{"id":814,"date":"2018-03-20T16:08:52","date_gmt":"2018-03-20T16:08:52","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/?post_type=chapter&p=814"},"modified":"2018-08-08T15:11:13","modified_gmt":"2018-08-08T15:11:13","slug":"9-2-concentration","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/chapter\/9-2-concentration\/","title":{"raw":"9.2 Concentration","rendered":"9.2 Concentration"},"content":{"raw":"
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<\/div>\r\n
9.2<\/span> Concentration<\/span><\/div>\r\n<\/div>\r\n
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Learning Objectives<\/h3>\r\n
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  1. Express the amount of solute in a solution in various concentration units.<\/li>\r\n \t
  2. Use molarity to determine quantities in chemical reactions.<\/li>\r\n \t
  3. Determine the resulting concentration of a diluted solution.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

    To define a solution precisely, we need to state its concentration<\/span><\/span>: how much solute is dissolved in a certain amount of solvent. Words such as dilute<\/em> or concentrated<\/em> are used to describe solutions that have a little or a lot of dissolved solute, respectively, but these are relative terms whose meanings depend on various factors.<\/p>\r\n\r\n

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    Solubility<\/h2>\r\n

    There is usually a limit to how much solute will dissolve in a given amount of solvent. This limit is called the solubility<\/span><\/span>\u00a0of the solute. Some solutes have a very small solubility, while other solutes are soluble in all proportions. Table 9.2 \"Solubilities of Various Solutes in Water at 25\u00b0C (Except as Noted)\"<\/a> lists the solubilities of various solutes in water. Solubilities vary with temperature, so Table 9.2 \"Solubilities of Various Solutes in Water at 25\u00b0C (Except as Noted)\"<\/a> includes the temperature at which the solubility was determined.<\/p>\r\n\r\n

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    Table 9.2<\/span> Solubilities of Various Solutes in Water at 25\u00b0C (Except as Noted)<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Substance<\/th>\r\nSolubility (g in 100 mL of H2<\/sub>O)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    AgCl(s)<\/td>\r\n0.019<\/td>\r\n<\/tr>\r\n
    C6<\/sub>H6<\/sub>(\u2113) (benzene)<\/td>\r\n0.178<\/td>\r\n<\/tr>\r\n
    CH4<\/sub>(g)<\/td>\r\n0.0023<\/td>\r\n<\/tr>\r\n
    CO2<\/sub>(g)<\/td>\r\n0.150<\/td>\r\n<\/tr>\r\n
    CaCO3<\/sub>(s)<\/td>\r\n0.058<\/td>\r\n<\/tr>\r\n
    CaF2<\/sub>(s)<\/td>\r\n0.0016<\/td>\r\n<\/tr>\r\n
    Ca(NO3<\/sub>)2<\/sub>(s)<\/td>\r\n143.9<\/td>\r\n<\/tr>\r\n
    C6<\/sub>H12<\/sub>O6<\/sub> (glucose)<\/td>\r\n120.3 (at 30\u00b0C)<\/td>\r\n<\/tr>\r\n
    KBr(s)<\/td>\r\n67.8<\/td>\r\n<\/tr>\r\n
    MgCO3<\/sub>(s)<\/td>\r\n2.20<\/td>\r\n<\/tr>\r\n
    NaCl(s)<\/td>\r\n36.0<\/td>\r\n<\/tr>\r\n
    NaHCO3<\/sub>(s)<\/td>\r\n8.41<\/td>\r\n<\/tr>\r\n
    C12<\/sub>H22<\/sub>O11<\/sub> (sucrose)<\/td>\r\n204.0 (at 20\u00b0C)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n

    If a solution contains so much solute that its solubility limit is reached, the solution is said to be saturated<\/span>A solution whose solute is at its solubility limit.<\/span><\/span>, and its concentration is known from information contained in Table 9.2 \"Solubilities of Various Solutes in Water at 25\u00b0C (Except as Noted)\"<\/a>. If a solution contains less solute than the solubility limit, it is unsaturated<\/span>A solution whose solute is less than its solubility limit.<\/span><\/span>. Under special circumstances, more solute can be dissolved even after the normal solubility limit is reached; such solutions are called supersaturated<\/em> and are not stable. If the solute is solid, excess solute can easily recrystallize. If the solute is a gas, it can bubble out of solution uncontrollably, like what happens when you shake a soda can and then immediately open it.<\/p>\r\n\r\n

