong><\/em><\/span><\/a><\/p>\n<hr \/>\n<h3 id=\"disease\"><strong><span style=\"color: #ff0000\">8.2 Disease as a Homeostatic Imbalance<\/span><\/strong><\/h3>\n<h4><span style=\"color: #ff0000\"><strong>What Is Disease?<\/strong><\/span><\/h4>\n<p><span style=\"color: #000000\">Disease is any failure of normal physiological function that leads to negative symptoms. While disease is often a result of infection or injury, most diseases involve the disruption of normal homeostasis. Anything that prevents positive or negative feedback system from working correctly could lead to disease if the mechanisms of disruption become strong enough.<\/span><\/p>\n<p><span style=\"color: #000000\">Aging is a general example of\u00a0disease as a result of homeostatic imbalance. As an organism ages, weakening of feedback loops gradually results in an unstable internal environment. This lack of homeostasis increases the risk for illness and is responsible for the physical changes associated with aging. Heart failure is the result of negative feedback mechanisms that become overwhelmed, allowing destructive positive feedback mechanisms to compensate for the failed feedback mechanisms. This leads to high blood pressure and enlargement of the heart, which eventually becomes too stiff to pump blood effectively, resulting in heart failure. Severe heart failure can be fatal.<\/span><\/p>\n<h4><strong><span style=\"color: #ff0000\">Diabetes: A Disease of Failed Homeostasis<\/span><\/strong><\/h4>\n<p><span style=\"color: #000000\">Diabetes, a metabolic disorder caused by excess blood glucose levels, is a key example of disease caused by failed homeostasis. In ideal circumstances, homeostatic control mechanisms should prevent this imbalance from occurring. However, in some people, the mechanisms do not work efficiently enough or the amount of blood glucose is too great to be effectively managed. In these cases, medical intervention is necessary to restore homeostasis and prevent permanent organ damage.<\/span><\/p>\n<h4><span style=\"color: #ff0000\"><strong>Normal Blood Sugar Regulation<\/strong><\/span><\/h4>\n<p><span style=\"color: #000000\">The human body maintains constant levels of glucose throughout the day. After a meal, blood glucose levels rise, as glucose is transported from the small intestine into the blood stream. In response to this, the pancreas (the sensor) releases insulin into the bloodstream where it acts as a <strong><em>hormone<\/em><\/strong>. As you learned in Chapter 6, <strong><em>hormones<\/em><\/strong> are molecules that are made in one part of the body, secreted into the bloodstream and are transported to a distant part of the body, where they mediate an effect or reaction at that secondary target. Insulin is a peptide hormone that is released by the pancreas in response to elevated levels of blood glucose. Insulin binds with high efficiency to receptor proteins on the surface of liver cells, where it turns on signaling within the liver to increase the uptake of glucose from the bloodstream (Figure 8.4). Other body cells, such as skeletal muscle, adipose tissue, and brain cells are also activated by insulin. When a molecule has multiple different effects on the body, these multiple effects are called<strong><em> pleiotropic effects.<\/em><\/strong> These other cell types will also take up glucose to use as an energy source.\u00a0 This lowers blood glucose levels back to normal levels. The liver can take up more glucose than other tissue types and convert it into a large carbohydrate molecule called glycogen, that you learned about in Chapter 6. It is stored as this carbohydrate until glucose is needed when it can then be broken down to released back into the blood stream. Up to 10% of the volume in liver cells is in the form of glycogen.<\/span><\/p>\n<p><a href=\"https:\/\/wou.edu\/chemistry\/files\/2019\/01\/insulin-glucagon-model.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/wou.edu\/chemistry\/files\/2019\/01\/insulin-glucagon-model.png\" alt=\"\" class=\"alignnone wp-image-4561\" width=\"695\" height=\"460\" srcset=\"https:\/\/wou.edu\/chemistry\/files\/2019\/01\/insulin-glucagon-model.png 923w, https:\/\/wou.edu\/chemistry\/files\/2019\/01\/insulin-glucagon-model-300x199.png 300w, https:\/\/wou.edu\/chemistry\/files\/2019\/01\/insulin-glucagon-model-768x508.png 768w\" sizes=\"(max-width: 695px) 100vw, 695px\" \/><\/a><\/p>\n<p><span style=\"color: #000000\"><strong>Figure 8.4 Glucose Homeostasis.<\/strong> When blood sugar rises due to a meal (Path 1), the pancreas senses the increase in blood glucose levels. In response, it releases the peptide hormone, insulin. Insulin interacts with downstream target cells in the body, including liver and muscle tissue, where it causes the uptake of glucose from the blood stream into the cell. The excess glucose is stored as the carbohydrate, glycogen. This returns blood glucose levels back to normal.\u00a0 If it has been several hours after eating a mean, blood glucose levels will begin to fall (Path 2). This signals liver cells to breakdown glycogen into glucose monomers. The glucose can then be realeased back into the bloodstream.<\/span><\/p>\n<p><span style=\"color: #000000\">Figure is by:<\/span> <a href=\"https:\/\/www.biologycorner.com\/\" rel=\"cc:attributionURL\">Shannan Muskopf<\/a> <span style=\"color: #000000\">from<\/span> <a href=\"https:\/\/www.biologycorner.com\/\">Biologycorner.com<\/a><\/p>\n<hr \/>\n<p><span style=\"color: #000000\">In between meals or during times of fasting, blood glucose levels begin to drop.\u00a0 This activates the pancreas to secrete a different hormone, called glucagon. Glucagon signaling activates the liver to begin breaking down the glycogen storage molecule into free glucose. The glucose is then released back into the blood stream, increasing blood glucose levels (Figure 8.4).<\/span><\/p>\n<p><span style=\"color: #000000\">Over the course of a day, blood glucose levels will fluctuate modestly around the homeostatic set point (Figure 8.5). As meals are eaten, this triggers a rise in blood glucose that is counteracted by the secretion of insulin. In between meals, blood glucose levels fall, and glucagon is released by the pancreas to signal to the liver to release glucose back into the blood stream.<\/span><\/p>\n<div class=\"wp-caption aligncenter\">\n<div class=\"figure-cont\">\n<p><img decoding=\"async\" src=\"https:\/\/textimgs.s3.amazonaws.com\/boundless-anatomy-and-physiology\/glucose-day-english.svg#fixme\" alt=\"Graph depicting glucose metabolism over the course of a day. The x-axis is time, with labels for each hour of the day (and indications of when breakfast, lunch, and dinner occurred). The y-axis is a measure of blood glucose in milligrams per deciliter. The trend lines show a spike in blood glucose immediately after each meal.\" \/><\/p>\n<p class=\"wp-caption-text\"><span style=\"color: #000000\"><strong>Figure 8.5. Homeostasis of Glucose Metabolism<\/strong>: This image illustrates glucose metabolism over the course of a day. Homeostasis may become imbalanced if the pancreas is overly stressed, making it unable to balance glucose metabolism. This can lead to diabetes.<\/span><\/p>\n<hr \/>\n<h4><span style=\"color: #ff0000\"><strong>Causes of Homeostatic Disruption<\/strong><\/span><\/h4>\n<p><span style=\"color: #000000\">People with type 1 diabetes do not produce insulin due to auto-immune destruction of the insulin producing cells, while people with type 2 diabetes have chronic high blood glucose levels that cause insulin resistance to develop.\u00a0 With diabetes, blood glucose is increased by normal glucagon activity, but the lack of or resistance to insulin means that blood sugar levels are unable to return to normal. This causes metabolic changes that result in diabetes symptoms like weakened blood vessels and frequent urination. Diabetes is normally treated with insulin injections, which replaces the missing negative feedback of normal insulin secretions. If diabetes is left untreated or becomes resistant to treatment, more serious side effects are seen, including peripheral neuropathy (loss of feeling in the extremities), loss of circulation in the extremities, blurred vision and\/or blindness.<\/span><\/p>\n<\/div>\n<\/div>\n<\/header>\n<header>\n<p class=\"paragraph\"><span style=\"color: #000000\">Overall,<strong> Diabetes<\/strong> is a disease caused by a broken feedback loop involving the hormone insulin. The broken feedback loop makes it difficult or impossible for the body to bring high blood sugar down to a healthy level.<\/span><\/p>\n<p><a href=\"#title\"><span><em><strong><span style=\"color: #ff0000\">(Back to the Top)<\/span><\/strong><\/em><\/span><\/a><\/p>\n<hr \/>\n<h3 id=\"7intro\"><span><strong>8.3. Measuring Homeostasis to Evaluate Health<br \/>\n<\/strong><\/span><\/h3>\n<p><span style=\"color: #000000\">Since homeostatic imbalances can lead to disease states or even death, homeostasis has been identified as one of the eight core concepts of biology. The American Association of Medical Colleges reports that a physicians ability to identify and apply knowledge about homeostasis should be regarded as one of their key competencies. Thus, physicians need a way to evaluate the homeostatic health of their patients. They need to be able to evaluate the mixtures of compounds that are found within the human body.<br \/>\n<\/span><\/p>\n<p><span style=\"color: #000000\"><span>Recall that in Chapter 2, 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. Also recall that mixtures can be of two types: <strong><em>Homogeneous and Heterogeneous<\/em><\/strong>, where <strong><em>homogeneous mixtures<\/em><\/strong> combine so intimately that they are observed as a single substance, even though they are not. <strong><em>Heterogeneous mixtures<\/em><\/strong>, on the other hand, are non-uniform and have regions of the mixture that look different from other regions of the mixture. <strong><em>Homogeneous mixtures<\/em><\/strong> can be further broken down into two classifications: <strong><em>Colloids and Solutions<\/em><\/strong>. A <strong><em>colloid<\/em> <\/strong>is a mixture that contains particles with diameters ranging from 2 to 500 nm. <strong><em>Colloids<\/em><\/strong> appear uniform in nature and have the same composition throughout but are cloudy or opaque. Blood is a good example of a colloid. <strong><em>True solutions<\/em><\/strong>, on the other hand, 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. The remaining sections of this chapter will focus on the characteristics of true solutions.<\/span><\/span><\/p>\n<p><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\"><span>The major component of the <strong><em>solution<\/em><\/strong> is called the <em><strong>solvent<\/strong><\/em>, and the minor component(s) are called the<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 <strong><em>solute<\/em><\/strong>. 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><span>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<h4><a href=\"#title\"><span><em><strong><span style=\"color: #ff0000\">(Back to the Top)<\/span><\/strong><\/em><\/span><\/a><\/h4>\n<div class=\"mt-section\" id=\"section_1\">\n<hr \/>\n<h4 id=\"typessolutions\" class=\"editable\"><strong>Types of Solutions<\/strong><span><\/span><\/h4>\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><span style=\"color: #ff0000\">(Back to the Top)<\/span><\/strong><\/em><\/span><\/a><\/h4>\n<\/div>\n<div class=\"mt-section\" id=\"section_2\">\n<hr \/>\n<h3 id=\"solubility2\" class=\"editable\"><strong>8.4 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><span style=\"color: #000000\"><a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/solution-process.png\" style=\"color: #000000\"><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><\/span><\/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-4362-1\" 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><em><span style=\"color: #000000\">Video showing the crystallization of a supersaturated solution of sodium acetate. Video by :<\/span> <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>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 style=\"color: #000000\"><strong>Figure 8.6 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><\/p>\n<hr \/>\n<p><span><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> <\/span><\/p>\n<h4><span><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><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 8.7 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.<\/span><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><span style=\"color: #000000\"><strong>Figure 8.7: 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<\/span> <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. <a href=\"https:\/\/wou.edu\/chemistry\/files\/2017\/05\/solubility_rules_2.png\" style=\"color: #000000\"><\/a><\/span><\/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<p><a href=\"#title\"><span><em><strong><span style=\"color: #ff0000\">(Back to the Top)<\/span><\/strong><\/em><\/span><\/a><\/p>\n<\/div>\n<hr \/>\n<h3 id=\"solutionconcentration\"><strong><span>8.5 Solution Concentration<\/span><\/strong><\/h3>\n<p class=\"para editable block\" id=\"averill_1.0-ch04_s02_p01\"><span><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\" style=\"color: #000000\">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.<\/span><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>8.5.1 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\" style=\"color: #000000\"><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>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><strong>Preparation of Solutions<\/strong><\/span><\/h4>\n<\/div>\n<p><span style=\"color: #000000\">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:<\/span><\/p>\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<p><span style=\"color: #000000\">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.<\/span><\/p>\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 8.8 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><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\" style=\"color: #000000\"><strong>Figure 8.8: 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. \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>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\">2<\/sub>\u00b72H<sub class=\"subscript\">2<\/sub>O, in enough ethanol to make exactly 500 mL of solution. What is the molar concentration of CoCl<sub class=\"subscript\">2<\/sub>\u00b72H<sub class=\"subscript\">2<\/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\">2<\/sub>\u00b72H<sub class=\"subscript\">2<\/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<p><span style=\"color: #000000\">\u00a0\u00a0\u00a0 3. Plug values into the Molarity equation:<\/span><\/p>\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><a href=\"#title\"><span><em><strong><span style=\"color: #ff0000\">(Back to the Top)<\/span><\/strong><\/em><\/span><\/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<hr \/>\n<\/div>\n<h4 id=\"partspermillion\"><span style=\"color: #000000\"><strong>8.5.2 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>. 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><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><span style=\"color: #000000\"><strong>Solution = Solute + Solvent<\/strong><\/span><\/p>\n<p><span style=\"color: #000000\">Thus, we can fill in the values and then solve for the unknown.<\/span><\/p>\n<p><span style=\"color: #000000\"><strong>100 mL = 7 mL + X mL of Solvent (in this case water)<\/strong><\/span><\/p>\n<p><span style=\"color: #000000\">shifting the 7 over to the other side, we can see that:<\/span><\/p>\n<p><span style=\"color: #000000\"><strong>100 mL &#8211; 7 mL\u00a0 = 93 mL H<sub>2<\/sub>O<\/strong><\/span><\/p>\n<h4><span><strong><em>Example 2<\/em> <\/strong><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><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><a href=\"https:\/\/wou.edu\/chemistry\/files\/2018\/02\/example-3.png\"><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 8.9).\u00a0<\/span><\/p>\n<figure id=\"CNX_Chem_03_05_faucet\"><figcaption>\n<div class=\"wp-caption alignnone\">\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 8.9.