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    Note<\/h3>\r\n

    Recrystallization of excess solute from a supersaturated solution usually gives off energy as heat. Commercial heat packs containing supersaturated sodium acetate (NaC2<\/sub>H3<\/sub>O2<\/sub>) take advantage of this phenomenon. You can probably find them at your local drugstore.<\/p>\r\n\r\n<\/div>\r\n

    Most solutions we encounter are unsaturated, so knowing the solubility of the solute does not accurately express the amount of solute in these solutions. There are several common ways of specifying the concentration of a solution.\u00a0<\/span><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    Percent Composition<\/h2>\r\n<\/div>\r\n<\/div>\r\n
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    There are several ways of expressing the concentration of a solution by using a percentage. The mass\/mass percent<\/span><\/span>\u00a0(% m\/m) is defined as the mass of a solute divided by the mass of a solution times 100:<\/p>\r\n [latex]\\%\\,m\/m\\,=\\,\\frac{mass\\,of\\,solute}{mass\\,of\\,solution}\\,\u00d7\\,100\\%[\/latex] <\/span>\r\n

    If you can measure the masses of the solute and the solution, determining the mass\/mass percent is easy. Each mass must be expressed in the same units to determine the proper concentration.<\/p>\r\n\r\n

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    Example 2<\/h3>\r\n

    A saline solution with a mass of 355 g has 36.5 g of NaCl dissolved in it. What is the mass\/mass percent concentration of the solution?<\/p>\r\n

    Solution<\/p>\r\n

    We can substitute the quantities given in the equation for mass\/mass percent:<\/p>\r\n [latex]\\%\\,\u2009m\/m\\,=\\,\\frac{36.5\\,g}{355\\,g}\\,\u00d7\\,100%\\,=\\,10.3\\%[\/latex] <\/span><\/span>\r\n\r\n<\/div>\r\n

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    Skill-Building Exercise<\/h3>\r\n
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    1. \r\n
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      A dextrose (also called D-glucose, C6<\/sub>H12<\/sub>O6<\/sub>) solution with a mass of 2.00 \u00d7 102<\/sup> g has 15.8 g of dextrose dissolved in it. What is the mass\/mass percent concentration of the solution?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

      For gases and liquids, volumes are relatively easy to measure, so the concentration of a liquid or a gas solution can be expressed as a volume\/volume percent<\/span><\/span>\u00a0(% v\/v): the volume of a solute divided by the volume of a solution times 100:<\/p>\r\n [latex]\\%\\,\u2009v\/v\\,=\\,\\frac{volume\\,of\\,solute}{volume\\,of\\,solution}\\,\u00d7\\,100\\%[\/latex] <\/span>\r\n

      Again, the units of the solute and the solution must be the same. A hybrid concentration unit, mass\/volume percent<\/span><\/span>\u00a0(% m\/v), is commonly used for intravenous (IV) fluids (Figure 9.2 \"Mass\/Volume Percent\"<\/a>). It is defined as the mass in grams of a solute, divided by volume in milliliters of solution times 100:<\/p>\r\n [latex]\\%\\,\u2009m\/v\\,=\\,\\frac{mass\\,of\\,solute\\,(g)}{volume\\,of\\,solution\\,(mL)}\\,\u00d7\\,100\\%[\/latex] <\/span>\r\n

      Each percent concentration can be used to produce a conversion factor between the amount of solute, the amount of solution, and the percent. Furthermore, given any two quantities in any percent composition, the third quantity can be calculated, as the following example illustrates.<\/p>\r\n\r\n

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      Example 3<\/h3>\r\n

      A sample of 45.0% v\/v solution of ethanol (C2<\/sub>H5<\/sub>OH) in water has a volume of 115 mL. What volume of ethanol solute does the sample contain?<\/p>\r\n