<\/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<p class=\"para editable block\" id=\"averill_1.0-ch04_s02_s02_p13\"><span><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><br \/>\n<\/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 id=\"equivalents\" class=\"editable\"><span style=\"color: #000000\"><strong>8.5.3 Equivalents<\/strong><\/span><\/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><span style=\"color: #ff0000\">(Back to the Top)<\/span><\/strong><\/em><\/span><\/a><\/h4>\n<hr \/>\n<h3 id=\"dilutions\" class=\"para editable block\"><span><strong>8.6 Dilutions<\/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><span style=\"color: #000000\">The procedure for preparing a solution of known concentration from a stock solution is shown in Figure 8.10. 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:<\/span><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><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><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 8.10<\/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\">s<\/sub>) containing the desired amount of solute (M<sub class=\"subscript\">s<\/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\">s<\/sub>)(M<sub class=\"subscript\">s<\/sub>) = (<em class=\"emphasis\">V<\/em><sub class=\"subscript\">d<\/sub>)(M<sub class=\"subscript\">d<\/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><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 style=\"color: #ff0000\"><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><strong>8.7 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<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup> ions and one Cr<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">7<\/sub><sup class=\"superscript\">2\u2212<\/sup> ion per formula unit. Like other ionic compounds, it is a strong electrolyte that dissociates in aqueous solution to give hydrated NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup> and Cr<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">7<\/sub><sup class=\"superscript\">2\u2212<\/sup> 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 NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup> ions and the one 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 8.11).\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.<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 8.11 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><span><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\">4<\/sub>)<sub class=\"subscript\">2<\/sub>Cr<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">7<\/sub>, then the concentration of Cr<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">7<\/sub><sup class=\"superscript\">2\u2212<\/sup> must also be 1.43 M because there is one Cr<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">7<\/sub><sup class=\"superscript\">2\u2212<\/sup> ion per formula unit. However, there are two NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup> ions per formula unit, so the concentration of NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup> ions is 2\u00a0\u00d7\u00a01.43 M = 2.86 M. Because each formula unit of (NH<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub>Cr<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">7<\/sub> produces <em class=\"emphasis\">three<\/em> ions when dissolved in water (2NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup>\u00a0+\u00a01Cr<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">7<\/sub><sup class=\"superscript\">2\u2212<\/sup>), 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 (NH<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub>Cr<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">7<\/sub> can then be calculated by dividing 1.43 M by 4.29 M, yielding 0.333 equivalents. Thus, for (NH<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub>Cr<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">7<\/sub>, dissolving 0.333 moles of the compound will yield 1 mole of ions in the solution.<\/span><br \/>\n<\/span><\/p>\n<div class=\"exercises block\" id=\"averill_1.0-ch04_s02_s03_n01\">\n<h4 class=\"title\"><strong><em><span>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\">3<\/sub>)CHOH<\/span><\/li>\n<li><span style=\"color: #000000\">0.032 M In(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">3<\/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><\/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\"><span style=\"color: #000000\"><strong>1.\u00a0\u00a0 0.21 M NaOH<\/strong><\/span><\/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\">+<\/sup> ion and one OH<sup class=\"superscript\">\u2212<\/sup> ion, the concentration of each ion is the same as the concentration of NaOH: [Na<sup class=\"superscript\">+<\/sup>] = 0.21 M and [OH<sup class=\"superscript\">\u2212<\/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\">3<\/sub>)<sub class=\"subscript\">2<\/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\">3<\/sub>)<sub class=\"subscript\">2<\/sub>CHOH molecules, so [(CH<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/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\">3+<\/sup> ions and NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup> 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\">3<\/sub>)<sub class=\"subscript\">3<\/sub> produces one In<sup class=\"superscript\">3+<\/sup> ion and three NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup> ions, so a 0.032 M In(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">3<\/sub> solution contains 0.032 M In<sup class=\"superscript\">3+<\/sup> and 3\u00a0\u00d7\u00a00.032 M = 0.096 M NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2013<\/sup>\u2014that is, [In<sup class=\"superscript\">3+<\/sup>] = 0.032 M and [NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>] = 0.096 M<\/span><\/p>\n<h4><a href=\"#title\"><span><em><strong><span style=\"color: #ff0000\">(Back to the Top)<\/span><\/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=\"movement\"><strong>8.8 Movement of Molecules Across the Membrane<\/strong><\/h3>\n<section id=\"fs-id1516545\" class=\" focusable\">\n<p id=\"fs-id1332770\"><span style=\"color: #000000\">One of the great wonders of the cell membrane is its ability to regulate the concentration of substances inside the cell. These substances include ions such as Ca<sup>++<\/sup>, Na<sup>+<\/sup>, K<sup>+<\/sup>, and Cl<sup>\u2013<\/sup>; nutrients including sugars, fatty acids, and amino acids; and waste products, particularly carbon dioxide (CO<sub>2<\/sub>), which must leave the cell.<\/span><\/p>\n<p id=\"fs-id1667456\"><span style=\"color: #000000\">The membrane\u2019s lipid bilayer structure provides the first level of control. The phospholipids are tightly packed together, and the membrane has a hydrophobic interior. This structure causes the membrane to be selectively permeable. A membrane that has <em><strong>selective permeability<\/strong> <\/em>allows only substances meeting certain criteria to pass through it unaided. In the case of the cell membrane, only relatively small, nonpolar materials can move through the lipid bilayer (remember, the lipid tails of the membrane are nonpolar). Some examples of these are other lipids, oxygen and carbon dioxide gases, and alcohol. However, water-soluble materials\u2014like glucose, amino acids, and electrolytes\u2014need some assistance to cross the membrane because they are repelled by the hydrophobic tails of the phospholipid bilayer. All substances that move through the membrane do so by one of two general methods, which are categorized based on whether or not energy is required. <em><strong>Passive transport<\/strong><\/em> is the movement of substances across the membrane without the expenditure of cellular energy. In contrast, <em><strong>active transport<\/strong><\/em> is the movement of substances across the membrane using energy from adenosine triphosphate (ATP). You have seen examples of these types of transport mechanisms in Chapter 4, where we learned about the generation of an action potential within a neuron.