      Solution<\/p>\r\n

      A percentage concentration is simply the number of parts of solute per 100 parts of solution. Thus, the percent concentration of 45.0% v\/v implies the following:<\/p>\r\n [latex]45.0\\%\\,v\/v\\,\u2192\\,\\frac{45\\,mL\\,C_2H_5OH}{100\\,mL\\,solution}[\/latex] <\/span>\r\n

      That is, there are 45 mL of C2<\/sub>H5<\/sub>OH for every 100 mL of solution. We can use this fraction as a conversion factor to determine the amount of C2<\/sub>H5<\/sub>OH in 115 mL of solution:<\/p>\r\n [latex]115\\,\\rlap{\\text{---------------}} mL\\,solution\\,\u00d7\\,\\frac{45\\,mL\\,C_2H_5OH}{100\\,\\rlap{\\text{---------------}}mL\\,solution}\\,=\\,51.8\\,mL\\,C_2H_5OH[\/latex] <\/span>\r\n\r\n<\/div>\r\n

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      Note<\/h3>\r\n

      The highest concentration of ethanol that can be obtained normally is 95% ethanol, which is actually 95% v\/v.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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      Skill-Building Exercise<\/h3>\r\n
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      1. \r\n
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        What volume of a 12.75% m\/v solution of glucose (C6<\/sub>H12<\/sub>O6<\/sub>) in water is needed to obtain 50.0 g of C6<\/sub>H12<\/sub>O6<\/sub>?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

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        Example 4<\/h3>\r\n

        A normal saline IV solution contains 9.0 g of NaCl in every liter of solution. What is the mass\/volume percent of normal saline?<\/p>\r\n

        Solution<\/p>\r\n

        We can use the definition of mass\/volume percent, but first we have to express the volume in milliliter units:<\/p>\r\n1 L = 1,000 mL<\/span><\/span>\r\n

        Because this is an exact relationship, it does not affect the significant figures of our result.<\/p>\r\n [latex]\\%\\,m\/v\\,=\\,\\frac{9.0\\,g\\,NaCl}{1,000\\,mL\\,solution}\\,\u00d7\\,100\\%\\,=\\,0.90\\%\\,m\/v[\/latex] <\/span>\r\n\r\n<\/div>\r\n

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        Skill-Building Exercise<\/h3>\r\n
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        1. \r\n
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          The chlorine bleach that you might find in your laundry room is typically composed of 27.0 g of sodium hypochlorite (NaOCl), dissolved to make 500.0 mL of solution. What is the mass\/volume percent of the bleach?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

          In addition to percentage units, the units for expressing the concentration of extremely dilute solutions are parts per million (ppm)\u00a0<\/span><\/span>and parts per billion (ppb)<\/span><\/span>. Both of these units are mass based and are defined as follows:<\/p>\r\n [latex]ppm\\,=\\,\\frac{mass\\,of\\,solute}{mass\\,of\\,solution}\\,\u00d7\\,1,000,000[\/latex] <\/span>\r\n [latex]ppb\\,=\\,\\frac{mass\\,of\\,solute}{mass\\,of\\,solution}\\,\u00d7\\,1,000,000,000[\/latex]<\/span>\r\n

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          Note<\/h3>\r\n

          Similar to parts per million and parts per billion, related units include parts per thousand (ppth) and parts per trillion (ppt).<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

          Concentrations of trace elements<\/em> in the body\u2014elements that are present in extremely low concentrations but are nonetheless necessary for life\u2014are commonly expressed in parts per million or parts per billion. Concentrations of poisons and pollutants are also described in these units. For example, cobalt is present in the body at a concentration of 21 ppb, while the State of Oregon\u2019s Department of Agriculture limits the concentration of arsenic in fertilizers to 9 ppm.<\/p>\r\n\r\n

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          Note<\/h3>\r\n

          In aqueous solutions, 1 ppm is essentially equal to 1 mg\/L, and 1 ppb is equivalent to 1 \u00b5g\/L.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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          Example 5<\/h3>\r\n