<\/span><\/p>\n<section id=\"fs-id1497364\" class=\" focusable\">\n<h4><span style=\"color: #ff0000\"><strong>Passive Transport<\/strong><\/span><\/h4>\n<p id=\"fs-id1435471\"><span style=\"color: #000000\">In order to understand <em>how <\/em>substances move passively across a cell membrane, it is necessary to understand concentration gradients and diffusion. A <em><strong>concentration gradient<\/strong> <\/em>is the difference in concentration of a substance across a space. Molecules (or ions) will spread\/diffuse from where they are more concentrated to where they are less concentrated until they are equally distributed in that space. (When molecules move in this way, they are said to move<em> down<\/em> their concentration gradient.) <em><strong>Diffusion<\/strong><\/em> is the movement of particles from an area of higher concentration to an area of lower concentration. A couple of common examples will help to illustrate this concept. Imagine being inside a closed bathroom. If a bottle of perfume were sprayed, the scent molecules would naturally diffuse from the spot where they left the bottle to all corners of the bathroom, and this diffusion would go on until no more concentration gradient remains. Another example is a spoonful of sugar placed in a cup of tea. Eventually the sugar will diffuse throughout the tea until no concentration gradient remains. In both cases, if the room is warmer or the tea hotter, diffusion occurs even faster as the molecules are bumping into each other and spreading out faster than at cooler temperatures. Having an internal body temperature around 98.6<sup>\u00b0 <\/sup>F thus also aids in diffusion of particles within the body.<\/span><\/p>\n<div id=\"fs-id1240097\" class=\"note anatomy interactive\">\n<figure class=\"wp-caption aligncenter\"><figcaption class=\"wp-caption-text\">Visit this <a href=\"http:\/\/openstaxcollege.org\/l\/diffusion\">link<\/a> to see diffusion and how it is propelled by the kinetic energy of molecules in solution.\u00a0<\/figcaption><\/figure>\n<\/div>\n<p id=\"fs-id1502280\"><span style=\"color: #000000\">Whenever a substance exists in greater concentration on one side of a semipermeable membrane, such as the cell membranes, any substance that can move down its concentration gradient across the membrane will do so. Consider substances that can easily diffuse through the lipid bilayer of the cell membrane, such as the gases oxygen (O<sub>2<\/sub>) and CO<sub>2<\/sub>. O<sub>2<\/sub> generally diffuses into cells because it is more concentrated outside of them, and CO<sub>2<\/sub> typically diffuses out of cells because it is more concentrated inside of them. Neither of these examples requires any energy on the part of the cell, and therefore they use passive transport to move across the membrane.<\/span><\/p>\n<p><span style=\"color: #000000\">Before moving on, you need to review the gases that can diffuse across a cell membrane. Because cells rapidly use up oxygen during metabolism, there is typically a lower concentration of O<sub>2<\/sub> inside the cell than outside. As a result, oxygen will diffuse from the interstitial fluid directly through the lipid bilayer of the membrane and into the cytoplasm within the cell. On the other hand, because cells produce CO<sub>2<\/sub> as a byproduct of metabolism, CO<sub>2<\/sub> concentrations rise within the cytoplasm; therefore, CO<sub>2<\/sub> will move from the cell through the lipid bilayer and into the interstitial fluid, where its concentration is lower. This mechanism of molecules moving across a cell membrane from the side where they are more concentrated to the side where they are less concentrated is a form of passive transport called simple diffusion (Figure 8.12).<\/span><\/p>\n<p class=\"wp-caption alignnone\" style=\"width: 700px\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/opentextbc.ca\/anatomyandphysiology\/wp-content\/uploads\/sites\/142\/2016\/03\/0305_Simple_Diffusion_Across_Plasma_Membrane.jpg\" alt=\"This figure shows the simple diffusion of small non-polar molecules across the plasma membrane. A red horizontal arrow pointing towards the right indicates the progress of time. The nonpolar molecules are shown in blue and are present in higher numbers in the extracellular fluid. There are a few nonpolar molecules in the cytoplasm and their number increases with time.\" class=\"\" width=\"700\" height=\"293\" \/><span style=\"color: #000000\"><strong><\/strong><\/span><\/p>\n<p class=\"wp-caption aligncenter\"><span style=\"color: #000000\"><strong>Figure 8.12. Simple Diffusion across the Cell (Plasma) Membrane.<\/strong> The structure of the lipid bilayer allows small, uncharged substances such as oxygen and carbon dioxide, and hydrophobic molecules such as lipids, to pass through the cell membrane, down their concentration gradient, by simple diffusion.<br \/>\n<\/span><\/p>\n<hr \/>\n<p id=\"fs-id1126733\"><span style=\"color: #000000\">Large polar or ionic molecules, which are hydrophilic, cannot easily cross the phospholipid bilayer. Very small polar molecules, such as water, can cross via simple diffusion due to their small size. Charged atoms or molecules of any size cannot cross the cell membrane via simple diffusion as the charges are repelled by the hydrophobic tails in the interior of the phospholipid bilayer. Solutes dissolved in water on either side of the cell membrane will tend to diffuse down their concentration gradients, but because most substances cannot pass freely through the lipid bilayer of the cell membrane, their movement is restricted to protein channels and specialized transport mechanisms in the membrane. <em><strong>Facilitated diffusion<\/strong><\/em> is the diffusion process used for those substances that cannot cross the lipid bilayer due to their size, charge, and\/or polarity (Figure 8.13). A common example of facilitated diffusion is the movement of glucose into the cell, where it is used to make ATP. Although glucose can be more concentrated outside of a cell, it cannot cross the lipid bilayer via simple diffusion because it is both large and polar. To resolve this, a specialized carrier protein called the glucose transporter will transfer glucose molecules into the cell to facilitate its inward diffusion.<\/span><\/p>\n<figure id=\"fig-ch03_01_05\">\n<div class=\"title\"><\/div><figcaption><\/figcaption><p class=\"wp-caption alignnone\" style=\"width: 699px\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/opentextbc.ca\/anatomyandphysiology\/wp-content\/uploads\/sites\/142\/2016\/03\/0306_Facilitated_Diffusion.jpg\" alt=\"This diagram shows the different means of facilitated diffusion across the plasma membrane. In the top panel, a channel protein is shown to allow the transport of solutes across the membrane. In the bottom panel, the membrane contains carrier proteins in addition to channel proteins.\" class=\"\" width=\"699\" height=\"883\" \/><\/p><figcaption class=\"wp-caption-text\"><span style=\"color: #000000\"><strong>Figure 8.13. Facilitated Diffusion.<\/strong> (a) Facilitated diffusion of substances crossing the cell (plasma) membrane takes place with the help of proteins such as channel proteins and carrier proteins. Channel proteins are less selective than carrier proteins, and usually mildly discriminate between their cargo based on size and charge. (b) Carrier proteins are more selective, often only allowing one particular type of molecule to cross.<\/span><\/p>\n<hr \/>\n<\/figcaption><\/figure>\n<p id=\"fs-id1535725\"><span style=\"color: #000000\">As an example, even though sodium ions (Na<sup>+<\/sup>) are highly concentrated outside of cells, these electrolytes are charged and cannot pass through the nonpolar lipid bilayer of the membrane. Their diffusion is facilitated by membrane proteins that form sodium channels (or \u201cpores\u201d), so that Na<sup>+<\/sup> ions can move down their concentration gradient from outside the cells to inside the cells. There are many other solutes that must undergo facilitated diffusion to move into a cell, such as amino acids, or to move out of a cell, such as wastes. Because facilitated diffusion is a passive process, it does not require energy expenditure by the cell.<\/span><\/p>\n<p id=\"fs-id1506843\"><span style=\"color: #000000\">Water also can move freely across the cell membrane of all cells, either through protein channels or by slipping between the lipid tails of the membrane itself. <em><strong>Osmosis<\/strong> <\/em>is the diffusion of water through a semipermeable membrane (Figure 8.14).