          If the concentration of cobalt in a human body is 21 ppb, what mass in grams of Co is present in a body having a mass of 70.0 kg?<\/p>\r\n

          Solution<\/p>\r\n

          A concentration of 21 ppb means \u201c21 g of solute per 1,000,000,000 g of solution.\u201d Written as a conversion factor, this concentration of Co is as follows:<\/p>\r\n[latex]21\\,ppb\\,Co\\,\u2192\\,\\frac{21\\,g\\,Co}{1,000,000,000\\,g\\,solution}[\/latex] <\/span>\r\n

          We can use this as a conversion factor, but first we must convert 70.0 kg to gram units:<\/p>\r\n [latex]70.0\\,kg\\,\u00d7\\,\\frac{1,000\\,g}{1\\,kg}\\,=\\,7.00\\,\u00d7\\,10^4\\,g[\/latex] <\/span>\r\n

          Now we determine the amount of Co:<\/p>\r\n [latex]7.00\\,\u00d7\\,10^4\\,\\rlap{\\text{--------------}}g\\,solution\\,\u00d7\\,\\frac{21\\,g\\,Co}{1,000,000,000\\,\\rlap{\\text{--------------}}g\\,solution}\\,=\\,0.0015\/,g\\,Co[\/latex] <\/span>\r\n

          This is only 1.5 mg.<\/p>\r\n\r\n<\/div>\r\n

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          Skill-Building Exercise<\/h3>\r\n
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          1. \r\n
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            An 85 kg body contains 0.012 g of Ni. What is the concentration of Ni in parts per million?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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            Molarity<\/h2>\r\n

            Another way of expressing concentration is to give the number of moles of solute per unit volume of solution. Such concentration units are useful for discussing chemical reactions in which a solute is a product or a reactant. Molar mass can then be used as a conversion factor to convert amounts in moles to amounts in grams.<\/p>\r\n

            Molarity<\/span><\/span>\u00a0is defined as the number of moles of a solute dissolved per liter of solution:<\/p>\r\n [latex]molarity\\,=\\,\\frac{number\\,of\\,moles\\,of\\,solute}{number\\,of\\,liters\\,of\\,solution}[\/latex] <\/span>\r\n

            Molarity is abbreviated M (often referred to as \u201cmolar\u201d), and the units are often abbreviated as mol\/L. It is important to remember that \u201cmol\u201d in this expression refers to moles of solute and that \u201cL\u201d refers to liters of solution. For example, if you have 1.5 mol of NaCl dissolved in 0.500 L of solution, its molarity is therefore<\/p>\r\n [latex]\\frac{1.5\\,mol\\,NaCl}{0.500\\,L\\,solution}\\,=\\,3.0\\,M\\,NaCl[\/latex] <\/span>\r\n

            which is read as \u201cthree point oh molar sodium chloride.\u201d Sometimes (aq) is added when the solvent is water, as in \u201c3.0 M NaCl(aq).\u201d<\/p>\r\n

            Before a molarity concentration can be calculated, the amount of the solute must be expressed in moles, and the volume of the solution must be expressed in liters, as demonstrated in the following example.<\/p>\r\n\r\n

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            Example 6<\/h3>\r\n

            What is the molarity of an aqueous solution of 25.0 g of NaOH in 750 mL?<\/p>\r\n

            Solution<\/p>\r\n

            Before we substitute these quantities into the definition of molarity, we must convert them to the proper units. The mass of NaOH must be converted to moles of NaOH. The molar mass of NaOH is 40.00 g\/mol:<\/p>\r\n [latex]25.0\\,\\rlap{\\text{-----------}}g\\,NaOH\\,\u00d7\\,\\frac{1\\,mol\\,NaOH}{40.00\\,\\rlap{\\text{-----------}}g\\,NaOH}\\,=\\,0.625\\,mol\\,NaOH[\/latex] <\/span>\r\n