<\/span><\/p>\n<figure id=\"fig-ch03_01_06\">\n<div class=\"title\"><\/div><figcaption><\/figcaption><p class=\"wp-caption alignnone\" style=\"width: 704px\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/opentextbc.ca\/anatomyandphysiology\/wp-content\/uploads\/sites\/142\/2016\/03\/0307_Osmosis.jpg\" alt=\"This figure shows the diffusion of water through osmosis. The left panel shows a beaker with water and different solute concentrations. A semipermeable membrane is present in the middle of the beaker. In the right panel, the water concentration is higher to the right of the semipermeable membrane.\" class=\"\" width=\"704\" height=\"400\" \/><\/p><figcaption class=\"wp-caption-text\"><span style=\"color: #000000\"><strong>Figure 8.14. Osmosis.<\/strong> Osmosis is the diffusion of water through a semipermeable membrane down its concentration gradient. If a membrane is permeable to water, though not to a solute, water will equalize its own concentration by diffusing to the side of lower water concentration (and thus the side of higher solute concentration). In the beaker on the left, the solution on the right side of the membrane is hypertonic.<\/span><\/p>\n<hr \/>\n<\/figcaption><\/figure>\n<p><span style=\"color: #000000\">The movement of water molecules is not itself regulated by cells, so it is important that cells are exposed to an environment in which the concentration of solutes outside of the cells (in the extracellular fluid) is equal to the concentration of solutes inside the cells (in the cytoplasm). <em><strong class=\"b ph\">T<\/strong><strong class=\"b ph\">onicity <\/strong><\/em>is used to describe the variations of solute in a solution with the solute inside the cell.\u00a0 Three terms\u2014<strong><em>hypotonic, isotonic, and hypertonic<\/em><\/strong>\u2014are used to compare the relative solute concentration of a cell to that of the extracellular fluid surrounding the cells. <\/span><\/p>\n<p><span style=\"color: #000000\">In a <strong class=\"b ph\"><em>hypotonic<\/em>\u00a0<\/strong><em><strong>solution<\/strong><\/em>, such as tap water, the extracellular fluid has a lower concentration of solutes than the fluid inside the cell, and water enters the cell. (Note that water is moving down its concentration gradient) If this occurs in an animal cell, the cell may burst, or <strong><em>lyse<\/em><\/strong>.<\/span><\/p>\n<p class=\"p\"><span style=\"color: #000000\">In a <strong><em>hypertonic\u00a0solution<\/em><\/strong> (the prefix <em>hyper<\/em>&#8211; refers to the extracellular fluid having a higher concentration of solutes than the cell\u2019s cytoplasm), the fluid contains less water than the cell does. Because the cell has a lower concentration of solutes, the water will leave the cell. In effect, the solute is drawing the water out of the cell. This may cause an animal cell to shrivel, or <strong><em>crenate<\/em><\/strong>.<\/span><\/p>\n<p class=\"p\"><span style=\"color: #000000\">In an <strong><em>isotonic\u00a0solution<\/em><\/strong>, the extracellular fluid has the same solute concentration as the cell. If the concentration of solutes of the cell matches that of the extracellular fluid, there will be no net movement of water into or out of the cell. <\/span><\/p>\n<p class=\"p\"><span style=\"color: #000000\">Blood cells in hypertonic, isotonic, and hypotonic solutions take on characteristic appearances as shown in Figure 8.15. A critical aspect of homeostasis in living things is to create an internal environment in which all of the body\u2019s cells are in an isotonic solution. Various organ systems, particularly the kidneys, work to maintain this homeostasis.<\/span><\/p>\n<figure id=\"fig-ch03_01_07\">\n<div class=\"title\"><\/div><figcaption><\/figcaption><p class=\"wp-caption alignnone\" style=\"width: 701px\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/opentextbc.ca\/anatomyandphysiology\/wp-content\/uploads\/sites\/142\/2016\/03\/0346_Concentration_of_Solutions.jpg\" alt=\"This image shows how a red blood cell responds to the tonicity of solution. The left panel shows the hypertonic case, the middle panel shows the isotonic case and the right panel shows the hypotonic case.\" class=\"\" width=\"701\" height=\"363\" \/><\/p><figcaption class=\"wp-caption-text\"><span style=\"color: #000000\"><strong>Figure 8.15. States of Tonicity.<\/strong> A hypertonic solution has a solute concentration higher than another solution. An isotonic solution has a solute concentration equal to another solution. A hypotonic solution has a solute concentration lower than another solution.<\/span><\/p>\n<hr \/>\n<\/figcaption><\/figure>\n<p class=\"p\"><span style=\"color: #000000\">Some organisms, such as plants, fungi, bacteria, and some protists, have cell walls that surround the plasma membrane and prevent cell lysis. The plasma membrane can only expand to the limit of the cell wall, so the cell will not lyse. In fact, the cytoplasm in plants is always slightly hypertonic compared to the cellular environment, and water will always enter a cell if water is available. This influx of water produces turgor pressure, which stiffens the cell walls of the plant (Figure 8.16). In nonwoody plants, turgor pressure supports the plant. If the plant cells become hypertonic, as occurs in drought or if a plant is not watered adequately, water will leave the cell. Plants lose turgor pressure in this condition and wilt.<\/span><\/p>\n<div class=\"wrapper-div\">\n<div class=\"fig fignone\" id=\"fig_d41e71\">\n<div class=\"imagearea\"><img loading=\"lazy\" decoding=\"async\" alt=\"media\/image26.png\" src=\"http:\/\/www.opentextbooks.org.hk\/system\/files\/resource\/34\/34471\/34633\/media\/image26.png\" class=\"\" width=\"703\" height=\"243\" \/><\/div>\n<p class=\"figcap\"><strong><span style=\"color: #000000\"><span class=\"fignum\">Figure 8.16 <\/span>The turgor pressure within a plant cell depends on the tonicity of the surrounding solution.<\/span><\/strong><\/p>\n<hr \/>\n<\/div>\n<\/div>\n<p id=\"fs-id1307738\"><span style=\"color: #000000\">Another mechanism besides diffusion to passively transport materials between compartments is filtration. Unlike diffusion of a substance from where it is more concentrated to less concentrated, filtration uses a hydrostatic pressure gradient that pushes the fluid\u2014and the solutes within it\u2014from a higher pressure area to a lower pressure area. Filtration is an extremely important process in the body. For example, the circulatory system uses filtration to move plasma and substances across the endothelial lining of capillaries and into surrounding tissues, supplying cells with the nutrients. Furthermore, filtration pressure in the kidneys provides the mechanism to remove wastes from the bloodstream.<\/span><\/p>\n<p><a href=\"#title\"><span><em><strong><span style=\"color: #ff0000\">(Back to the Top)<\/span><\/strong><\/em><\/span><\/a><\/p>\n<hr \/>\n<\/section>\n<section id=\"fs-id850990\" class=\" focusable\">\n<h4><span style=\"color: #ff0000\"><strong>Active Transport<\/strong><\/span><\/h4>\n<p id=\"fs-id1666810\"><span style=\"color: #000000\">For all of the transport methods described above, the cell expends no energy. Membrane proteins that aid in the passive transport of substances do so without the use of ATP. During <em><strong>active transport,<\/strong><\/em> ATP is required to move a substance across a membrane, often with the help of protein carriers, and usually <strong><em>against<\/em><\/strong> its concentration gradient.<\/span><\/p>\n<p id=\"fs-id1465738\"><span style=\"color: #000000\">One of the most common types of active transport involves proteins that serve as pumps. The word \u201cpump\u201d probably conjures up thoughts of using energy to pump up the tire of a bicycle or a basketball. Similarly, energy from ATP is required for these membrane proteins to transport substances\u2014molecules or ions\u2014across the membrane, usually against their concentration gradients (from an area of low concentration to an area of high concentration).