            Next, we convert the volume units from milliliters to liters:<\/p>\r\n [latex]750\\,\\rlap{\\text{-----}}mL\\,\u00d7\\,\\frac{1\\,L}{1,000\\,\\rlap{\\text{-----}}mL}\\,=\\,0.750\\,L[\/latex] <\/span>\r\n

            Now that the quantities are expressed in the proper units, we can substitute them into the definition of molarity:<\/p>\r\n [latex]M\\,=\\,\\frac{0.625\\,mol\\,NaOH}{0.750\\,L}\\,=\\,0.833\\,M\\,NaOH[\/latex] <\/span>\r\n\r\n<\/div>\r\n

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            Skill-Building Exercise<\/h3>\r\n
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            1. \r\n
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              If a 350 mL cup of coffee contains 0.150 g of caffeine (C8<\/sub>H10<\/sub>N4<\/sub>O2<\/sub>), what is the molarity of this caffeine solution?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

              The definition of molarity can also be used to calculate a needed volume of solution, given its concentration and the number of moles desired, or the number of moles of solute (and subsequently, the mass of the solute), given its concentration and volume. The following example illustrates this.<\/p>\r\n\r\n

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              Example 7<\/h3>\r\n
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              1. What volume of a 0.0753 M solution of dimethylamine [(CH3<\/sub>)2<\/sub>NH] is needed to obtain 0.450 mol of the compound?<\/li>\r\n \t
              2. Ethylene glycol (C2<\/sub>H6<\/sub>O2<\/sub>) is mixed with water to make auto engine coolants. How many grams of C2<\/sub>H6<\/sub>O2<\/sub> are in 5.00 L of a 6.00 M aqueous solution?<\/li>\r\n<\/ol>\r\n

                Solution<\/p>\r\n

                In both parts, we will use the definition of molarity to solve for the desired quantity.<\/p>\r\n\r\n

                  \r\n \t
                1. [latex]0.0753\\,M\\,=\\,\\frac{0.450\\,mol\\,(CH_3)_2NH}{volume\\,of\\,solution}[\/latex] <\/span>\r\n

                  To solve for the volume of solution, we multiply both sides by volume of solution and divide both sides by the molarity value to isolate the volume of solution on one side of the equation:<\/p>\r\n [latex]volume\\,of\\,solution\\,=\\,\\frac{0.450\\,mol\\,(CH_3)_2NH}{0.0753\\,M}\\,=\\,5.98\\,L[\/latex] <\/span>\r\n

                  Note that because the definition of molarity is mol\/L, the division of mol by M yields L, a unit of volume.<\/p>\r\n<\/li>\r\n \t

                2. \r\n

                  The molar mass of C2<\/sub>H6<\/sub>O2<\/sub> is 62.08 g\/mol., so<\/p>\r\n [latex]6.00\\,M\\,=\\,\\frac{moles\\,of\\,solute}{5.00\\,L}[\/latex] <\/span>\r\n

                  To solve for the number of moles of solute, we multiply both sides by the volume:<\/p>\r\nmoles of solute = (6.00 M)(5.00 L) = 30.0 mol<\/span><\/span>\r\n

                  Note that because the definition of molarity is mol\/L, the product M \u00d7 L gives mol, a unit of amount. Now, using the molar mass of C3<\/sub>H8<\/sub>O3<\/sub>, we convert mol to g:<\/p>\r\n [latex]30.0\\,\\rlap{\\text{------}}mol\\,\u00d7\\,\\frac{62.08\\,g}{\\rlap{\\text{------}}mol}\\,=\\,1,860\\,g[\/latex] <\/span>\r\n

                  Thus, there are 1,860 g of C2<\/sub>H6<\/sub>O2<\/sub> in the specified amount of engine coolant.<\/p>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n

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                  Note<\/h3>\r\n