<\/span><\/p>\n<p id=\"fs-id2056316\"><span style=\"color: #000000\">The <em><strong>sodium-potassium pump<\/strong><\/em>, which is also called Na<sup>+<\/sup>\/K<sup>+<\/sup> ATPase, transports sodium out of a cell while moving potassium into the cell. The Na<sup>+<\/sup>\/K<sup>+<\/sup> pump is an important ion pump found in the membranes of many types of cells. These pumps are particularly abundant in nerve cells, which are constantly pumping out sodium ions and pulling in potassium ions to maintain an electrical gradient across their cell membranes. An <em><strong>electrical gradient<\/strong> <\/em>is a difference in electrical charge across a space. In the case of nerve cells, for example, the electrical gradient exists between the inside and outside of the cell, with the inside being negatively-charged (at around -70 mV) relative to the outside. The negative electrical gradient is maintained because each Na<sup>+<\/sup>\/K<sup>+<\/sup> pump moves three Na<sup>+<\/sup> ions out of the cell and two K<sup>+<\/sup> ions into the cell for each ATP molecule that is used (Figure 8.17). This process is so important for nerve cells that it accounts for the majority of their ATP usage.<\/span><\/p>\n<figure id=\"fig-ch03_01_08\">\n<div class=\"title\"><\/div><figcaption><\/figcaption><p class=\"wp-caption alignnone\" style=\"width: 706px\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/opentextbc.ca\/anatomyandphysiology\/wp-content\/uploads\/sites\/142\/2016\/03\/0308_Sodium_Potassium_Pump.jpg\" alt=\"This diagram shows many sodium potassium pumps embedded in the membrane. Potassium is pumped into the cytoplasm and sodium is pumped out of the cytoplasm.\" class=\"\" width=\"706\" height=\"335\" \/><\/p><figcaption class=\"wp-caption-text\"><span style=\"color: #000000\"><strong>Figure 8.17. Sodium-Potassium Pump.<\/strong> The sodium-potassium pump is found in many cell (plasma) membranes. Powered by ATP, the pump moves sodium and potassium ions in opposite directions, each against its concentration gradient. In a single cycle of the pump, three sodium ions are extruded from and two potassium ions are imported into the cell.<\/span><\/p>\n<hr \/>\n<\/figcaption><\/figure>\n<p id=\"eip-152\"><span style=\"color: #000000\">Active transport pumps can also work together with other active or passive transport systems to move substances across the membrane. For example, the sodium-potassium pump maintains a high concentration of sodium ions outside of the cell. Therefore, if the cell needs sodium ions, all it has to do is open a passive sodium channel, as the concentration gradient of the sodium ions will drive them to diffuse into the cell. In this way, the action of an active transport pump (the sodium-potassium pump) powers the passive transport of sodium ions by creating a concentration gradient. When active transport powers the transport of another substance in this way, it is called secondary active transport.<\/span><\/p>\n<p id=\"eip-214\"><span style=\"color: #000000\"><strong><em>Symporters<\/em><\/strong> are secondary active transporters that move two substances in the same direction. For example, the sodium-glucose symporter uses sodium ions to \u201cpull\u201d glucose molecules into the cell. Because cells store glucose for energy, glucose is typically at a higher concentration inside of the cell than outside. However, due to the action of the sodium-potassium pump, sodium ions will easily diffuse into the cell when the symporter is opened. The flood of sodium ions through the symporter provides the energy that allows glucose to move through the symporter and into the cell, against its concentration gradient.<\/span><\/p>\n<p id=\"eip-677\"><span style=\"color: #000000\">Conversely, <strong><em>antiporters<\/em><\/strong> are secondary active transport systems that transport substances in opposite directions. For example, the sodium-hydrogen ion antiporter uses the energy from the inward flood of sodium ions to move hydrogen ions (H+) out of the cell. The sodium-hydrogen antiporter is used to maintain the pH of the cell\u2019s interior.<\/span><\/p>\n<p id=\"fs-id1500925\"><span style=\"color: #000000\">Other forms of active transport do not involve membrane carriers. <em><strong>Endocytosis<\/strong><\/em> (bringing \u201cinto the cell\u201d) is the process of a cell ingesting material by enveloping it in a portion of its cell membrane, and then pinching off that portion of membrane (Figure 8.18). Once pinched off, the portion of membrane and its contents becomes an independent, intracellular vesicle. A <em><strong>vesicle<\/strong><\/em> is a membranous sac\u2014a spherical and hollow organelle bounded by a lipid bilayer membrane. Endocytosis often brings materials into the cell that must to be broken down or digested. <em><strong>Phagocytosis<\/strong><\/em> (\u201ccell eating\u201d) is the endocytosis of large particles. Many immune cells engage in phagocytosis of invading pathogens. Like little Pac-men, their job is to patrol body tissues for unwanted matter, such as invading bacterial cells, phagocytize them, and digest them. In contrast to phagocytosis, <em><strong>pinocytosis<\/strong><\/em> (\u201ccell drinking\u201d) brings fluid containing dissolved substances into a cell through membrane vesicles.<\/span><\/p>\n<figure id=\"fig-ch03_01_09\" class=\"span-all\">\n<div class=\"title\"><\/div><figcaption><\/figcaption><p class=\"wp-caption alignnone\" style=\"width: 701px\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/opentextbc.ca\/anatomyandphysiology\/wp-content\/uploads\/sites\/142\/2016\/03\/0309_Three_Forms_of_Endocytosis.jpg\" alt=\"This image shows the three different types of endocytosis. The left panel shows phagocytosis, where a large particle is seen to be engulfed by the membrane into a vacuole. In the middle panel, pinocytosis is shown, where a small particle is engulfed into a vesicle. In the right panel, receptor-mediated endocytosis is shown; the ligand binds to the receptor and is then engulfed into a coated vesicle.\" class=\"\" width=\"701\" height=\"319\" \/><\/p><figcaption class=\"wp-caption-text\"><span style=\"color: #000000\"><strong>Figure 8.18. Three Forms of Endocytosis.<\/strong> Endocytosis is a form of active transport in which a cell envelopes extracellular materials using its cell membrane. (a) In phagocytosis, which is relatively nonselective, the cell takes in a large particle. (b) In pinocytosis, the cell takes in small particles in fluid. (c) In contrast, receptor-mediated endocytosis is quite selective. When external receptors bind a specific ligand, the cell responds by endocytosing the ligand.<\/span><\/p>\n<hr \/>\n<\/figcaption><\/figure>\n<p id=\"fs-id1097508\"><span style=\"color: #000000\">Phagocytosis and pinocytosis take in large portions of extracellular material, and they are typically not highly selective in the substances they bring in. Cells regulate the endocytosis of specific substances via receptor-mediated endocytosis. <strong>Receptor-mediated endocytosis<\/strong> is endocytosis by a portion of the cell membrane that contains many receptors that are specific for a certain substance. Once the surface receptors have bound sufficient amounts of the specific substance (the receptor\u2019s ligand), the cell will endocytose the part of the cell membrane containing the receptor-ligand complexes. Iron, a required component of hemoglobin, is endocytosed by red blood cells in this way. Iron is bound to a protein called transferrin in the blood. Specific transferrin receptors on red blood cell surfaces bind the iron-transferrin molecules, and the cell endocytoses the receptor-ligand complexes.<\/span><\/p>\n<p id=\"fs-id1171152\"><span style=\"color: #000000\">In contrast with endocytosis, <strong>exocytosis<\/strong> (taking \u201cout of the cell\u201d) is the process of a cell exporting material using vesicular transport (Figure 8.19). Many cells manufacture substances that must be secreted, like a factory manufacturing a product for export. These substances are typically packaged into membrane-bound vesicles within the cell. When the vesicle membrane fuses with the cell membrane, the vesicle releases it contents into the interstitial fluid. The vesicle membrane then becomes part of the cell membrane. Cells of the stomach and pancreas produce and secrete digestive enzymes through exocytosis (Figure 8.20). Endocrine cells produce and secrete hormones that are sent throughout the body, and certain immune cells produce and secrete large amounts of histamine, a chemical important for immune responses.