                  Dimethylamine has a \u201cfishy\u201d odor. In fact, organic compounds called amines cause the odor of decaying fish. (For more information about amines, see Chapter 15 \"Organic Acids and Bases and Some of Their Derivatives\"<\/a>, Section 15.1 \"Functional Groups of the Carboxylic Acids and Their Derivatives\" and Section 15.11 \"Amines: Structures and Names\" through Section 15.13 \"Amines as Bases\".)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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                  Skill-Building Exercise<\/h3>\r\n<\/div>\r\n
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                    \r\n \t
                  1. \r\n
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                    What volume of a 0.0902 M solution of formic acid (HCOOH) is needed to obtain 0.888 mol of HCOOH?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                  2. \r\n
                    \r\n

                    Acetic acid (HC2<\/sub>H3<\/sub>O2<\/sub>) is the acid in vinegar. How many grams of HC2<\/sub>H3<\/sub>O2<\/sub> are in 0.565 L of a 0.955 M solution?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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                    Using Molarity in Stoichiometry Problems<\/h2>\r\n

                    Of all the ways of expressing concentration, molarity is the one most commonly used in stoichiometry problems because it is directly related to the mole unit. Consider the following chemical equation:<\/p>\r\nHCl(aq) + NaOH(s) \u2192 H2<\/sub>O(\u2113) + NaCl(aq)<\/span><\/span>\r\n

                    Suppose we want to know how many liters of aqueous HCl solution will react with a given mass of NaOH. A typical approach to answering this question is as follows:<\/p>\r\n\r\n

                    \"image\"<\/div>\r\n

                    In itself, each step is a straightforward conversion. It is the combination of the steps that is a powerful quantitative tool for problem solving.<\/p>\r\n\r\n

                    \r\n

                    Example 8<\/h3>\r\n

                    How many milliliters of a 2.75 M HCl solution are needed to react with 185 g of NaOH? The balanced chemical equation for this reaction is as follows:<\/p>\r\nHCl(aq) + NaOH(s) \u2192 H2<\/sub>O(\u2113) + NaCl(aq)<\/span><\/span>\r\n

                    Solution<\/p>\r\n

                    We will follow the flowchart to answer this question. First, we convert the mass of NaOH to moles of NaOH using its molar mass, 40.00 g\/mol:<\/p>\r\n [latex]185\\,\\rlap{\\text{-----------}}g\\,NaOH\\,\u00d7\\,\\frac{1\\,mol\\,NaOH}{40.00\\rlap{\\text{-----------}}g\\,NaOH}\\,=\\,4.63\\,mol\\,NaOH[\/latex] <\/span>\r\n

                    Using the balanced chemical equation, we see that there is a one-to-one ratio of moles of HCl to moles of NaOH. We use this to determine the number of moles of HCl needed to react with the given amount of NaOH:<\/p>\r\n[latex]4.63\\,\\rlap{\\text{--------------}}mol\\,NaOH\\,\u00d7\\,\\frac{1\\,mol\\,HCl}{1\\rlap{\\text{--------------}}mol\\,NaOH}\\,=\\,4.63\\,mol\\,HCl[\/latex] <\/span>\r\n

                    Finally, we use the definition of molarity to determine the volume of 2.75 M HCl needed:<\/p>\r\n [latex]2.75\\,M\\,HCl\\,=\\,\\frac{4.63\\,mol\\,HCl}{volume\\,of\\,HCl\\,solution}[\/latex]\u00a0<\/span>\r\n [latex]volume\\,of\\,HCl\\,=\\,\\frac{4.63\\,mol\\,HCl}{2.75\\,M\\,HCl}\\,=\\,1.68\\,\\rlap{\\text{---}}L\\,\u00d7\\,\\frac{1,000\\,mL}{1\\,\\rlap{\\text{---}}L}\\,=\\,1,680\\,mL[\/latex] <\/span>\r\n

                    We need 1,680 mL of 2.75 M HCl to react with the NaOH.<\/p>\r\n\r\n<\/div>\r\n

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                    Skill-Building Exercise<\/h3>\r\n
                      \r\n \t
                    1. \r\n
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                      How many milliliters of a 1.04 M H2<\/sub>SO4<\/sub> solution are needed to react with 98.5 g of Ca(OH)2<\/sub>? The balanced chemical equation for the reaction is as follows:<\/p>\r\nH2<\/sub>SO4<\/sub>(aq) + Ca(OH)2<\/sub>(s) \u2192 2H2<\/sub>O(\u2113) + CaSO4<\/sub>(aq)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