<\/span><\/p>\n<figure id=\"fig-ch03_01_10\">\n<div class=\"title\"><\/div><figcaption><\/figcaption><p class=\"wp-caption alignnone\" style=\"width: 700px\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/opentextbc.ca\/anatomyandphysiology\/wp-content\/uploads\/sites\/142\/2016\/03\/0310_Exocytosis.jpg\" alt=\"This figure shows the process of exocytosis. A vesicle is shown fusing with the membrane and then releasing its contents into the extracellular fluid.\" class=\"\" width=\"700\" height=\"787\" \/><\/p><figcaption class=\"wp-caption-text\"><span style=\"color: #000000\"><strong>Figure 8.19. Exocytosis.<\/strong> Exocytosis is much like endocytosis in reverse. Material destined for export is packaged into a vesicle inside the cell. The membrane of the vesicle fuses with the cell membrane, and the contents are released into the extracellular space.<\/span><\/p>\n<hr \/>\n<\/figcaption><\/figure>\n<figure id=\"fig-ch03_01_11\">\n<div class=\"title\"><\/div><figcaption><\/figcaption><p class=\"wp-caption alignnone\" style=\"width: 703px\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/opentextbc.ca\/anatomyandphysiology\/wp-content\/uploads\/sites\/142\/2016\/03\/0311_Pancreatic_Cells_Micrograph.jpg\" alt=\"This micrograph shows the structure of a pancreatic acinar cell and the location of secretory vesicles.\" class=\"\" width=\"703\" height=\"333\" \/><\/p><figcaption class=\"wp-caption-text\"><span style=\"color: #000000\"><strong>Figure 8.20. Pancreatic Cells\u2019 Enzyme Products.<\/strong> The pancreatic acinar cells produce and secrete many enzymes that digest food. The tiny black granules in this electron micrograph are secretory vesicles filled with enzymes that will be exported from the cells via exocytosis. LM \u00d7 2900. (Micrograph provided by the Regents of University of Michigan Medical School \u00a9 2012)<\/span><\/figcaption><span style=\"color: #000000\">View the<\/span><span style=\"color: #ff0000\"> <a href=\"http:\/\/virtualslides.med.umich.edu\/Histology\/EMsmallCharts\/3%20Image%20Scope%20finals\/226%20-%20Pancreas_001.svs\/view.apml\" style=\"color: #ff0000\">University of Michigan WebScope<\/a> <\/span><span style=\"color: #000000\">to explore the tissue sample in greater detail.<\/span><\/p>\n<p><a href=\"#title\"><span><em><strong><span style=\"color: #ff0000\">(Back to the Top)<\/span><\/strong><\/em><\/span><\/a><figcaption class=\"wp-caption-text\"><span style=\"color: #000000\"><\/span><\/p>\n<hr \/>\n<\/figcaption><\/figure>\n<\/section>\n<\/section>\n<h3 id=\"7summary\" class=\"para\"><span><strong>8.9 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\"><span style=\"color: #000000\">\u00a0<\/span><\/p>\n<section id=\"fs-id1863415\" class=\"summary focusable\">\n<p id=\"fs-id1449977\"><span style=\"color: #000000\">The cell membrane provides a barrier around the cell, separating its internal components from the extracellular environment. It is composed of a phospholipid bilayer, with hydrophobic internal lipid \u201ctails\u201d and hydrophilic external phosphate \u201cheads.\u201d Various membrane proteins are scattered throughout the bilayer, both inserted within it and attached to it peripherally. The cell membrane is selectively permeable, allowing only a limited number of materials to diffuse through its lipid bilayer. All materials that cross the membrane do so using passive (non energy-requiring) or active (energy-requiring) transport processes. During passive transport, materials move by simple diffusion or by facilitated diffusion through the membrane, down their concentration gradient. Water passes through the membrane in a diffusion process called osmosis. During active transport, energy is expended to assist material movement across the membrane in a direction against their concentration gradient. Active transport may take place with the help of protein pumps or through the use of vesicles.<\/span><\/p>\n<\/section>\n<section id=\"fs-id746702\" class=\"interactive-exercise focusable\">\n<h1><span style=\"color: #000000\">Practice Questions<\/span><\/h1>\n<\/section>\n<section id=\"fs-id1495386\" class=\"multiple-choice focusable\">\n<div class=\"bcc-box bcc-info\">\n<h3>Review Questions<\/h3>\n<p><span style=\"color: #000000\">1. Because they are embedded within the membrane, ion channels are examples of ________.<\/span><\/p>\n<ol type=\"A\">\n<li><span style=\"color: #000000\">receptor proteins<\/span><\/li>\n<li><span style=\"color: #000000\">integral proteins<\/span><\/li>\n<li><span style=\"color: #000000\">peripheral proteins<\/span><\/li>\n<li><span style=\"color: #000000\">glycoproteins<\/span><\/li>\n<\/ol>\n<p><span style=\"color: #000000\">2. The diffusion of substances within a solution tends to move those substances ________ their ________ gradient.<\/span><\/p>\n<ol type=\"A\">\n<li><span style=\"color: #000000\">up; electrical<\/span><\/li>\n<li><span style=\"color: #000000\">up; electrochemical<\/span><\/li>\n<li><span style=\"color: #000000\">down; pressure<\/span><\/li>\n<li><span style=\"color: #000000\">down; concentration<\/span><\/li>\n<\/ol>\n<p><span style=\"color: #000000\">3. Ion pumps and phagocytosis are both examples of ________.<\/span><\/p>\n<ol type=\"A\">\n<li><span style=\"color: #000000\">endocytosis<\/span><\/li>\n<li><span style=\"color: #000000\">passive transport<\/span><\/li>\n<li><span style=\"color: #000000\">active transport<\/span><\/li>\n<li><span style=\"color: #000000\">facilitated diffusion<\/span><\/li>\n<\/ol>\n<p><span style=\"color: #000000\">4. Choose the answer that best completes the following analogy: Diffusion is to ________ as endocytosis is to ________.<\/span><\/p>\n<ol type=\"A\">\n<li><span style=\"color: #000000\">filtration; phagocytosis<\/span><\/li>\n<li><span style=\"color: #000000\">osmosis; pinocytosis<\/span><\/li>\n<li><span style=\"color: #000000\">solutes; fluid<\/span><\/li>\n<li><span style=\"color: #000000\">gradient; chemical energy<\/span><\/li>\n<\/ol>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3><span style=\"color: #000000\">Critical Thinking Questions<\/span><\/h3>\n<p><span style=\"color: #000000\">1. What materials can easily diffuse through the lipid bilayer, and why?<\/span><\/p>\n<p><span style=\"color: #000000\">2. Why is receptor-mediated endocytosis said to be more selective than phagocytosis or pinocytosis?<\/span><\/p>\n<p><span style=\"color: #000000\">3. What do osmosis, diffusion, filtration, and the movement of ions away from like charge all have in common? In what way do they differ?<\/span><\/p>\n<\/div>\n<\/section>\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\">\n<li><span style=\"color: #000000\">NH<sub class=\"subscript\">3<\/sub><\/span><\/li>\n<li><span style=\"color: #000000\">HF<\/span><\/li>\n<li><span style=\"color: #000000\">CH<sub class=\"subscript\">3<\/sub>CH<sub class=\"subscript\">2<\/sub>CH<sub class=\"subscript\">2<\/sub>OH<\/span><\/li>\n<li>\n<p class=\"para\"><span style=\"color: #000000\">Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub><\/span><\/p>\n<div class=\"informalfigure medium\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/2012books.lardbucket.org\/books\/principles-of-general-chemistry-v1.0\/section_08\/4b8d2b1f623560048a737b8eb7a3ade4.jpg\" class=\"\" width=\"698\" height=\"365\" \/><\/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\">\n<li><span style=\"color: #000000\">CH<sub class=\"subscript\">3<\/sub>CO<sub class=\"subscript\">2<\/sub>H<\/span><\/li>\n<li><span style=\"color: #000000\">NaCl<\/span><\/li>\n<li><span style=\"color: #000000\">Na<sub class=\"subscript\">2<\/sub>S<\/span><\/li>\n<li><span style=\"color: #000000\">Na<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/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\">2<\/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) 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.<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\">\n<li><span style=\"color: #000000\">0.2593 M NaBrO<sub class=\"subscript\">3<\/sub><\/span><\/li>\n<li><span style=\"color: #000000\">1.592 M KNO<sub class=\"subscript\">3<\/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\">\n<li><span style=\"color: #000000\">0.1065 M BaI<sub class=\"subscript\">2<\/sub><\/span><\/li>\n<li><span style=\"color: #000000\">1.135 M Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub><\/span><\/li>\n<li><span style=\"color: #000000\">1.428 M NH<sub class=\"subscript\">4<\/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\">\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\">\n<li><span style=\"color: #000000\">0.489 mol of NiSO<sub class=\"subscript\">4<\/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\">3<\/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\">\n<li><span style=\"color: #000000\">0.324 mol of K<sub class=\"subscript\">2<\/sub>MoO<sub class=\"subscript\">4<\/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\">3<\/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\">\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\">12<\/sub>H<sub class=\"subscript\">22<\/sub>O<sub class=\"subscript\">11<\/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\">\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\">4<\/sub>H<sub class=\"subscript\">6<\/sub>O<sub class=\"subscript\">6<\/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\">\n<li><span style=\"color: #000000\">BaCl<sub class=\"subscript\">2<\/sub>(aq)\u00a0+\u00a0Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) \u2192<\/span><\/li>\n<li><span style=\"color: #000000\">Ca(OH)<sub class=\"subscript\">2<\/sub>(aq)\u00a0+\u00a0H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub>(aq) \u2192<\/span><\/li>\n<li><span style=\"color: #000000\">Al(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">3<\/sub>(aq)\u00a0+\u00a0H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) \u2192<\/span><\/li>\n<li><span style=\"color: #000000\">Pb(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq)\u00a0+\u00a0CuSO<sub class=\"subscript\">4<\/sub>(aq) \u2192<\/span><\/li>\n<li><span style=\"color: #000000\">Al(CH<sub class=\"subscript\">3<\/sub>CO<sub class=\"subscript\">2<\/sub>)<sub class=\"subscript\">3<\/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\">2<\/sub>CrO<sub class=\"subscript\">4<\/sub>. A stock solution of Na<sub class=\"subscript\">2<\/sub>CrO<sub class=\"subscript\">4<\/sub> containing 20.0% solute by mass with a density of 1.19 g\/cm<sup class=\"superscript\">3<\/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\">2<\/sub>CrO<sub class=\"subscript\">4<\/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\">2<\/sub>] is an effective disinfectant for clothing and bedding. If a solution has a Ca(OCl)<sub class=\"subscript\">2<\/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\">6<\/sub>H<sub class=\"subscript\">5<\/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\">8<\/sub>H<sub class=\"subscript\">10<\/sub>N<sub class=\"subscript\">4<\/sub>O<sub class=\"subscript\">2<\/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\">\u2212<\/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\">\u22123<\/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<div class=\"mt-section\" id=\"section_6\">\n<h4><a href=\"#title\"><em><strong><span id=\"Contributors\"><span style=\"color: #ff0000\">(Back to the Top)<\/span><\/span><\/strong><\/em><\/a><\/h4>\n<hr \/>\n<h3 id=\"7refs\" class=\"editable\"><strong>8.10 References<\/strong><\/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\"><span style=\"color: #ff0000\">https:\/\/chem.libretexts.org\/Core\/Inorganic_Chemistry\/Chemical_Reactions\/Chemical_Reactions_1\/Solutions<\/span><\/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\"><span style=\"color: #ff0000\">https:\/\/chem.libretexts.org\/Textbook_Maps\/Introductory_Chemistry_Textbook_Maps\/Map%3A_The_Basics_of_GOB_Chemistry_(Ball_et_al.)<\/span><\/a><\/li>\n<li><span style=\"color: #000000\">Averill, B.A., Eldredge, P. (2012) <em>The Principles of Chemistry<\/em>. Libre Texts. Available at:\u00a0<\/span><a href=\"https:\/\/2012books.lardbucket.org\/books\/principles-of-general-chemistry-v1.0\/index.html\"><span style=\"color: #ff0000\">https:\/\/2012books.lardbucket.org\/books\/principles-of-general-chemistry-v1.0\/index.html<\/span> <\/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\"><span style=\"color: #ff0000\">https:\/\/en.wikipedia.org\/w\/index.php?title=Hydrate&amp;oldid=798015169<\/span><\/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:\u00a0<\/span><a href=\"http:\/\/www.chem1.com\/acad\/webtext\/solut\/solut-1.html\"><span style=\"color: #ff0000\"> http:\/\/www.chem1.com\/acad\/webtext\/solut\/solut-1.html<\/span><\/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> <span style=\"color: #ff0000\"><a href=\"http:\/\/cnx.org\/contents\/85abf193-2bd2-4908-8563-90b8a7ac8df6@9.311\" style=\"color: #ff0000\">http:\/\/cnx.org\/contents\/85abf193-2bd2-4908-8563-90b8a7ac8df6@9.311<\/a>.<\/span><\/span><\/div>\n<\/li>\n<li><span style=\"color: #000000\">Open Textbooks for Hong Kong. (2015) Tonicity chapter from Concepts in Biology. retrieved on Dec 31st, 2018 from<\/span> <a href=\"http:\/\/www.opentextbooks.org.hk\/ditatopic\/34633\"><span style=\"color: #ff0000\">http:\/\/www.opentextbooks.org.hk\/ditatopic\/34633<\/span><\/a><\/li>\n<li><span style=\"color: #000000\">Rice University () Section 3.1 The Cell Membrane from Anatomy and Physiology. BC Open Textbook Collection. Retreived on Dec 31st, 2018 from<\/span> <span style=\"color: #ff0000\"><a href=\"https:\/\/opentextbc.ca\/anatomyandphysiology\/chapter\/the-cell-membrane\/\" style=\"color: #ff0000\">https:\/\/opentextbc.ca\/anatomyandphysiology\/chapter\/the-cell-membrane\/<\/a><\/span><\/li>\n<li><span style=\"color: #000000\">Kahn Academy (2019) Homeostasis. In Human Body Systems.\u00a0 Retrieved on Jan 2nd, 2019 from<\/span> <a href=\"https:\/\/www.khanacademy.org\/science\/high-school-biology\/hs-human-body-systems\/hs-body-structure-and-homeostasis\/a\/homeostasis\"><span style=\"color: #ff0000\">https:\/\/www.khanacademy.org\/science\/high-school-biology\/hs-human-body-systems\/hs-body-structure-and-homeostasis\/a\/homeostasis<\/span><\/a><\/li>\n<li><span style=\"color: #000000\">Lumen Learning (2019) Homeostasis from Boundless Anatomy and Physiology. Retrieved on Jan 2nd, 2019 from<\/span> <a href=\"https:\/\/courses.lumenlearning.com\/boundless-ap\/chapter\/homeostasis\/\"><span style=\"color: #ff0000\">https:\/\/courses.lumenlearning.com\/boundless-ap\/chapter\/homeostasis\/<\/span><\/a><\/li>\n<li><span style=\"color: #000000\">Wikipedia contributors. (2019, February 9). Hypothermia. In <i>Wikipedia, The Free Encyclopedia<\/i>. Retrieved 18:47, February 25, 2019, from<\/span><span style=\"color: #ff0000\"> <a class=\"external free\" href=\"https:\/\/en.wikipedia.org\/w\/index.php?title=Hypothermia&amp;oldid=882437028\" style=\"color: #ff0000\">https:\/\/en.wikipedia.org\/w\/index.php?title=Hypothermia&amp;oldid=882437028<\/a><\/span><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<footer class=\"mt-content-footer\"><\/footer>\n<\/section>\n","protected":false},"excerpt":{"rendered":"<p>Chapter 8: Homeostasis and Cellular Function This text is published under creative commons licensing. For referencing this work, please click here. 8.1 The Concept of Homeostasis 8.2\u00a0Disease as a Homeostatic Imbalance 8.3 Measuring Homeostasis to Evaluate Health 8.4 Solubility 8.5 Solution Concentration 8.5.1 Molarity 8.5.2 Parts Per Solutions 8.5.3 Equivalents 8.6 Dilutions 8.7 Ion Concentrations [&hellip;]<\/p>\n","protected":false},"author":280,"featured_media":0,"parent":4261,"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-4362","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/wou.edu\/chemistry\/wp-json\/wp\/v2\/pages\/4362","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=4362"}],"version-history":[{"count":0,"href":"https:\/\/wou.edu\/chemistry\/wp-json\/wp\/v2\/pages\/4362\/revisions"}],"up":[{"embeddable":true,"href":"https:\/\/wou.edu\/chemistry\/wp-json\/wp\/v2\/pages\/4261"}],"wp:attachment":[{"href":"https:\/\/wou.edu\/chemistry\/wp-json\/wp\/v2\/media?parent=4362"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}