                      The general steps for performing stoichiometry problems such as this are shown in Figure 9.3 \"Diagram of Steps for Using Molarity in Stoichiometry Calculations\"<\/a>. You may want to consult this figure when working with solutions in chemical reactions. The double arrows in Figure 9.3 \"Diagram of Steps for Using Molarity in Stoichiometry Calculations\"<\/a> indicate that you can start at either end of the chart and, after a series of simple conversions, determine the quantity at the other end.<\/p>\r\n\r\n

                      \r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"720\"]\"image\" Figure 9.3 Diagram of Steps for Using Molarity in Stoichiometry Calculations <\/em>[\/caption]\r\n

                      When using molarity in stoichiometry calculations, a specific sequence of steps usually leads you to the correct answer.<\/p>\r\n\r\n<\/div>\r\n

                      Many of the fluids found in our bodies are solutions. The solutes range from simple ionic compounds to complex proteins. Table 9.3 \"Approximate Concentrations of Various Solutes in Some Solutions in the Body*\"<\/a> lists the typical concentrations of some of these solutes.<\/p>\r\n\r\n

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                      Table 9.3<\/span> Approximate Concentrations of Various Solutes in Some Solutions in the Body*<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                      Solution<\/th>\r\nSolute<\/th>\r\nConcentration (M)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
                      blood plasma<\/td>\r\nNa+<\/sup><\/td>\r\n0.138<\/td>\r\n<\/tr>\r\n
                      K+<\/sup><\/td>\r\n0.005<\/td>\r\n<\/tr>\r\n
                      Ca2+<\/sup><\/td>\r\n0.004<\/td>\r\n<\/tr>\r\n
                      Mg2+<\/sup><\/td>\r\n0.003<\/td>\r\n<\/tr>\r\n
                      Cl\u2212<\/sup><\/td>\r\n0.110<\/td>\r\n<\/tr>\r\n
                      HCO3<\/sub>\u2212<\/sup><\/td>\r\n0.030<\/td>\r\n<\/tr>\r\n
                      stomach acid<\/td>\r\nHCl<\/td>\r\n0.10<\/td>\r\n<\/tr>\r\n
                      urine<\/td>\r\nNaCl<\/td>\r\n0.15<\/td>\r\n<\/tr>\r\n
                      PO4<\/sub>3\u2212<\/sup><\/td>\r\n0.05<\/td>\r\n<\/tr>\r\n
                      NH2<\/sub>CONH2<\/sub> (urea)<\/td>\r\n0.30<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n
                      *Note: Concentrations are approximate and can vary widely.<\/th>\r\n<\/tr>\r\n<\/tfoot>\r\n<\/table>\r\n<\/div>\r\n
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                      Looking Closer: The Dose Makes the Poison<\/h3>\r\n

                      Why is it that we can drink 1 qt of water when we are thirsty and not be harmed, but if we ingest 0.5 g of arsenic, we might die? There is an old saying: the dose makes the poison<\/em>. This means that what may be dangerous in some amounts may not be dangerous in other amounts.<\/p>\r\n

                      Take arsenic, for example. Some studies show that arsenic deprivation limits the growth of animals such as chickens, goats, and pigs, suggesting that arsenic is actually an essential trace element in the diet. Humans are constantly exposed to tiny amounts of arsenic from the environment, so studies of completely arsenic-free humans are not available; if arsenic is an essential trace mineral in human diets, it is probably required on the order of 50 ppb or less. A toxic dose of arsenic corresponds to about 7,000 ppb and higher, which is over 140 times the trace amount that may be required by the body. Thus, arsenic is not poisonous in and of itself. Rather, it is the amount that is dangerous: the dose makes the poison.<\/p>\r\n

                      Similarly, as much as water is needed to keep us alive, too much of it is also risky to our health. Drinking too much water too fast can lead to a condition called water intoxication, which may be fatal. The danger in water intoxication is not that water itself becomes toxic. It is that the ingestion of too much water too fast dilutes sodium ions, potassium ions, and other salts in the bloodstream to concentrations that are not high enough to support brain, muscle, and heart functions. Military personnel, endurance athletes, and even desert hikers are susceptible to water intoxication if they drink water but do not replenish the salts lost in sweat. As this example shows, even the right substances in the wrong amounts can be dangerous!<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

                      Equivalents<\/h2>\r\n<\/div>\r\n<\/div>\r\n
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                      Concentrations of ionic solutes are occasionally expressed in units called equivalents (Eq)<\/span><\/span>. One equivalent equals 1 mol of positive or negative charge. Thus, 1 mol\/L of Na+<\/sup>(aq) is also 1 Eq\/L because sodium has a 1+ charge. A 1 mol\/L solution of Ca2+<\/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. (For more information about the ions present in blood plasma, see Chapter 3 \"Ionic Bonding and Simple Ionic Compounds\"<\/a>, Section 3.3 \"Formulas for Ionic Compounds\".)<\/p>\r\n\r\n<\/div>\r\n

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                      Dilution<\/h2>\r\n

                      When solvent is added to dilute a solution, the volume of the solution changes, but the amount of solute does not change. Before dilution, the amount of solute was equal to its original concentration times its original volume:<\/p>\r\namount in moles = (concentration \u00d7 volume)initial<\/sub><\/span><\/span>\r\n

                      After dilution, the same amount of solute is equal to the final concentration times the final volume:<\/p>\r\namount in moles = (concentration \u00d7 volume)final<\/sub><\/span><\/span>\r\n

                      To determine a concentration or amount after a dilution, we can use the following equation:<\/p>\r\n(concentration \u00d7 volume)initial<\/sub> = (concentration \u00d7 volume)final<\/sub><\/span><\/span>\r\n

                      Any units of concentration and volume can be used, as long as both concentrations and both volumes have the same unit.<\/p>\r\n\r\n

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                      Example 9<\/h3>\r\n

                      A 125 mL sample of 0.900 M NaCl is diluted to 1,125 mL. What is the final concentration of the diluted solution?<\/p>\r\n

                      Solution<\/p>\r\n

                      Because the volume units are the same, and we are looking for the molarity of the final solution, we can use (concentration \u00d7 volume)initial<\/sub> = (concentration \u00d7 volume)final<\/sub>:<\/p>\r\n(0.900 M \u00d7 125 mL) = (concentration \u00d7 1,125 mL)<\/span><\/span>\r\n

                      We solve by isolating the unknown concentration by itself on one side of the equation. Dividing by 1,125 mL gives<\/p>\r\n [latex]concentration\\,=\\,\\frac{0.900\\,M\\,\u00d7\\,125\\,\\rlap{\\text{-----}}mL}{1,125\\,\\rlap{\\text{-----}}mL}\\,=\\,0.100\\,M[\/latex]<\/span>\r\n

                      as the final concentration.<\/p>\r\n\r\n<\/div>\r\n

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                      Skill-Building Exercise<\/h3>\r\n
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                      1. \r\n
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                        A nurse uses a syringe to inject 5.00 mL of 0.550 M heparin solution (heparin is an anticoagulant drug) into a 250 mL IV bag, for a final volume of 255 mL. What is the concentration of the resulting heparin solution?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

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                        Concept Review Exercises<\/h3>\r\n
                          \r\n \t
                        1. \r\n
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                          What are some of the units used to express concentration?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                        2. \r\n
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                          Distinguish between the terms solubility<\/em> and concentration<\/em>.<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n

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                          Answers<\/h3>\r\n
                          <\/div>\r\n
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                            \r\n \t
                          1. \r\n
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                            % m\/m, % m\/v, ppm, ppb, molarity, and Eq\/L (answers will vary)<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                          2. \r\n
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                            Solubility is typically a limit to how much solute can dissolve in a given amount of solvent. Concentration is the quantitative amount of solute dissolved at any concentration in a solvent.<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

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                            Key Takeaways<\/h3>\r\n
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